On the Question

ON THE CALCULATION OF BRIDGES

of the Résal system.

Engineer N. Nekrasov.

Tomsk.

Steam Typo-Lithography of P. I. Makushin, Blagoveshchensky Lane, own house.

1903.

Printed with the permission of the Director of the Tomsk Technological Institute of Emperor Nicholas II.

On the Calculation of Bridges of the Résal System

One of the structures built specially for the Paris Exposition of 1900 was a pedestrian bridge across the Seine between the Alma and Jena bridges. Its system belongs to the well-known engineer Résal and has the aim of spanning a very considerable total opening (120 meters), without constructing bulky abutment piers and abutments. The ordinary method of taking up the thrust by a tie at the level of the heels was inapplicable, in view of the necessity of avoiding obstruction to navigation. Therefore Résal resorted to a very ingenious device: to balance the thrust of the middle arch (see drawing 1) by thrusts of opposite direction from the lateral semi-arches, the ends of which rest on rocking supports and are connected to one another by a tie.

drawing 1

of the lateral semi-arches, the ends of which rest on rocking supports and are connected to one another by a tie.

The calculation of the middle part, which represents an ordinary two-hinged statically indeterminate arch, open in the upper parts and of solid section in the lower parts, presents nothing special.

Conversely, a system of two side half-arches connected by a tie also constitutes a statically indeterminate system, the calculation of which is further complicated by the fact that both side arches are of solid cross-section, so that the usual method of determining the statically indeterminate quantity from the deformations of the separate rectilinear elements of the system is not applicable.

Therefore the authors of the above-named design, engineers Alby and Lion, who worked under the direct supervision of Résal, applied the following method for resolving the static indeterminacy. (See Génie Civil, 1900, No. 937, May 26.) It was assumed temporarily that, when, for example, the left half-arch is loaded, the left rocking support does not exist. Under this assumption the moments and longitudinal compressive forces in various sections of the left half-arch were determined for loading by arbitrary loads. Then the deformations of the separate elements of the half-arch, into which it was divided for this purpose, were computed; the calculation was made by an approximate formula for curved members with a very large radius of curvature.

Next, the horizontal force acting on the tie and the corresponding elongation of the latter were determined.

Having these data, one could determine the lowering of the point \(C\), at which the omitted support had been located, under the action of various forces, including also the force \(V\) applied at this point \(C\).

But the force \(V\) produces in the tie a force \(g = \dfrac{Vl}{h}\), which is transmitted by means of the tie to the right half-arch. At the same time the point \(C\) of the left half-arch is lowered still further, owing to the displacement of the whole tie to the left by the amount of the horizontal projection of the shortening of the right half-arch.

Denoting this projection by \(a\), and the horizontal projection of the shortening of the right half-arch and the elongation of the tie by \(b\) and \(c\), we obtain the lowering of the point \(C\)

\[ y=\frac{a+b+c}{h}\cdot l \]

Determining, in addition, the lowering of the point \(C\) for the case of the action of the force \(V\), applied at the point \(C\) and equal to \(10^m\), from the pro-

from the proportionality of the lowerings one can determine the reactions of the rocking supports, and then also the force in the tie.

In the present article an attempt is made to give yet another exact method of calculation, based on the theorem on the derivative of the work of deformation.

We shall first derive an approximate formula for the preliminary calculation of the sections of the side semi-arches, the tie, and the rocking supports.

Let us first consider separately the side semi-arch, neglecting the elongation of the tie and of the rocking support; in other words, a parabolic arch with two hinges, one of which is at the vertex of the parabola.

To make it possible to verify the formulas, let us first take the general case, when the second hinge is not situated at the vertex of the parabola, but at some other point of the curve at a height \(n\) above the level of the first hinge (see Fig. 2).

Fig. 2

We shall derive the general expression for the rise for this type of arch; then we shall apply it to the case \(n=f,\ l=g\), and, for verification, to the case \(n=0,\ l=2g\). The equation of the parabola referred to rectangular axes with origin at \(A\) will be:

\[ (g-x)^2=(f-y)\,2p, \]

where the parameter \(p\) is determined from the condition that at \(x=0,\ y=0\),

\[ 2p=\frac{g^2}{f}. \]

The equation assumes the form:

\[ (g-x)^2=\frac{g^2}{f}(f-y), \]

or

\[ (a)\ldots\ldots y=x\left(\frac{2g-x}{g^2}\right)f, \]

whence

\[ \frac{dy}{dx}=\frac{g-x}{p}=\left(\frac{g-x}{g^2}\right)f\ldots\ldots (b) \]

We express the vertical support reactions as functions of the thrust \(H\), under the action of a load \(P\) at a distance \(a\) from the left support,

\[ \text{c) }\ldots\ldots\quad \begin{aligned} V &= P\frac{l-a}{l}+H\frac{n}{l}\quad &&\text{for support } A,\\ V'&=P\frac{a}{l}-H\frac{n}{l}\quad &&\text{for support } D. \end{aligned} \]

Next, the moment for a section at a distance \(x\) from the left support will be expressed as:

\[ \text{(x) }\ldots\ldots\quad Mx=P\left(\frac{l-a}{l}\right)x+H\frac{nx}{l}-Hy \quad \text{from } x=0 \text{ to } x=a, \]

\[ Mx=P\frac{l-a}{l}x+H\frac{nx}{l}-Hy-P(x-a) \quad \text{from } x=a \text{ to } x=l. \]

To determine the expression for the thrust \(H\), we shall use the known formulas for the deformation elements of an arch:

\[ \text{(l) }\ldots\ldots \left\{ \begin{aligned} \Delta x&=-\int \Delta\varphi_x\cdot dy+\int \frac{N_x}{EW}\,dx,\\ \Delta y&=\int \Delta\varphi_x\cdot dx+\int \frac{N_x}{EW}\,dx,\\ \Delta\varphi&=\int \frac{M_x\,ds}{EY}. \end{aligned} \right. \]

Integrating the expression for \(\Delta\varphi\) over the limits from \(0\) to \(x\), we have

\[ \Delta\varphi_x-\Delta\varphi_0 = \frac{1}{EY}\int_0^x M_x\,dx\cdot\frac{ds}{dx}. \]

Taking approximately, for a flat parabolic arch,

\[ \frac{dx}{ds}=\Cos\varphi=const, \]

we obtain

\[ \Delta\varphi_x-\Delta\varphi_0 = \frac{1}{EY\Cos\varphi}\int_0^x M_x\,dx. \]

\[ \Delta\varphi_x = \Delta\varphi_0+ \frac{1}{EY\Cos\varphi} \left[ P\frac{l-a}{l}\cdot\frac{x^2}{2} + \frac{Hnx^2}{2l} + \right. \]

\[ -\frac{Hgx^2}{2p}+\left.\frac{Hnx^3}{6p}\right|_{\substack{x=x\\ x=0}} -\left.P\left(\frac{x^2}{2}-ax\right)\right|_{\substack{x=x\\ x=a}} = \Delta\varphi_0+\frac{1}{2EY\operatorname{Cos}\varphi} \left\{ \frac{P(l-a)}{l}x^2 +H\left(\frac{nx^2}{l}-\frac{gx^2}{p}+\frac{x^3}{3p}\right) -P(x-a)^2 \right\} \ldots\ldots (f) \]

Next, for an approximate calculation we shall neglect the effect of the longitudinal compressive force on the deformation of the arch and determine:

\[ \left|\Delta x\right|_0^x = -\int_0^x \Delta\varphi_x\,dy = -\Delta\varphi_0\,y -\frac{1}{2EY\operatorname{Cos}\varphi} \int_0^x \left\{ \frac{P(l-a)}{l}x^2 +H\left(\frac{nx^2}{l}-\frac{gx^2}{p}+\frac{x^3}{3p}\right) \right\} \cdot \frac{dy}{dx}\cdot dx - \int_a^x P(x-a)^2\frac{dy}{dx}\,dx \]

Substituting, instead of the derivative, its value from (b), we obtain after integration:

\[ \left|\Delta x\right|_0^x = -\Delta\varphi_0\,y -\frac{1}{2EY\operatorname{Cos}\varphi} \left\{ \frac{P(l-a)}{lp} \left(\frac{x^3g}{3}-\frac{x^4}{4}\right) + \frac{H}{p} \left[ \frac{gnx^3}{3l} -\frac{g^2x^3}{3p} +\frac{gx^4}{12p} -\frac{nx^4}{4l} +\frac{gx^4}{4p} -\frac{x^5}{15p} \right] -\frac{P}{12p}(x-a)^3(4g-3x-a) \right\} \ldots\ldots (g). \]

