D. E. Romanov, Eng.

A New Method for Calculating Bent-Up Bars in Beams under Moving Loads

Tomsk
1928

D. E. Romanov, Eng.

A New Method for Calculating Bent-Up Bars in Beams under Moving Loads.

(with VII tables of drawings).

A brief survey of existing methods of calculation.

The role of bent-up bars in reinforced-concrete flexural structures has already been sufficiently studied¹². Their principal purpose is to resist those principal or, as they are often called, oblique tensile stresses which arise in beams from the combination of tangential, or shearing, stresses with normal stresses, according to the formula:

\[ \sigma_{\max}=\frac{\sigma}{2}+\sqrt{\frac{\sigma^{2}}{4}+\tau^{2}} \]

and are inclined at an angle \(\varphi\), for which

\[ \operatorname{tg} 2\varphi=\frac{2\tau}{\sigma}. \]

The entire calculation of reinforced-concrete structures is usually carried out for the second stage, when the concrete no longer works in tension and all tensile stresses are taken up solely by the steel. In this case, as is known, the shearing stresses, attaining their greatest value at the neutral axis, have a constant magnitude within the limits of the whole tensile zone (Fig. 1-d). Consequently, the principal stresses in this zone will be equal to the shearing stresses, having a direction at an angle of \(45^\circ\), constant within the limits of the entire tensile zone, owing to the fact that when \(\sigma=0\)

\[ \sigma_{\max}=\tau \quad\text{and}\quad \operatorname{tg}2\varphi=\frac{2\tau}{0}=\infty, \]

therefore, \(\varphi=45^\circ\).

Thus it follows that, for the calculation of bent-up bars, it is quite sufficient to find the sum of all shearing stresses of one direction acting along the neutral layer, by constructing the corresponding diagram, taking its area, etc.

The points of bend are also determined from the \(\tau\)-diagram, by dividing its area by straight lines into parts proportional to the cross-sections of the bent-up bars. In the case of a diagram of complex form, the method of division by Foerster’s summation curves³ may be successfully applied.

This is how the calculation of bent-up bars is most often performed.

Further, owing to the fact that in beams of constant depth the magnitude of the shearing stresses, apart from the dimensions of the section, depends only on the magnitude of the shearing force \((Q)\), since

\[ \tau=\frac{Q}{b\cdot z},\ \text{where} \]

\(b\)—the width of the section, and \(z\)—the lever arm of the internal forces, the calculation just mentioned by diagrams \(\tau\) gains complete determinacy. For example, under a moving load it is possible to introduce into the calculation the greatest shearing forces, found from influence lines, and, ultimately, to construct the diagram of the greatest tangential stresses, provided that such an exact allowance for the action of the load is justified by necessity (for example, in railway bridges).

But in beams of variable outline, when the magnitude of the tangential stresses, besides the shearing force, also depends on the bending moment, such a calculation becomes indeterminate, since the limits of loading for obtaining \(\max Q\) and \(\max M\) are different, owing to the fact that the form of their influence lines is entirely different.

Of course, here also, in pursuit of accuracy, one could create determinacy in the loading by constructing the influence line of the tangential stresses according to the known formula:

\[ \tau = \frac{Q}{b \cdot z} - \frac{M \cdot \operatorname{tg}\alpha}{h \cdot b \cdot z}, \quad \text{where} \]

\(h\)—is the effective height of the section, and \(\alpha\)—the angle between the tangent to the outer generatrix of the beam in the section \(^{11}\) (Fig. 1-a). It can be constructed quite simply: if the quantity

\[ \frac{1}{b \cdot z} \]

is taken as a constant multiplier for the entire line \(\tau\), then from the ordinates of the influence line of the shearing force \((Q)\) one need only subtract the ordinates of the line of the bending moment \((M)\), reduced by multiplication by the value

\[ \frac{\operatorname{tg}\alpha}{h}, \]

i.e., superpose one line on the other.

From such an influence line we would be able to determine \(\max \tau\) quite exactly in this case as well.