Correspondingly:

\[ \left|\Delta y\right|_0^y = x\Delta\varphi_0 +\frac{1}{2EY\operatorname{Cos}\varphi} \int_0^x \left\{ \frac{P(l-a)}{l}x^2 + H\left(\frac{nx^2}{l}-\frac{gx^2}{p}+\frac{x^3}{3p}\right) \right\}dx + \]

\[ -\int_a^x P(x-a)^2\,dx = x\Delta\varphi_0 +\frac{1}{2EY\operatorname{Cos}\varphi} \left\{ P\frac{l-a}{l}\cdot\frac{x^3}{3} + H\left(\frac{nx^3}{3l}-\frac{gx^3}{3p}+\frac{x^4}{12p}\right) -\frac{P(x-a)^3}{3} \right\} \ldots\ldots (h) \]

Let us apply formulas \((g)\) and \((h)\) to particular cases; let us note that, in view of the immobility of the support hinges,

\[ \left|\Delta x\right|_{0}^{l}=0 \]

and

\[ \left|\Delta y\right|_{0}^{n}=0 \]

and that, for \(x=l\), \(y=n\). Substituting the indicated limits, we obtain

\[ \Delta \varphi_{0}\cdot n = -\frac{1}{2EY\operatorname{Cos}\varphi} \left\{ \frac{P(l-a)}{lp} \left( \frac{l^{3}q}{3}-\frac{l^{4}}{4} \right) + \right. \]

\[ \left. + \frac{H}{p} \left( \frac{gnl^{3}}{3l} -\frac{g^{2}l^{3}}{3p} +\frac{gl^{4}}{3p} -\frac{nl^{4}}{4l} -\frac{l^{5}}{15p} \right) + \right. \]

\[ \left. -\frac{P}{12p}(l-a)^{3}(4g-3l-a) \right\} \ldots\ldots (I) \]

\[ \Delta \varphi_{0}\cdot l = -\frac{1}{2EY\operatorname{Cos}\varphi} \left\{ \frac{P(l-a)}{l}\cdot\frac{l^{3}}{3} + H \left( \frac{nl^{3}}{3l} -\frac{gl^{3}}{3p} +\frac{l^{4}}{12p} \right) + \right. \]

\[ \left. -\frac{P(l-a)^{3}}{3} \right\} \ldots\ldots (II) \]

Eliminating \(\Delta \varphi_{0}\) from these two equations and taking, respectively, \(P\) and \(H\) outside the brackets, after reduction by \(\dfrac{1}{2EY\operatorname{Cos}\varphi}\) we obtain:

\[ P(l-a) \left\{ \frac{1}{12p} \left( -4l^{3}g +3l^{4} +4l^{3}g -8agl^{2} +4a^{2}gl \right. \right. \]

\[ \left. \left. -3l^{4} +6al^{3} -3a^{2}l^{2} -al^{3} +2a^{2}l^{2} -a^{3}l \right) + \right. \]

\[ \left. +\frac{n}{3}(l^{2}-l^{2}+2al-a^{2}) \right\} - H \left\{ -\frac{nl^{4}}{3p} +\frac{2gnl^{3}}{3p} + \right. \]

\[ \left. -\frac{g^{2}l^{4}}{3p^{2}} +\frac{gl^{5}}{3p^{2}} -\frac{l^{6}}{15p^{2}} -\frac{n^{2}l^{2}}{3} \right\} =0 \]

Bringing like terms together:

\[ Hl \left\{ nl(2g-l) -\frac{l^{2}}{p} \left( g^{2}-gl+\frac{l^{2}}{5} \right) -n^{2}p \right\} = \]

\[ = P\left(\frac{l-a}{4}\right) \left\{ -8gal +4a^{2}g +5al -2a^{2}l -a^{3} +\frac{4npa}{l}(2l-a) \right\} \;-\;(III) \]

This equation gives the expression of the thrust for the general case of an arch with the springings not at the same level. In application to the case of the Résal bridge, when we have:

\[ g=l;\ h=f;\ p=\frac{l^2}{2f}; \]

equation (III) becomes the following:

\[ Hl\left[fl^2-2f\left(l^2-l^2+\frac{l^2}{5}\right)-\frac{fl^2}{2}\right]= \]

\[ =P\frac{l-a}{4}\left\{-8al^2+4a^2l+5al^2-a^2l-a^3+4al^2-2a^2l\right\}, \]

whence

\[ H=\frac{5}{2}\frac{P}{fl^3}(l-a)a(l^2+al-a^2)*) \ldots\ldots (IV) \]

This formula is the desired approximate expression of the thrust for a flat parabolic arch, when one of the hinges is located at the vertex of the parabola.

Let us apply, for verification, formula (III) to an arch with springings at the same level. For this case:

\[ l=2g;\ n=0;\ p=\frac{l^2}{8f} \]

substituting, we obtain

\[ H=\frac{5}{8}\frac{P(l-a)l(l^2-al+a^2)}{fl^3}, \]

which represents the well-known formula for a complete parabolic arch.

Into the formula obtained for the lateral half-arch it is easy to introduce corrections for the elongation of the tie, the shortening of the rocking support, and temperature; but in the subsequent exposition similar formulas will easily be obtained from the general formulas for the whole system.

*) This formula was derived by the author during the development of a student bridge design at the request of the deeply respected Professor L. F. Nikolai.

To derive exact formulas, we shall use a method based on the theorem on the derivative of the work of deformation.

First of all, we note that the work of deformation of the entire system under the action of some load, under the assumption of ideal hinged connections, will be the sum of the works of the separate parts of the system, i.e.
\(A = A_{BC} + A_{AC} + A_{CD} + A_{DE} + A_{DF}\)
(the designations correspond to the schematic drawing 3).

Taking the derivative with respect to the statically indeterminate quantity being determined, by the theorem on the derivative of the work of deformation, we have:

\[ \frac{dA}{dH} = \frac{dA_{BC}}{dH} + \frac{dA_{AC}}{dH} + \frac{dA_{CD}}{dH} + \frac{dA_{DE}}{dH} + \frac{dA_{DF}}{dH} = 0 \ldots \tag{6} \]

since the work of deformation of the support reactions, in view of the immobility of the points \(B, A, E, F\), must be taken equal to zero.

Fig. 3

Taking as the origin of coordinates, for the parabola \(AC\), the point \(C\), and for the parabola \(DE\), the point \(D\), and directing the \(y\)-axis vertically downward and the \(x\)-axis along the tie, the equations of both parabolas will have one and the same expression:

\[ x^2 = 2py, \quad \text{where } 2p = \frac{l^2}{f}, \text{ i.e.} \]

\[ x^2 = \frac{l^2}{f} y. \]

It is easy, from the foregoing, to determine the support reactions:

\[ (1)\ldots \begin{cases} V_C=P\dfrac{l-a}{l}-\dfrac{Hf}{l},\\[6pt] V_A=P\dfrac{a}{l}+\dfrac{Hf}{l}, \end{cases} \]

and then to compose expressions for the moments and longitudinal compressive forces for the arch \(AC\) under the influence of a load acting on it at a distance \(a\) from the support \(C\).

Let the angle made by the tangent to the axis of the arch in the section situated at a distance \(x\) from the left support with the \(X\)-axis be denoted by \(\varphi\).