However, in all practical calculations, influence lines of \(\tau\) are not used, probably because of a certain cumbersomeness of their own. Unfortunately, we have not even seen their direct application in those cases where the limits of loading for obtaining \(\max \tau\) were specially investigated and where these influence lines could have given an exhaustive answer \(^{2}\).

In practice, in cases of variable heights, certain approximate rules are usually used, for example Mörsch’s rule \(^{2}\), according to which the influence of the bending moment on the magnitude of the shearing stresses is taken only as some part of its greatest value, at maximum \(Q\), etc. To neglect the influence of the moment altogether is by no means always possible, because in some cases this creates an excessive exaggeration of the margin of strength in the bent-up bars, while in others, on the contrary, an underestimation of it.

But it is perfectly clear that approximate rules of this kind, although perhaps sufficient for practical purposes in many cases, do not constitute a complete solution of the problem, and only influence lines of \(\tau\) can give an entirely exact result.

Below we shall see some other possibilities, but for the present let us note only that, apart from the indicated possible indeterminacy, calculation of bent-up bars by diagrams of tangential stresses is in general practically inconvenient, since it burdens the designer with a mass of unnecessary computations: it is necessary without fail to find the shearing forces, compute the tangential stresses, the area of their diagram, etc.; and, to crown it all, the magnitude of the tangential stresses itself is of no practical interest, since the allowable limit of 15 kg/cm\(^2\) \(^{6}\) \(^{7}\) is very rarely exceeded.

Further, it must be said that the very form of the diagram of shearing stresses, i.e. the law of their distribution along the span, is often not needed in practical calculations, taking into account the circumstance that

in many practical cases one may quite freely assign the locations of the bent-up portions along the span on the basis of structural considerations.

Thus, in calculations for shear, in most cases what is important is not the magnitude of the greatest tangential stresses in individual sections, but their total sum over a part of the span, which makes it possible to find the total required amount of transverse and inclined reinforcement.

From this point of view, methods of calculation based on the direct determination of the sum of the tangential stresses—i.e., the area of their diagram—by means of bending moments deserve special attention, entirely without knowing either the magnitude of the tangential stresses themselves or even the magnitude of the shearing forces.

Prof. Löser⁴ initiated such methods, giving a relation between the increment of the bending moment within a portion of a beam of constant depth and the magnitude of the sum of the tangential stresses on that portion; for a portion of variable depth he derived only a certain approximate relation.

Further, Prof. Löser showed that in those cases where it is impossible to assign the locations of the bent-up portions structurally, there is no need at all to resort to finding them from the diagram of shearing stresses; rather, they can be found graphically from the diagram of bending moments, since the difference between the ordinates of this diagram at the boundaries of any portion expresses the increment of moments needed by us for the given position of the load.

Löser’s method, highly valuable in its idea, is unfortunately, in its exact sense, suitable for calculating bent-up bars only under a fixed load, and then only in beams of constant depth.

It does not give us the possibility of taking into account the influence of the mobility of the load, of finding the limits of loading, and, in general, of determining the truly greatest required amount of bent-up reinforcement under the action of this load, still less in beams of variable depth.

An attempt to resolve all these questions and to deepen the method of calculating bent-up bars by bending moments for all cases of beams under the action of moving loads is the aim of the present work.

Fundamental equations and formulas for calculation.

The author has already had occasion¹¹ to prove that the sum of the shearing forces acting along the neutral layer over a portion of a reinforced-concrete beam of variable depth is expressed by the following simple relation:

\[ \sum_1^2 T=\frac{M_2}{z_2}-\frac{M_1}{z_1}, \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

where \(M_2\) and \(M_1\) are the bending moments at the boundaries of the portion \((1—2)\), and \(z_2\) and \(z_1\) are the corresponding lever arms of the internal forces.

The equation given, as the result of integrating the tangential stresses along the neutral layer, was obtained for a number of particular cases of variation of the bending moment and of the effective depth along a beam subjected to simple bending; this made it possible to draw the logical conclusion that it is valid for a beam with any variation of the depth \(h\) and of the moment \(M\).

But at the same time the most general derivation is also possible, for arbitrary variation of \(M\) and \(h\) along the portion; in the interests of completeness it is given below, and in all our subsequent solutions we shall have in view exclusively only a plane system.

Let us take a horizontal bar of variable outline, subjected to simple bending by certain vertical forces (Fig. 1), understanding

under simple bending, the action is only that of a moment, without a longitudinal force. In such a bar the neutral axis will coincide with the axis passing through the centers of gravity of the transformed sections, i.e., with the geometric axis.

In any vertical section the shearing stresses are distributed, as was already noted, nonuniformly over the section, having their greatest value on the geometric axis and below it:

\[ \tau=\frac{dM}{b\cdot z\cdot dx}-\frac{M\cdot dz}{b\cdot z^2\cdot dx} =\frac{Q}{b\cdot z}-\frac{M\cdot \tg\alpha}{h\cdot z\cdot b}{}^{2)} \ldots \ldots \ldots \quad (2) \]

Let us denote, as before \(^{11)}\), the shearing force per unit length by \(T\); then the sum of these shearing forces on any segment \((n-m)\) will be equal to

\[ \sum_n^m T=\int_n^m T\cdot dx =\int_n^m \tau\cdot b\cdot dx =\int_n^m\left(\frac{dM}{b\cdot z\cdot dx} -\frac{M\cdot dz}{b\cdot z^2\cdot dx}\right)b\cdot dx = \]

\[ =\int_n^m \frac{dM}{z}-\int_n^m \frac{M\cdot dz}{z^2}{}^{11)} \]

The first integral can easily be taken by parts.

Indeed, integrating, we have:

\[ \sum_n^m T=\frac{1}{z}\cdot M\Big|_n^m -\int_n^m M\cdot d\left(\frac{1}{z}\right) -\int_n^m \frac{M\cdot dz}{z^2}. \]

After substituting the limits in the first term and transforming the second, we obtain:

\[ \sum_n^m T=\frac{M_m}{z_m}-\frac{M_n}{z_n} +\int_n^m \frac{M\cdot dz}{z^2} -\int_n^m \frac{M\cdot dz}{z^2}. \]

Canceling equal terms of opposite sign, we finally find:

\[ \boxed{\sum_n^m T=\frac{M_m}{z_m}-\frac{M_n}{z_n}} \ldots \ldots \ldots \ldots \quad (3) \]

The derivation given is valid for any bent bar, of arbitrary shape and with arbitrary variation of \(M\) over the segment, and frees us from the necessity of proving the validity of equation (1) for particular cases. \(^{11)}\)

Equation (3) may be rewritten in the following form:

\[ \boxed{\sum_n^m T=Z_m-Z_n=\Delta Z_{n,m}}, \quad \text{where} \]

\(Z_m\) and \(Z_n\) are internal tensile forces, and \(\Delta Z_{n,m}\) is their increment.

All our further reasoning will be based on the 3rd equation, and therefore we shall agree to call it the fundamental equation.

Let us proceed to establishing the dependence between the amount of required reinforcement and the magnitude of the shearing forces.

In view of the fact that, in the case of simple bending, the neutral axis coincides with the geometric axis, the greatest principal tensile stresses will likewise be on the geometric axis, and, as is known, being equal in magnitude to the shearing stresses, they act on planes inclined at an angle of \(45^\circ\) to the axis of the beam.

Consequently, their sum, which we need for calculating the bent-up bars, i.e., in other words, the sum of the oblique tensile forces on the segment \((n-m)\),

\[ \boxed{\sum_{n}^{m} U} = \sum_{n}^{m} T \cdot \cos 45^\circ = \boxed{0.707\left(\frac{M_m}{z_m}-\frac{M_n}{z_n}\right)} \ . . . \ (4) \]

These forces must be taken up entirely by the stirrups and bent-up bars, since, in the stage of behavior of a reinforced-concrete beam adopted for the calculation, the concrete offers no resistance to tension.

Let us introduce the following additional notation:

\(F_b\)—the cross-sectional area of all the stirrups on the segment under consideration,

\(F_s\)—the analogous area of the bent-up bars,

\(\sigma_{e_1}\)—the allowable tensile stress for the stirrups,

\(\sigma_e\)—the allowable stress for the bent-up bars.