Then:

\[ (2)\ldots \begin{cases} M_x=V_Cx+Hy-\text{ from }\ldots\ldots x=0\text{ to }x=a,\\ M_x=V_Cx+Hy-P(x-a)\ldots x=a\ldots x=l,\\ N_x=N\Cos\varphi+V_C\sin\varphi\ldots\ldots\ldots x=0\ldots x=a,\\ N_x=-N\Cos\varphi+V_C\sin\varphi-P\sin\varphi\; x=a\ldots x=l. \end{cases} \]

or, substituting the value of \(V_C\):

\[ (3) \begin{cases} M_x=P\dfrac{l-a}{l}x-H\left(\dfrac{f}{l}x-y\right)\ldots\ldots\ldots\ldots\ldots\text{ from }x=0\text{ to }x=a,\\[8pt] M_x=P\dfrac{l-a}{l}x-H\left(\dfrac{f}{l}x-y\right)-P(x-a)\ldots\ldots x=a\ldots x=l,\\[8pt] M_x=-H\left(\Cos\varphi+\dfrac{f}{l}\sin\varphi\right)+P\dfrac{l-a}{l}\sin\varphi\ldots\ldots x=0\ldots x=a,\\[8pt] N_x=-H\left(\Cos\varphi+\dfrac{f}{l}\sin\varphi\right)+P\dfrac{l-a}{l}\sin\varphi-P\sin\varphi\; x=a\ldots x=l. \end{cases} \]

The thrust \(H\) along the tie is transmitted to the arch \(DE\) and causes the vertical support reactions:

\[ (4)\ldots \begin{cases} V_D=H\dfrac{f}{l},\\[6pt] V_E=-H\dfrac{f}{l}. \end{cases} \]

and, respectively, the moments and longitudinal forces:

\[ (5)\ldots \left\{ \begin{aligned} M_x&=-H\left(\frac{f}{l}x-y\right)\\ N_x&=-H\left(\cos\varphi+\frac{f}{l}\sin\varphi\right) \end{aligned} \right. \]

Let us now determine the expressions for the individual terms of equation (0).

In what follows we shall adhere to the following notation:

\(W\) — area of the transverse section of the rocking supports.
\(W_1\) — area of the section of the tie.
\(F\) — area of the section of the half-arches (in the general case a variable quantity).
\(Y\) — moment of inertia of the section of the half-arches.
\(r\) — radius of curvature of the parabolic axis of the half-arches.

The work of deformation of the support \(BC\) will be expressed as:

\[ A_{BC}=\frac{1}{2}V_C\cdot\triangle f, \quad\text{where the change in length} \]

of the support is:

\[ \triangle f=\frac{V_C\cdot f}{EW}, \quad\text{whence:} \]

\[ A_{BC}=\frac{f}{2EW}\left(P\frac{l-a}{l}-H\frac{f}{l}\right)^2, \quad\text{and} \]

\[ (6)\ldots\quad \frac{dA_{BC}}{dH} = -\frac{f^2}{lEW}\left(P\frac{l-a}{l}-H\frac{f}{l}\right) = \frac{f^2}{EWl^2}\left[Hf-P(l-a)\right] \]

Accordingly, for the support \(FD\):

\[ A_{FD} = \frac{1}{2}V_D\cdot\triangle f = \frac{1}{2}V_D^2\frac{f}{EW} = \frac{H^2f^3}{2EWl^2}, \]

since

\[ \triangle f=\frac{V_D\cdot f}{EW} \]

\[ (7)\ldots\quad \frac{dA_{DF}}{dH} = \frac{Hf^3}{EWl^2} \]

For the tie \(CD\)

\[ A_{CD}=\frac{1}{2}H\cdot \Delta Z=\frac{1}{2}H\cdot \frac{HL}{EW} =\frac{H^{2}L}{2W,E} \]

\[ (8)\ldots\quad \frac{dA_{CD}}{dH}=\frac{HL}{EW}, \]

In proceeding to determine the terms of equation (0) corresponding to the arches \(AC\) and \(DE\), we shall use the formula expressing the work of deformation of a curved bar in bending, taking into account the finite dimensions of the radius of curvature of the bar. (See, for example, “New Methods of Structural Mechanics,” Müller-Breslau; translation by Mitinsky, ed. 1898, p. 247).

For the time being we shall not take into account the influence of changes in temperature, but shall introduce them subsequently into the final formula.

The work of deformation of a curved bar is

\[ A_i=\int \frac{R^{2}\,ds}{2EF}+\int \frac{M^{2}\,ds}{2EY},\quad \text{where} \]

\[ R=N-\frac{M}{r}; \]
substituting \(R\) and differentiating with respect to \(H\), we have:

\[ \frac{dA_i}{dH} =\int \frac{N}{EF}\cdot \frac{dN}{dH}\,ds -\int \frac{N}{EFr}\cdot \frac{dM}{dH}\,ds +\]
\[ -\int \frac{M}{EFr}\frac{dN}{dH}\,ds +\int \frac{M}{EFr^{2}}\frac{dM}{dH}\,ds +\int \frac{M}{EY}\cdot \frac{dM}{dH}\,ds\ldots(9). \]

The expressions for \(M\) and \(N\) we have in formulas (3) and (5); differentiating them, we obtain:

\[ \text{for the arch } AC:\quad \frac{dM}{dH}=-\left(\frac{f}{l}x-y\right) \]

\[ \frac{dN}{dH}=-\left(\mathrm{Cos}\,\varphi+\frac{f}{l}\mathrm{Sin}\,\varphi\right) \]

\[ \text{for the arch } DE:\quad \frac{dM}{dH}=-\left(\frac{f}{l}x-y\right) \]

\[ \frac{dN}{dH}=-\left(\mathrm{Cos}\,\varphi+\frac{f}{l}\mathrm{Sin}\,\varphi\right) \]

In applying formula (9) to the arc \(DE\), it gives, noting that \(ds=dx\cdot\dfrac{1}{\Cos\varphi}\):

\[ \frac{dA_{DE}}{dH} = \int_{0}^{l} \frac{H\left(\Cos\varphi+\dfrac{f}{l}\Sin\varphi\right)^{2}} {EF\Cos\varphi}\,dx - \int_{0}^{l} \frac{2H\left(\dfrac{f}{l}x-y\right)\left(\Cos\varphi+\dfrac{f}{l}\Sin\varphi\right)} {EF_r\Cos\varphi}\,dx + \int_{0}^{l} \frac{H\left(\dfrac{f}{l}x-y\right)^{2}} {EF_r^{2}\Cos\varphi}\,dx + \int_{0}^{l} \frac{H\left(\dfrac{f}{l}x-y\right)^{2}} {EY\Cos\varphi}\,dx \ldots\ldots\ldots (10). \]

We note that, by the property of the parabola:

\[ \tg\varphi=\frac{2y}{x}=\frac{2fx}{l^{2}} \]

\[ y=\frac{f}{l^{2}}x^{2} \]

Substituting, we have:

\[ \frac{dA_{DE}}{dH} = \frac{H}{E} \int_{0}^{l} \frac{\Cos\varphi}{F} \left( 1+\frac{2f}{l}\tg\varphi+\frac{f^{2}}{l^{2}}\tg^{2}\varphi \right)\,dx + \]

\[ -\frac{2H}{E} \int_{0}^{l} \frac{1}{F_r} \left( \frac{f}{l}x-y \right) \left( 1+\frac{f}{l}\tg\varphi \right)\,dx + \frac{H}{E} \int_{0}^{l} \frac{1}{F_r\Cos\varphi} \left( \frac{f^{2}}{l^{2}}x^{2} + \right. \]

\[ \left. -\frac{2f}{l}xy+y^{2} \right)\,dx + \frac{H}{E} \int_{0}^{l} \frac{1}{Y\Cos\varphi} \left( \frac{f}{l}x-y \right)^{2}\,dx = \]

\[ = \frac{H}{E} \int_{0}^{l} \frac{\Cos\varphi}{F} \left( 1+\frac{4f^{2}}{l^{3}}x+\frac{4f^{4}}{l^{6}}x^{2} \right)\,dx - \frac{2H}{E} \int_{0}^{l} \frac{1}{F_r} \left( \frac{f}{l}x-\frac{2f^{3}}{l^{4}}x^{2} + \right. \]

\[ \left. -\frac{f}{l^{2}}x^{2} -\frac{2f^{3}x^{3}}{l^{5}} \right)\,dx + \frac{H}{E} \int_{0}^{l} \left( \frac{1}{F_r^{2}\Cos\varphi} + \frac{1}{Y\Cos\varphi} \right) \left( \frac{f^{2}}{l^{2}}x^{2} + \right. \]

\[ \left. -\frac{2f^{2}}{l^{3}}x^{3} + \frac{f^{2}}{l^{4}}x^{4} \right)\,dx \ldots\ldots (11). \]

We apply the same formula (9) to the arc \(AC\); here we note that, since the load \(P\) is applied in the section \(x=a\), the additional terms in the expressions for \(M_x\) and \(N_x\): \([-P(x-a)]\), \((-P\Sin\varphi)\), must be introduced in the integration only between the limits \(a\) and \(l\).

Thus we obtain:

\[ \begin{aligned} \frac{dA_{AC}}{dH} ={}& \int_{0}^{l} \frac{\left[H\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right) -P\frac{l-a}{l}\Sin\varphi\right] \left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right)} {EF\Cos\varphi}\,dx \\[4pt] &+\int_{0}^{l} \frac{P\Sin\varphi\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right)} {EF\Cos\varphi}\,dx \\[4pt] &-\int_{0}^{l} \frac{\left(\frac{f}{l}x-y\right) \left[H\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right) -P\frac{l-a}{l}\Sin\varphi\right]} {EF r\Cos\varphi}\,dx \\[4pt] &-\int_{a}^{l} \frac{\left(\frac{f}{l}x-y\right)P\Sin\varphi} {EF r\Cos\varphi}\,dx \\[4pt] &+\int_{0}^{l} \frac{\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right) \left[P\frac{l-a}{l}x-H\left(\frac{f}{l}x-y\right)\right]} {EF r\Cos\varphi}\,dx \\[4pt] &-\int_{a}^{l} \frac{\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right)P(x-a)} {EF r\Cos\varphi}\,dx \\[4pt] &-\int_{0}^{l} \frac{\left(\frac{f}{l}x-y\right) \left[P\frac{l-a}{l}x-H\left(\frac{f}{l}x-y\right)\right]} {EF r^{2}\Cos\varphi}\,dx \\[4pt] &+\int_{a}^{l} \frac{\left(\frac{f}{l}x-y\right)P(x-a)} {EF r^{2}\Cos\varphi}\,dx \\[4pt] &-\int_{0}^{l} \frac{\left(\frac{f}{l}x-y\right) \left[P\frac{l-a}{l}x-H\left(\frac{f}{l}x-y\right)\right]} {EY\Cos\varphi}\,dx \\[4pt] &+\int_{a}^{l} \frac{\left(\frac{f}{l}x-y\right)P(x-a)} {EY\Cos\varphi}\,dx . \end{aligned} \]

Expanding the parentheses in the integrals with limits from \(0\) to \(l\), we separate the terms that contain the factor \(H\).

Then

\[ \frac{dA_{AC}}{dH} = \int_{0}^{l} \frac{H\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right)^{2}} {EF\Cos\varphi}\,dx + \]

\[ -\int_{0}^{l}\frac{2H\left(\frac{f}{l}x-y\right)\left(\cos\varphi+\frac{f}{l}\sin\varphi\right)}{EFr\cos\varphi}\,dx +\int_{0}^{l}\frac{H\left(\frac{f}{l}x-y\right)^2}{EFr^{2}\cos\varphi}\,dx + \]
\[ +\int_{0}^{l}\frac{H\left(\frac{f}{l}x-y\right)^2}{EY\cos\varphi}\,dx -\int_{0}^{l}\frac{\frac{P(l-a)}{l}\sin\varphi\left(\cos\varphi+\frac{f}{l}\sin\varphi\right)}{EF\cos\varphi}\,dx + \]
\[ +\int_{0}^{l}\frac{\frac{P(l-a)}{l}\sin\varphi\left(\frac{f}{l}x-y\right)}{EFr\cos\varphi}\,dx +\int_{0}^{l}\frac{\frac{P(l-a)}{l}x\left(\cos\varphi+\frac{f}{l}\sin\varphi\right)}{EFr\cos\varphi}\,dx + \]
\[ -\int_{0}^{l}\frac{\frac{P(l-a)}{l}x\left(\frac{f}{l}x-y\right)}{EFr^{2}\cos\varphi}\,dx -\int_{0}^{l}\frac{\frac{P(l-a)}{l}x\left(\frac{f}{l}x-y\right)}{EY\cos\varphi}\,dx + \]
\[ +\text{ the sum of integrals with limits from }a\text{ to }l \ldots\ldots\ldots\quad (12) \]

Comparing the first four terms of equality (12) and equality (10), we observe that they are completely identical. Substituting, respectively, the value \(\frac{dA_{DE}}{dH}\) into equality (12) and making certain transformations, we obtain:

\[ \frac{dA_{AC}}{dH} = \frac{dA_{DE}}{dH} -\frac{P(l-a)}{El} \left\{ \int_{0}^{l}\frac{\cos\varphi}{F}\tg\varphi\left(1+\frac{f}{l}\tg\varphi\right)\,dx + \right. \]
\[ \left. -\int_{0}^{l}\frac{1}{Fr}\tg\varphi\left(\frac{f}{l}x-y\right)\,dx -\int_{0}^{l}\frac{1}{Fr}\left(1+\frac{f}{l}\tg\varphi\right)\,dx + \right. \]
\[ \left. +\int_{0}^{l}\frac{x\left(\frac{f}{l}x-y\right)}{Fr^{2}\cos\varphi}\,dx +\int_{0}^{l}\frac{1}{Y\cos\varphi}x\left(\frac{f}{l}x-y\right)\,dx \right\} + \]
\[ +\frac{P}{E} \left\{ \int_{a}^{l}\frac{\cos\varphi}{F}\tg\varphi\left(1+\frac{f}{l}\tg\varphi\right)\,dx + \right. \]
\[ \left. -\int_{a}^{l}\frac{1}{Fr}\tg\varphi\left(\frac{f}{l}x-y\right)\,dx -\int_{a}^{l}\frac{1}{Fr}x\left(1+\frac{f}{l}\tg\varphi\right)\,dx + \right. \]
\[ \left. +\int_{a}^{l}\frac{1}{Fr^{2}\cos\varphi}x\left(\frac{f}{l}x-y\right)\,dx +\int_{a}^{l}\frac{1}{Y\cos\varphi}x\left(\frac{f}{l}x-y\right)\,dx + \right. \]
\[ \left. +\int_{a}^{l}\frac{a}{Fr}\left(1+\frac{f}{l}\tg\varphi\right)\,dx + \right. \]

\[ -\int_a^l \frac{a}{F r^2 \operatorname{Cos}\varphi}\left(\frac{f}{l}x-y\right)\,dx -\int_a^l \frac{a}{Y\operatorname{Cos}\varphi}\left(\frac{f}{l}x-y\right)\,dx \ldots\ldots (13) \]

Substituting here also \(tg\varphi=\dfrac{2fx}{l^2}\) and \(y=\dfrac{fx^2}{l^2}\), we transform (13):

\[ \begin{aligned} \frac{dA_{AC}}{dH} &= \frac{dA_{DE}}{dH} -\frac{P(l-a)}{El}\Biggl\{ \int_0^l \frac{\operatorname{Cos}\varphi}{F} \left(\frac{2fx}{l^2}+\frac{4f^3x^2}{l^5}\right)\,dx \\ &\quad +\int_0^l \frac{1}{Fr} \left(\frac{2f^2x^2}{l^3}-\frac{2f^2x^3}{l^4}\right)\,dx -\int_0^l \frac{1}{Fr} \left(x+\frac{2f^2x^2}{l^3}\right)\,dx \\ &\quad +\int_0^l \left( \frac{1}{Fr^2\operatorname{Cos}\varphi} + \frac{1}{Y\operatorname{Cos}\varphi} \right) \left( \frac{f}{l}x^2-\frac{fx^3}{l^2} \right)\,dx \Biggr\} \\ &\quad +\frac{P}{E}\Biggl\{ \int_a^l \frac{\operatorname{Cos}\varphi}{F} \left(\frac{2fx}{l^2}+\frac{4f^3x^2}{l^5}\right)\,dx +\int_a^l \frac{1}{Fr} \left(\frac{2f^2x^2}{l^3}-\frac{2f^2x^3}{l^4}\right)\,dx \\ &\quad -\int_a^l \frac{1}{Fr} \left(x+\frac{2f^2x^2}{l^3}\right)\,dx +\int_a^l \left( \frac{1}{Fr^2\operatorname{Cos}\varphi} + \frac{1}{Y\operatorname{Cos}\varphi} \right) \left( \frac{f}{l}x^2-\frac{fx^3}{l^2} \right)\,dx \\ &\quad +\int_a^l \frac{a}{Fr} \left(1+\frac{2f^2x}{l^3}\right)\,dx -\int_a^l \left( \frac{a}{Fr^2\operatorname{Cos}\varphi} + \frac{a}{Y\operatorname{Cos}\varphi} \right) \left( \frac{f}{l}x-\frac{fx^2}{l^3} \right)\,dx \Biggr\} \ldots\ldots (14) \end{aligned} \]

From consideration of expressions (11) and (14) it is evident that, in the integrand functions, besides the variable \(x\), there also enter the quantities \(F\), \(Y\), \(\operatorname{Cos}\varphi\), and \(r\); of these quantities \(\operatorname{Cos}\varphi\) and \(r\) can be expressed through \(x\) by virtue of the property of the parabola. As for \(F\) and \(Y\), their dependence on \(x\) is very indeterminate; it is hardly possible to express it by an algebraic equation, since their changes often occur by jumps. Under these conditions, exact calculation of the definite integrals of expressions (11) and (14) becomes impossible in general form.