Then

\[ \sum U = F_b \cdot \cos 45^\circ \cdot \sigma_{e_1} + F_s \cdot \sigma_e, \]

in view of the fact that the stirrups are usually placed perpendicular to the axis of the beam, i.e., at an angle of \(45^\circ\) to the direction of action of the oblique tensile forces.

Dividing both sides of the equality by \(\cos 45^\circ\), we obtain:

\[ \frac{\sum U}{\cos 45^\circ} = \sum T = F_b \cdot \sigma_{e_1} + \sqrt{2}\cdot F_s \cdot \sigma_e \ ^4) \]

Denoting

\[ \frac{\sigma_{e_1}}{\sigma_e}=\psi^*), \]

we finally find the design relation:

\[ \boxed{\frac{\sum T}{\sigma_e}=\psi \cdot F_b+\sqrt{2}\cdot F_s} \ . . . . . . . \ (5) \]

or

\[ \boxed{F_s=0.707\left(\frac{\sum T}{\sigma_e}-\psi \cdot F_b\right)} \ . . . . . . . \ (6) \]

Determination of the greatest value of \(\sum T\).

In calculating bent-up bars it is important to be able to find the greatest possible value of the magnitude of the sum of the shearing forces. The author has already touched upon this question in passing\(^{11}\), noting that in order to obtain \(\sum T_{\max}\) it is “necessary to take such a distribution of the load, i.e., of the magnitudes \(M\) at the boundaries of the segment, for which their difference will be greatest”... It is quite evident that for this it is necessary to construct the influence line of \(\sum T\) and, using it, by placing the load in the most unfavorable position, to compute \(\sum T_{\max}\).\(^{**}\)

* For example, according to the standards of Gosplan\(^{6}\), \(\sigma_{e_1}=\sigma_e=1200\ \text{kg}/\text{cm}^2\), i.e., \(\psi=1\); according to the standards of NKPS\(^{7}\) for reinforced-concrete bridges, \(\sigma_{e_1}=700\ \text{kg}/\text{cm}^2\), \(\sigma_e=900\ \text{kg}/\text{cm}^2\), i.e., \(\psi=7/9\).

** The importance of such influence lines was pointed out by Prof. G. V. Ulyaninsky in his review of the above-mentioned article by the author, where, among other things, he wrote: “......... unfortunately, however, the author did not draw the logical conclusion concerning the importance of determining influence lines for the expression:

\[ \sum_{1}^{2} T=\frac{M_2}{z_2}-\frac{M_1}{z_1}. \]

Only thanks to such influence lines can one determine the actual greatest stresses in the bent-up bars of one segment or another”...

As we have seen above, knowing the sum of the shearing forces, it is easy to find the required amount of transverse and inclined reinforcement; and conversely, the sufficiency of the quantity of this reinforcement, installed, for example, without calculation, for structural reasons, can be ascertained by constructing the influence line of \(\Sigma T\) for the section under consideration and, loading the max \(\Sigma T\), finding the actual stress from the formula:

\[ \sigma_e=\frac{\Sigma T}{\varphi\cdot F_b+\sqrt{2}\cdot F_s} \qquad \cdots \cdots \cdots \cdot (7) \]

The construction of the influence line of \(\Sigma T\) is very simple if the influence lines of the bending moments are known; i.e., it is only necessary to divide the ordinates of the influence line of the bending moment \(M_1\) by \(z_1\) and subtract from them the ordinates of the influence line of the moment \(M_2\), divided by its \(z_2\), that is, to superpose one line on the other.

For \(z_2=z_1=z\), the construction is simplified still more, and division by \(z\) may be applied only to the final result after loading, because in this case

\[ \sum_1^2 T=\frac{M_2-M_1}{z}=\frac{\Delta M_{1-2}}{z} \quad \text{(Löser’s formula \(^{4}\)).} \]

Let us construct and examine a number of influence lines.

Influence lines of \(\Sigma T\) in a simple beam.

a) With variable depth.