Let us first consider the simplest case, when the cross-section of the arch is constant or varies so little that, without significant error, certain mean values may be taken for \(F\) and \(Y\). In this case we shall note that, according to the conditions of construction, the side arches will always be very flat, so that \(\operatorname{Cos}\varphi\) and \(r\) will vary very slightly along the length of the arch.

For example, in the Paris bridge the span is \(l=22.5^{m}\), and the rise of the parabola is \(f=5^{m}\); consequently the angle \(\varphi\) varies from \(0^\circ\) to \(\operatorname{arctg}\dfrac{2.5}{22.5}=24^\circ\), and \(\operatorname{Cos}\varphi\) from 1 to 0.914. The change in \(r\) will be more noticeable, but in view of the fact that the terms in which it enters are comparatively small in magnitude, it is quite possible to take both \(\operatorname{Cos}\varphi\) and \(r\) as constant along the length of the arch, choosing for them certain mean values. It must also be noted that in the Paris bridge the side arches are taken with a very steep rise, which is hardly justified by necessity.

Having adopted the indicated assumptions, i.e. regarding \(E\), \(Y\), \(\operatorname{Cos}\varphi\), and \(r\) as constants, we take them out from under the integral signs; then the remaining integral algebraic functions of \(x\) are easily integrated.

From expression (11) we obtain:

\[ \begin{aligned} \frac{dA_{DE}}{dH} &= \frac{H\operatorname{Cos}\varphi}{EF} \int_{0}^{l} \left( 1+\frac{4f^{2}}{l^{3}}x+\frac{4f^{4}}{l^{6}}x^{2} \right)dx - \frac{2H}{EFr} \int_{0}^{l} \left( \frac{f}{l}x+ \right. \\ &\qquad\left. +\frac{2f^{3}}{l^{4}}x^{2} -\frac{f}{l^{2}}x^{2} -\frac{2f^{3}}{l^{5}}x^{3} \right)dx + \left( \frac{H}{EFr^{2}\operatorname{Cos}\varphi} + \frac{H}{EY\operatorname{Cos}\varphi} \right) \int_{0}^{l} \left( \frac{f^{2}}{l^{2}}x^{2} + \right. \\ &\qquad\left. -\frac{2f^{2}}{l^{3}}x^{3} +\frac{f^{2}}{l^{4}}x^{4} \right)dx = \frac{H\operatorname{Cos}\varphi}{EF} \left| x+\frac{2f^{2}}{l^{3}}x^{2} +\frac{4f^{4}}{3l^{6}}x^{3} \right|_{0}^{l} + \\ &\qquad -\frac{2H}{EFr} \left| \frac{f}{2l}x^{2} +\frac{2f^{3}}{3l^{4}}x^{3} -\frac{f}{3l^{2}}x^{3} -\frac{f^{3}}{2l^{5}}x^{4} \right|_{0}^{l} + \\ &\qquad + \left( \frac{H}{EFr^{2}\operatorname{Cos}\varphi} + \frac{H}{EY\operatorname{Cos}\varphi} \right) \left| \frac{f^{2}}{3l^{2}}x^{3} + \right. \\ &\qquad\left. -\frac{f^{2}}{2l^{3}}x^{4} +\frac{f^{2}}{5l^{4}}x^{5} \right|_{0}^{l} = \frac{H\operatorname{Cos}\varphi}{EF} \left( l+\frac{2f^{2}}{l} +\frac{4f^{4}}{3l^{3}} \right) - \frac{2H}{EFr} \left( \frac{fl}{2} + \right. \\ &\qquad\left. +\frac{2f^{3}}{3l} -\frac{fl}{3} -\frac{f^{3}}{2l} \right) + \left( \frac{H}{EFr^{2}\operatorname{Cos}\varphi} + \frac{H}{EY\operatorname{Cos}\varphi} \right) \left( \frac{f^{2}l}{3} -\frac{f^{2}l}{2} +\frac{f^{2}l}{5} \right). \end{aligned} \]

Reducing like terms, we finally obtain:

\[ \frac{dA_{DE}}{dH} = \frac{H}{E} \left\{ \frac{\operatorname{Cos}\varphi}{F} \left( l+\frac{2f^{2}}{l} +\frac{4f^{4}}{3l^{3}} \right) - \frac{2}{Fr} \left( \frac{fl}{6} +\frac{f^{3}}{6l} \right) + \left( \frac{1}{Fr^{2}\operatorname{Cos}\varphi} + \frac{1}{Y\operatorname{Cos}\varphi} \right) \frac{f^{2}l}{30} \right\} \ldots\ldots \tag{15} \]

Similarly, from (14):

\[ \begin{aligned} \frac{dA_{AC}}{dH} &= \frac{dA_{DE}}{dH} -\frac{P(l-a)}{El} \Bigg\{ \frac{\Cos\varphi}{F} \left|\frac{f x^{2}}{l^{2}}+\frac{4f^{3}x^{3}}{l^{5}}\right|_{0}^{l} \\ &\qquad -\frac{1}{Fr}\left|\frac{2f^{2}x^{3}}{3l^{3}}-\frac{f^{2}x^{4}}{2l^{4}}\right|_{0}^{l} -\frac{1}{Fr}\left|\frac{x^{2}}{2}+\frac{2f^{2}x^{3}}{3l^{3}}\right|_{0}^{l} \\ &\qquad +\left(\frac{1}{Fr^{2}\Cos\varphi}+\frac{1}{Y\Cos\varphi}\right) \left|\frac{f x^{3}}{3l}-\frac{f x^{4}}{4l^{2}}\right|_{0}^{l} \Bigg\} \\ &\quad +\frac{P}{E} \Bigg\{ \frac{\Cos\varphi}{F} \left|\frac{f x^{2}}{l^{2}}+\frac{4f^{3}x^{3}}{l^{5}}\right|_{a}^{l} \\ &\qquad -\frac{1}{Fr}\left|\frac{2f^{2}x^{3}}{3l^{3}}-\frac{f^{2}x^{4}}{2l^{4}}\right|_{a}^{l} -\frac{1}{Fr}\left|\frac{x^{2}}{2}+\frac{2f^{2}x^{3}}{3l^{3}}\right|_{a}^{l} \\ &\qquad +\left(\frac{1}{Fr^{2}\Cos\varphi}+\frac{1}{Y\Cos\varphi}\right) \left|\frac{f x^{3}}{3l}-\frac{f x^{4}}{4l^{2}}\right|_{a}^{l} +\frac{a}{Fr}\left|x+\frac{f^{2}x^{2}}{l^{3}}\right|_{a}^{l} \\ &\qquad +\frac{a}{Fr^{2}\Cos\varphi}-\frac{a}{Y\Cos\varphi} \left|\frac{f}{2l}x^{2}-\frac{f x^{3}}{3l^{2}}\right|_{a}^{l} \Bigg\} \\ &= \frac{dA_{DE}}{dH} -\frac{P(l-a)}{El} \Bigg\{ \frac{\Cos\varphi}{F}\left(f+\frac{4f^{3}}{l^{2}}\right) -\frac{1}{Fr}\left(\frac{5}{6}f^{2}+\frac{f^{2}}{2}\right) \\ &\qquad +\left(\frac{1}{Fr^{2}\Cos\varphi}+\frac{1}{Y\Cos\varphi}\right)\frac{fl^{2}}{12} \Bigg\} \\ &\quad +\frac{P}{E} \Bigg\{ \frac{\Cos\varphi}{F}\left(f+\frac{4f^{3}}{l^{2}}-\frac{fa^{2}}{l^{2}}-\frac{4f^{3}a^{3}}{l^{5}}\right) \\ &\qquad -\frac{1}{Fr}\left(\frac{5}{6}f^{2}+\frac{l^{2}}{2}-\frac{a^{2}}{2} -\frac{2f^{2}a^{3}}{3l^{3}}-\frac{2f^{2}a^{3}}{3l^{3}} +\frac{f^{2}a^{4}}{2l^{4}}\right) \\ &\qquad +\left(\frac{1}{Fr^{2}\Cos\varphi}+\frac{1}{Y\Cos\varphi}\right) \left(\frac{fl^{2}}{12}-\frac{fa^{3}}{3l}+\frac{fa^{4}}{4l^{2}}\right) \\ &\qquad -\frac{1}{Fr}\left(al+\frac{af^{2}}{l}-a^{2}-\frac{f^{2}a^{3}}{l^{3}}\right) \\ &\qquad -\left(\frac{1}{Fr^{2}\Cos\varphi}+\frac{1}{Y\Cos\varphi}\right) \left(\frac{fla}{6}-\frac{fa^{3}}{2l}+\frac{fa^{4}}{3l^{2}}\right) \Bigg\}. \end{aligned} \]