Take, in a beam \(AB\) of span \(l\), a section \((1—2)\) with abscissae \(a_1\) and \(a_2\) and with the lever arms of the internal forces, respectively, \(z_1\) and \(z_2\) (Fig. 3). The depth of the beam is variable and increases toward the middle of the span; therefore \(z_2>z_1\).

On the basis of equation (3), the influence line of \(\sum_1^2 T\) is obtained as the difference between the lines \(\dfrac{M_2}{z_2}\) and \(\dfrac{M_1}{z_1}\), each of which is a triangle with its vertex under the section, in accordance with the generally known properties of influence lines of moments. In this case, at the left support there will be support intercepts \(\dfrac{a_1}{z_1}\) and \(\dfrac{a_2}{z_2}\), and at the right, analogously, \(\dfrac{b_1}{z_1}\) and \(\dfrac{b_2}{z_2}\) (Fig. 3-b).

The difference between these two triangles will give us the form of the influence line of

\[ \sum_1^2 T. \]

In our example this line will consist of two portions, of opposite signs, owing to the fact that

\[ \frac{b_1}{z_1}>\frac{b_2}{z_2}. \]

The influence line obtained in Fig. (3-c), for clarity, is brought to a horizontal closing line; moreover, the left support intercept is equal to the positive value

\[ \left(\frac{a_2}{z_2}-\frac{a_1}{z_1}\right), \]

and the right one is negative,

\[ -\left(\frac{b_1}{z_1}-\frac{b_2}{z_2}\right). \]

We note that in all beam systems the line of influence of \(\Sigma T\) has a complete analogy with the line of influence of the horizontal projection of the force of the descending diagonal member of a truss (Fig. 4).

Indeed, projecting onto the horizontal axis all the forces applied to the left cut-off part of the truss, we obtain:

\[ O_{1-2}\cdot \cos \beta+U_{1-2}\cdot \cos \gamma+D_{1-2}\cdot \cos \alpha=0. \]

Taking into account that

\[ O_{1-2}=-\frac{M_{2}}{h_{2}\cdot \cos \beta},\qquad U_{1-2}=\frac{M_{1}}{h_{1}\cdot \cos \gamma}, \]

we find

\[ D_{1-2}\cdot \cos \alpha=\frac{M_{2}}{h_{2}}-\frac{M_{1}}{h_{1}}.\tag{8} \]

The analogy is complete, and we shall use it below.

The lengths of the platforms of the line of influence \(\Sigma T\), i.e. the quantities \(\lambda_a\) and \(\lambda_b\), we easily find by using the well-known method of determining the zero point of a diagonal member, shown in Fig. 4.

Having carried out an analogous construction for our beam (Fig. 3), from the similarity of the triangles \(AO_1u_1\) and \(AFK\), and also \(BO_2u_2\) and \(BFK\), we find that

\[ \frac{z_0}{\lambda_a}=\frac{z_1}{a_1};\qquad \frac{z_0}{\lambda_b}=\frac{z_2}{b_2}. \]

Moreover,

\[ \lambda_a+\lambda_b=l. \]

From these three equations, after eliminating the quantity \(z_0\), we obtain:

\[ \lambda_a=\frac{l\cdot a_1\cdot z_2}{a_1\cdot z_2+b_2\cdot z_1};\tag{9} \]

\[ \lambda_b=\frac{l\cdot b_2\cdot z_1}{a_1\cdot z_2+b_2\cdot z_1}.\tag{10} \]

The ordinates of the line of influence \(\Sigma T\) at the boundaries of the segment will be as follows:

\[ y_1=\frac{a_1}{l}\left(\frac{b_2}{z_2}-\frac{b_1}{z_1}\right);\tag{11} \]

\[ y_2=\frac{b_2}{l}\left(\frac{a_2}{z_2}-\frac{a_1}{z_1}\right).\tag{12} \]