Transforming, we have:

\[ \begin{aligned} \frac{dA_{AC}}{dH} &= \frac{dA_{DE}}{dH} +\frac{P}{E} \Bigg\{ \frac{\Cos\varphi}{F l^{5}}\,af(l-a)(l^{3}+4f^{2}l+4f^{2}a) \\ &\qquad +\frac{1}{6Frl}\,a(l-a)(3l^{4}+f^{2}l^{2}+f^{2}al+3f^{2}a^{2}) \\ &\qquad -\frac{1}{12l^{2}} \left(\frac{1}{Fr^{2}\Cos\varphi}+\frac{1}{Y\Cos\varphi}\right) af(l-a)(l^{2}+al-a^{2}) \Bigg\}\ldots\ldots \tag{16} \end{aligned} \]

Substituting now the results obtained by us from formulas (6), (7), (8), (15), (16) into equation (0):

\[ \begin{aligned} \frac{dA}{dH}={}& \frac{f^{2}}{EWl^{2}}\,[Hf-P(l-a)] +\frac{HL}{EW_{1}}+\frac{Hf^{3}}{EWl^{2}}+ \\ &+\frac{2H}{E}\left\{ \frac{\cos\varphi}{F}\left(l+\frac{2f}{l}+\frac{4f^{4}}{3l^{3}}\right) -\frac{2}{Fr}\left(\frac{fl}{6}+\frac{f^{3}}{6l}\right) +\left(\frac{1}{Fr^{2}\cos\varphi}+\right.\right.\\ &\left.\left.+\frac{1}{Y\cos\varphi}\right)\frac{f^{2}l}{30}\right\} +\frac{P}{E}\left\{ \frac{\cos\varphi}{Fl^{5}}\,af(l-a)(l^{3}+4f^{2}l+4f^{2}a)+\right.\\ &+\frac{1}{6Fr^{2}l^{4}}\,a(l-a)(3l^{4}+f^{2}l^{2}+f^{2}al+3f^{2}a^{2})+\\ &\left.-\frac{1}{12l^{2}}\left(\frac{1}{Fr^{2}\cos\varphi}+\frac{1}{Y\cos\varphi}\right) af(l-a)(l^{2}+al-a^{2})\right\}=0 \end{aligned} \]

Multiplying the whole equation by \(E\) and separating the terms with \(H\) into one part, and those with \(P\) into another, we obtain an equality from which the thrust \(H\) is determined:

\[ \begin{aligned} H\left\{ \frac{2f^{3}}{Wl^{2}}+\frac{L}{W_{1}} +\frac{2\cos\varphi}{3Fl^{3}}(3l^{4}+6f^{2}l^{2}+4f^{4})+\right.\\ \left.-\frac{2f}{3Frl}(l^{2}+f^{2}) +\frac{f^{2}l}{15}\left(\frac{1}{Fr^{2}\cos\varphi} +\frac{1}{Y\cos\varphi}\right)\right\} ={}& P\left\{\frac{l-a}{Wl^{2}}f^{2}+\right.\\ &-\frac{\cos\varphi}{Fl^{5}}\,af(l-a)(l^{3}+4f^{2}l+4f^{2}a)+\\ &-\frac{a}{6Frl^{4}}(l-a)(3l^{4}+f^{2}l^{2}+f^{2}al+3f^{2}a^{2})+\\ &\left.+\frac{1}{12l^{2}}\left(\frac{1}{Fr^{2}\cos\varphi} +\frac{1}{Y\cos\varphi}\right)af(l-a)(l^{2}+al-a^{2})\right\}. \end{aligned} \tag{17} \]

If one applies to the curved beam the laws of deformation of straight beams, as Résal does in his calculation, i.e., assuming that the dimensions of the section are extremely small in comparison with the radius of curvature, and putting \(r=\infty\) in formula (17), then it becomes:

\[ H\left\{\frac{2f^3}{Wl^2}+\frac{L}{W_1}+\frac{2\operatorname{Cos}\varphi}{3Fl^3}\left(3l^4+6f^2l^2+4f^4\right)+ \right. \]

\[ \left. +\frac{f^2l}{15Y\operatorname{Cos}\varphi} \right\} = P\left\{ \frac{l-a}{Wl^2}f^2+ \right. \]

\[ \left. -\frac{\operatorname{Cos}\varphi}{Fl^5}\,af(l-a)\left(l^3+4f^2l+4f^2a\right) +\frac{af(l-a)\left(l^2+al-a^2\right)}{12Yl^2\operatorname{Cos}\varphi} \right\}\ldots(18) \]

If in this equation we discard the terms expressing the influence of the tie, of the swinging supports, and of the second arch, and also neglect the direct influence of the longitudinal force on the deformation of the arch, then formula (18) gives:

\[ \frac{Hf^2l}{30Y\operatorname{Cos}\varphi} = \frac{P}{12l^2Y\operatorname{Cos}\varphi}\, af(l-a)\left(l^2+al-a^2\right), \quad \text{whence} \]

\[ H=\frac{5Pa(l-a)\left(l^2+al-a^2\right)}{2fl^3} \]
—the formula derived by us at the beginning of the article.

For the general case, when the cross-section of the arch varies sharply along its length, exact calculation of the definite integrals of expressions (11) and (14), as already mentioned, is impossible. In this case these integrals should be calculated approximately. The following method makes it possible to carry out the calculation with any desired accuracy. We divide the arch into \(n\) equal parts in such a way that, within each of them, the cross-section of the arch does not change appreciably. In most cases, the points of division may be the places where the posts transmitting the load are attached. Let there be a portion of the arch from \(x=(m-1)b\) to \(x=mb\), where \(b=\frac{l}{n}\). We integrate the above-mentioned integrals within these limits; suppose, moreover, that the load \(P\) is applied at a distance from the left support \(a=kb=\frac{kl}{n}\), and let \(m>k\).

The quantities \(\operatorname{Cos}\varphi\) and \(r\), on the basis of what was said above about their variation, may evidently be regarded as constant.

We first integrate expression (11)

\[ \left|\frac{d A_{DE}}{d H}\right| = \frac{mb}{(m-1)b}\,\frac{H}{E} \left\{ \frac{\Cos\varphi}{F} \int_{(m-1)b}^{mb} \left( 1+\frac{4f^{2}}{l^{3}}x+\frac{4f^{4}}{l^{6}}x^{2} \right)\,dx +\right. \]

\[ \left. -\frac{2}{Fr} \int_{(m-1)b}^{mb} \left( \frac{f}{l}x+\frac{2f^{3}}{l^{4}}x^{2} -\frac{f}{l^{2}}x^{2} -\frac{2f^{3}}{l^{5}}x^{3} \right)\,dx +\right. \]

\[ \left. +\left( \frac{1}{Fr^{2}\Cos\varphi} + \frac{1}{Y\Cos\varphi} \right) \int_{(m-1)b}^{mb} \left( \frac{f}{l^{2}}x^{2} -\frac{2f^{2}}{l^{3}}x^{3} +\frac{f^{2}}{l^{4}}x^{4} \right)\,dx \right\} \ldots (19) \]

Compute separately the definite integrals, taking into account:

\[ \left|x\right|_{(m-1)b}^{mb} = mb-(m-1)b=b \]

\[ \left|x^{2}\right|_{(m-1)b}^{mb} = m^{2}b^{2}-(m-1)^{2}b^{2} = b^{2}(2m-1) = b^{2}m_{1} \]

\[ \left|x^{3}\right|_{(m-1)b}^{mb} = m^{3}b^{3}-(m-1)^{3}b^{3} = b^{3}(3m^{2}-3m+1) = b^{3}m_{2} \]

\[ \left|x^{4}\right|_{(m-1)b}^{mb} = m^{4}b^{4}-(m-1)^{4}b^{4} = b^{4}(2m-1)(2m^{2}-2m+1) = b^{4}m_{3} \]

\[ \left|x^{5}\right|_{(m-1)b}^{mb} = m^{5}b^{5}-(m-1)^{5}b^{5} = b^{5}(5m^{4}-10m^{3}+10m^{2}-5m+1) = b^{5}m_{4} \]

Taking the indicated notations, we obtain:

\[ \int_{(m-1)b}^{mb} \left( 1+\frac{4f^{2}}{l^{3}}x^{2} +\frac{4f^{4}}{l^{6}}x^{2} \right)\,dx = \left| x+\frac{2f^{2}}{l^{3}}x^{2} +\frac{4f^{4}}{3l^{6}}x^{3} \right|_{(m-1)b}^{mb} = \]

\[ = \frac{l}{n} + \frac{2f^{2}}{n^{2}l^{2}}m_{1} + \frac{4f^{4}}{3n^{3}l^{3}}m_{2} \]

\[ \int_{(m-1)b}^{mb} \left( \frac{f}{l}x + \frac{2f^{3}}{l^{4}}x^{2} - \frac{f}{l^{2}}x^{2} - \frac{2f^{3}x^{3}}{l^{5}} \right) = \left| \frac{f}{2l}x^{2} + \frac{2f^{3}}{3l^{4}}x^{3} - \frac{fx^{3}}{3l^{2}} + \right. \]

\[ -\frac{f^{3}x^{4}}{2l^{5}}\Bigg|_{(m-1)b}^{mb} = \frac{lf}{2n^{2}}m_{1} + \left( \frac{2f^{3}}{3n^{3}l} - \frac{fl}{3n^{3}} \right)m_{2} - \frac{f^{3}}{2n^{4}l}m_{3} \]

\[ \int_{(m-1)b}^{mb} \left( \frac{f}{l^{2}}x^{2} - \frac{2f^{2}x^{3}}{l^{3}} + \frac{f^{2}}{l^{4}}x^{4} \right)dx = \left| \frac{f^{2}}{3l^{2}}x^{3} - \frac{f^{2}x^{4}}{2l^{3}} + \frac{f^{2}}{5l^{4}}x^{5} \right|_{(m-1)b}^{mb} = \]

\[ = \frac{f^{2}l}{3n^{3}}m_{2} - \frac{f^{2}l}{2n^{4}}m_{3} + \frac{f^{2}l}{5n^{5}}m_{4} \]

Substituting these results into expression (19) and summing over the entire length of the arch, we obtain:

\[ \frac{dA_{DE}}{dH} = \frac{H}{E} \sum_{m=1}^{m=n} \frac{\operatorname{Cos}\varphi}{F} \left\{ \frac{l}{n} + \frac{2f^{2}}{n^{2}l}m_{1} + \frac{4f^{4}}{3n^{3}l^{3}}m_{2} \right\} + \]

\[ - \frac{2}{Fr} \left\{ \frac{fl}{2n^{2}}m_{1} + \left( \frac{2f^{3}}{3n^{3}l} - \frac{3n^{3}}{fl} \right)m_{2} - \frac{f^{3}}{2n^{4}l}m_{3} + \right. \]

\[ \left. + \left( \frac{1}{Fr^{2}\operatorname{Cos}\varphi} + \frac{1}{Y\operatorname{Cos}\varphi} \right) \left( \frac{3n^{3}}{f^{2}l}m_{2} - \frac{f^{2}l}{2n^{4}}m_{3} + \frac{f^{2}l}{5n^{5}}m_{4} \right) \right\} \ldots \tag{20} \]

We integrate expression (14), first rewriting it in the following form:

\[ \left| \frac{dA_{AC}}{dH} \right|_{(m-1)b}^{mb} = \left| \frac{dA_{DE}}{dH} \right|_{(m-1)b}^{mb} + \frac{P(l-a)}{EL}M + \frac{P}{E}N \]

Here

\[ M = \frac{\operatorname{Cos}\varphi}{F} \int_{(m-1)b}^{mb} \left( \frac{2fx}{l^{2}} + \frac{4f^{3}x^{2}}{l^{5}} \right)dx - \frac{1}{Fr} \int_{(m-1)b}^{mb} \left( \frac{4f^{2}x^{2}}{l^{3}} - \frac{2f^{2}x^{3}}{l^{4}} + x \right)dx + \]

\[ + \left( \frac{1}{Fr^{2}\operatorname{Cos}\varphi} + \frac{1}{Y\operatorname{Cos}\varphi} \right) \int_{(m-1)b}^{mb} \left( \frac{f}{l}x^{2} - \frac{fx^{3}}{l^{2}} \right)dx \]

\[ N = M - \frac{a}{Fr} \int_{(m--)b}^{mb} \left( 1+ \frac{2f^{2}x}{l^{3}} \right)dx - \left( \frac{a}{Fr^{2}\operatorname{Cos}\varphi} + \frac{a}{\operatorname{Cos}\varphi} \right) \int_{(m--)b}^{mb} \left( \frac{f}{l}x - \frac{fx^{2}}{l^{2}} \right)dx \]

In this, the term \(\dfrac{P}{E}\cdot N\) enters only within the limits from \(a\) to \(l\), in other words for values of \(m\) from \(k\) to \(n\).

We compute the integrals, using the previous abbreviated notation:

\[ \int_{(m-1)b}^{mb} \left(\frac{2fx}{l^{2}}+\frac{4f^{3}x^{2}}{l^{5}}\right)\,dx = \left|\frac{fx^{2}}{2}+\frac{4f^{3}x^{3}}{3l^{5}}\right|_{(m-1)b}^{mb} = \]

\[ = \frac{f}{n^{2}}m_{1}+\frac{4f^{3}}{3n^{3}l^{2}}m_{2} \]

\[ \int_{(m-1)b}^{mb} \left(\frac{4f^{2}x^{2}}{l^{3}}-\frac{2f^{2}x^{3}}{l^{4}}+x\right)\,dx = \left|\frac{4f^{2}x^{3}}{3l^{3}}-\frac{f^{2}x^{4}}{2l^{4}}+\frac{x^{2}}{2}\right|_{(m-1)b}^{mb} = \]

\[ = \frac{4f^{2}}{3n^{3}}m_{2} - \frac{f^{2}}{2n^{4}}m_{3} + \frac{l^{2}}{2n^{2}}m_{1}. \]

\[ \int_{(m-1)b}^{mb} \left(\frac{f}{l}x^{2}-\frac{fx^{3}}{l^{2}}\right) = \left|\frac{fx^{3}}{3l^{2}}-\frac{fx^{4}}{4l^{2}}\right|_{(m-1)b}^{mb} = \]

\[ = \frac{fl}{3n^{3}}m_{2} - \frac{fl^{2}}{4n^{4}}m_{3} \]

\[ \int_{(m-1)b}^{mb} \left(1+\frac{2f^{2}x}{l^{3}}\right)\,dx = \left|x+\frac{f^{2}x^{2}}{l^{3}}\right|_{(m-1)b}^{mb} = \frac{l}{n}+\frac{f^{2}}{n^{2}l}m_{1} \]

\[ \int_{(m-1)b}^{mb} \left(\frac{f}{l}x-\frac{fx^{2}}{l^{2}}\right)\,dx = \left|\frac{fx^{2}}{2l}-\frac{fx^{3}}{3l^{2}}\right|_{(m-1)b}^{mb} = \]

\[ = \frac{fl}{2n^{2}}m_{1} - \frac{fl}{3n^{3}}m_{2} \]