Further, we easily find the expressions for the areas:

\[ \omega_a=\frac{\lambda_a\cdot y_1}{2} = -\frac{a_1^{2}}{2z_1}\cdot \frac{b_1\cdot z_2-b_2\cdot z_1}{a_1\cdot z_2+b_2\cdot z_1}; \tag{13} \]

\[ \omega_b=\frac{\lambda_b\cdot y_2}{2} = \frac{b_2^{2}}{2z_2}\cdot \frac{a_2\cdot z_1-a_1\cdot z_2}{a_1\cdot z_2+b_2\cdot z_1}. \tag{14} \]

In our example (Fig. 3) the i.l. \(\sum_1^2 T\) consisted of two segments of different signs. This was due to the simple circumstance that the depth of the beam was taken as increasing toward the middle of the span, and from formula (11) it is directly evident that in this case always

\[ \frac{b_2}{z_2}<\frac{b_1}{z_1}, \quad \text{i.e.} \]

\[ b_2<b_1,\quad \text{and}\quad z_2>z_1, \]

i.e. the ordinate \(y_1\) is obtained negative.

But the outline of the beam may also be decreasing (Fig. 5), i.e. \(z_2<z_1\), and in such a way that

\[ \frac{b_2}{z_2}>\frac{b_1}{z_1}. \]

The ordinate \(y_1\) will then be positive, and the i.l. \(\sum_1^2 T\) will be only of one sign. In Fig. 5 such an influence line is constructed.

The analogy with the horizontal projection of the web-member force of a truss, established above, fully explains to us the noted changes in the i.l. \(\sum T\), owing to the fact that in the first case (Fig. 3) the point of intersection of the lines joining the centers of compression (\(o_1\) and \(o_2\)) and the centers of tension (\(u_1\) and \(u_2\)) in the boundary sections was outside the span, whereas in the second case (Fig. 5) it was obtained between the supports*). In Figs. (3) and (5) this point is denoted by the letter \(R\). Using it, we can graphically obtain the entire i.l. \(\sum T\), by laying off only on the left support the segment

\[ \left(\frac{a_2}{z_2}-\frac{a_1}{z_1}\right). \]

c) On a segment with constant depth.

In this case:

\[ z_1=z_2=z; \]

\[ \frac{a_2}{z_2}-\frac{a_1}{z_1}=\frac{c}{z}; \]

\[ \frac{b_2}{z_2}-\frac{b_1}{z_1}=-\frac{c}{z}. \]

The construction of the influence line \(\sum T\) is considerably simplified (Fig. 2).

We obtain:

\[ \lambda_a=\frac{l\cdot a_1}{a_1+a_2}=a_1\cdot\frac{l}{\,l-c\,}; \qquad \ldots \ldots \ldots (15) \]

\[ \lambda_b=\frac{l\cdot b_2}{a_1+b_2}=b_2\cdot\frac{l}{\,l-c\,}; \qquad \ldots \ldots \ldots (16) \]

\[ y_1=-\frac{c\cdot a_1}{z\cdot l};\qquad y_2=\frac{c\cdot b_2}{z\cdot l}; \qquad \ldots \ldots \ldots (17) \]

\[ \omega_a=-\frac{a_1^2\cdot c}{2z(l-c)}; \qquad \ldots \ldots \ldots (18) \]

\[ \omega_b=\frac{b_2^2\cdot c}{2z(l-c)} \qquad \ldots \ldots \ldots (19) \]

*) Recall Ritter’s point for shear.

c) Loading by influence lines.

If we have only a uniformly distributed load, for example, a permanent load—$g$, a live load—$p$, and $g + p = q$, then we can formulate expressions for determining $\Sigma T_{\max}$ and $\Sigma T_{\min}$ that spare us the need to construct the influence lines $\Sigma T$.

Using the usual rule of loading, we obtain:

\[ \left. \begin{aligned} \Sigma T_{\max} &= q \cdot \omega_b + g \cdot \omega_a \\ \Sigma T_{\min} &= q \cdot \omega_a + g \cdot \omega_b \end{aligned} \right\} \qquad \ldots\ldots\ldots (20) \]

In these formulas, in place of $\omega_a$ and $\omega_b$, one must substitute their values, computed in the general case from formulas (13) and (14).