Substituting, we obtain:

\[ M= \frac{\operatorname{Cos}\varphi}{F} \left\{ \frac{f}{n^{2}}m_{1} + \frac{4f^{3}}{3n^{3}l^{2}}m_{2} \right\} - \frac{1}{Fr} \left\{ \frac{4f^{2}}{3n^{3}}m_{2} + \right. \]

\[ -\frac{f^{2}}{2n^{4}}m_{3}+\frac{l^{2}}{2n^{2}}m_{1}\Bigg\} +\left(\frac{1}{Fr^{2}\operatorname{Cos}\varphi}+\frac{1}{Y\operatorname{Cos}\varphi}\right) \left\{\frac{fl^{2}}{3n^{3}}m_{2}-\frac{4fl^{2}}{3n^{4}}m_{3}\right\} \]

\[ N=M+\frac{a}{Fr}\left\{\frac{l}{n}+\frac{f^{2}}{n^{2}l}m_{1}\right\} -\left(\frac{a}{Fr^{2}\operatorname{Cos}\varphi}+\frac{a}{Y\operatorname{Cos}\varphi}\right) \left\{\frac{fl}{2n^{2}}m_{1}-\frac{fl}{3n^{3}}m_{2}\right\} \]

Summing expressions (20) within the limits from \(0\) to \(l\) and observing that \(a=\dfrac{kl}{n}\), we combine the terms that contain the factor \(\dfrac{Pa}{E}\), and for this part of the formula we take the summation limits from \(m=k\) to \(m=n\). Then we obtain:

\[ \frac{dA_{AC}}{dH} = \frac{dA_{DE}}{dH} -\frac{P}{E}\sum_{m=1}^{m=k}\frac{\operatorname{Cos}\varphi}{F} \left\{\frac{f}{n^{2}}m_{1}+\frac{4f^{3}}{3n^{3}l^{2}}m_{2}\right\} + \]

\[ -\frac{1}{Fr} \left\{\frac{4f^{2}}{3n^{3}}m_{2}-\frac{f^{2}}{2n^{4}}m_{3} +\frac{l^{2}}{2n^{2}}m_{1}\right\} +\left(\frac{1}{Fr^{2}\operatorname{Cos}\varphi}+\frac{1}{Y\operatorname{Cos}\varphi}\right) \left\{\frac{fl^{2}}{3n^{3}}m^{2}+\right. \]

\[ \left. -\frac{4fl^{2}}{3n^{4}}m_{3}\right\} +\frac{Pk}{En}\sum_{m=1}^{m=n}\frac{\operatorname{Cos}\varphi}{F} \left\{\frac{f}{n^{2}}m_{1}+\frac{4f^{3}}{3n^{3}l^{2}}m_{2}\right\} + \]

\[ -\frac{1}{Fr} \left\{\frac{4f^{2}}{3n^{3}}m_{2}-\frac{f^{2}}{2n^{4}}m_{3} +\frac{l^{2}}{2n^{2}}m_{1}\right\} +\left(\frac{1}{Fr^{2}\operatorname{Cos}\varphi}+\frac{1}{Y\operatorname{Cos}\varphi}\right) \left\{\frac{fl^{2}}{3n^{3}}m_{2}+\right. \]

\[ \left. -\frac{4fl^{2}}{3n^{4}}m_{3}\right\} +\frac{Pkl}{En}\sum_{m=k}^{m=n}\frac{1}{Fr} \left\{\frac{l}{n}+\frac{f^{2}}{n^{2}l}m_{1}\right\} + \]

\[ -\left\{\frac{1}{Fr^{2}\operatorname{Cos}\varphi}+\frac{1}{Y\operatorname{Cos}\varphi}\right\} \left\{\frac{fl}{2n^{2}}m_{1}-\frac{fl}{3n^{3}}m_{2}\right\} \ldots \tag{21} \]

Thus, for the general case, the expressions needed for substitution into equation (б) have again been obtained. The complexity of these formulas is only apparent, since most of the coefficients are repeated; the computations, however, will by no means be more difficult than for an ordinary through arch with an average number of bars.

It remains now to introduce into the formulas derived by us the influence of temperature. We shall determine successively the additional terms, corresponding to each part of the system, in the equations of the derivative of the work of deformation.

1) For the support \(BC\), the work of deformation corresponding to its lengthening when it is shortened, or to a change of temperature by \(t_0^\circ\), will be

\[ A'_{BC}=V_C\cdot \triangle f,\quad \text{where } \triangle f=\alpha t_0 f,\ \text{and} \]

\[ V_C=\frac{P(l-a)}{l}-\frac{Hf}{l}. \]

\[ A'_{BC}=\left[\frac{P(l-a)}{l}-\frac{Hf}{l}\right]\alpha t_0 f \]

Differentiating with respect to \(H\), we then have:

\[ \frac{dA'_{BC}}{dH}=-\frac{\alpha t_0 f^2}{l} \]

For the support \(DE\) we have, correspondingly:

\[ A'_{DE}=V_D\cdot \triangle f;\quad \triangle f=\alpha t_0 f \]

\[ A'_{DE}=H\frac{\alpha f^2 t_0}{l};\quad V_D=H\frac{f}{l} \]

\[ \frac{dA'_{DE}}{dH}=\frac{\alpha f^2 t_0}{l} \]

For the tie \(CD\),

\[ A'_{CD}=H\cdot \triangle L,\quad \text{but } \triangle L=\alpha t_0 L \]

\[ A'_{CD}=\alpha Ht_0L \quad \text{and} \]

\[ \frac{dA'_{CD}}{dH}=\alpha t_0L \]

To compute the work of deformation of the arches \(DE\) and \(AC\), we use the formula:

\[ \frac{dA_t}{dH}=\int \alpha t_0 \frac{dN}{dH}\,ds \]

(see, for example, the cited work by Müller-Breslau, p. 247).

For the arc \(AC\) we have

\[ N_x=-\left[H\left(\Cos\varphi+\frac{f}{l}\sin\varphi\right)\right] -\frac{P(l-a)}{l}\Sin\varphi \quad \text{from } x=0 \text{ to } x=a \]

\[ N_x=-\left[H\left(\Cos\varphi+\frac{f}{l}\sin\varphi\right)\right] -\frac{P(l-a)}{l}\Sin\varphi-P\Sin\varphi \quad \text{from } x=a \text{ to } x=l \]

\[ \frac{dN_x}{dH}=-\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right) \]

\[ \frac{dA_{AC}}{dH} = -\int \alpha t_0\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right)\,ds. \]

Substituting here:

\[ ds\,\Cos\varphi=dx \]

\[ ds\,\Sin\varphi=dy \]

\[ \frac{dA'_{AC}}{dH} = -\int_0^l \alpha t_0\,dx -\int_0^f \alpha t_0\frac{f}{l}\,dy = -\alpha t_0 l-\frac{\alpha t_0 f^2}{l} = \]

\[ = -\alpha t_0\left(\frac{l^2+f^2}{l}\right) \]

Applying the same formula to the arc \(DE\):

\[ \frac{dN_x}{dH} = -\left(\Cos\varphi+\frac{f}{l}\Sin\varphi\right) \quad \text{and} \]

\[ \frac{dA'_{DE}}{dH} = -\alpha t_0\left(\frac{f^2+l^2}{l}\right) \]

Adding all the expressions obtained, we determine the value of the additional term in equation \((O)\):

\[ \frac{dA'}{dH} = -\frac{\alpha f^2 t_0}{l} +\frac{\alpha f^2 t_0}{l} +\alpha t_0 L - 2\left(\frac{f^2+l^2}{l}\right)\alpha t_0 \]

Combining like terms:

\[ \frac{dA'}{dH} = \alpha t_0\left[ L-\frac{2(f^2+l^2)}{l} \right] \qquad (22) \]

Examining this expression, it is easy to see that the expression in brackets is always positive. Indeed, the quantity

\(\dfrac{f^2 + l^2}{l}\) differs only insignificantly from \(l\), while, meanwhile, the full span of the bridge is always considerably greater than twice the span of the side arch.

Since this additional term enters into the numerator of the fraction expressing \(H\), it is clear that a lowering of temperature acts in the direction of increasing the thrust, and a rise in temperature in the direction of decreasing it; whence it follows that the magnitude of the tie should be established at temperatures lower than the average for the given locality.