For a segment of constant height, making the substitution from formulas (18) and (19), we find:

\[ \boxed{ \begin{aligned} \sum_{n}^{m} T_{\max} &= \frac{c}{2z(l-c)}\left(q \cdot b_m^{\,2} - g \cdot a_n^{\,2}\right) \\ \sum_{n}^{m} T_{\min} &= \frac{c}{2z(l-c)}\left(g \cdot b_m^{\,2} - q \cdot a_n^{\,2}\right) \end{aligned} } \qquad \ldots\ldots (21) \]

Let us find the limiting position of that section, with abscissa $a_m = x_0$, to the right of which there will be no positive $\Sigma T$.

Equating

\[ \sum_{n}^{m} T_{\max} = \frac{c}{2z(l-c)} \left(q \cdot b_m^{\,2} - g \cdot a_n^{\,2}\right) =0, \]

we obtain

\[ q \cdot b_m^{\,2} - g \cdot a_n^{\,2}=0. \]

Let us denote $q = m \cdot g$, and, taking into account that

\[ a_m = a_n + c, \]

\[ b_m = l - a_m, \]

we find

\[ x_0 = \frac{l\sqrt{m}+c}{\sqrt{m}+1} \qquad \ldots\ldots\ldots (22) \]

If the live load is in the form of a series of concentrated loads (railway train, convoy of automobiles, etc.), then the loading must be carried out in the usual manner adopted in the theory of structures for moving concentrated loads (see the example below).

Influence of the length of the segment on the accuracy of the final results.

If we sum the individual values $\Sigma T_{\max}$*) obtained for successive segments within some part of the beam, then we shall find a certain fictitious total value of the shearing forces, possible

*) One may proceed in exactly the same way with the values $\Sigma T_{\min}$.

for different positions of the live load. Let us call it \(\sum T_{f\max}\). It is determined by the equality:

\[ \sum_{0}^{n} T_f = \sum_{0}^{1} T + \sum_{1}^{2} T + \cdots + \sum_{n-1}^{n} T \ldots\ldots\ldots \tag{23} \]

If each of the terms is taken for one and the same position of the load on the span, then it is clear that

\[ \sum_{0}^{n} T_f = \sum_{0}^{n} T. \]

But we have agreed to sum the greatest values of the individual terms; therefore always

\[ \sum_{0}^{n} T_{f\max} > \sum_{0}^{n} T_{\max}, \]

with the exception, of course, of that particular case when the length of the segment is chosen equal to the length of the very part of the beam under consideration, in which case

\[ \sum_{0}^{n} T_{f\max} = \sum_{0}^{n} T_{\max}. \]

It is quite obvious that the most exact result for \(\sum T_f\) on a part of the span is obtained only for infinitely small lengths of the segments into which the part under consideration is divided; all other results, for \(c > dx\), will be smaller in their numerical magnitude.

Thus, in order to determine the influence of the length of the segment, it is necessary to find \(\sum T_{f\max}\) for \(c = dx\) and compare it with \(\sum T_{f\max}\) found for \(c > dx\).

But before passing to this comparison, let us clarify what practical significance the quantity \(\sum T_f\) has. In the given unfavorable position of the load, on some segment \((0—1)\) there acts a force \(\sum_{0}^{1} T_{\max}\), having a quite real value, for the carrying of which on this segment a definite number of stirrups and bent-up bars is required.

On the neighboring segment \((1—2)\) there will be its own design value \(\sum_{1}^{2} T_{\max}\), under another unfavorable position of the load, requiring the placing of its own corresponding stirrups and bends, etc.

Consequently, over the entire part of the beam \((0—n)\) there will act in total

\[ \sum_{0}^{n} T_{f\max}, \]

by the magnitude of which one must determine the total number of stirrups and bent-up bars on the part \((0—n)\), while from the character of the increase of \(\sum T_f\) one may, as we shall see below, find the locations of the bends.

Let us take a simple beam \(AB\) (Fig. 6-a), of constant section, with uniformly distributed live and dead load. Let us find for it the value of \(\sum T_{f\max}\) for the part of the span from \(0\) to \(x\), with segment lengths equal to \(dx\).

For this purpose, substituting in equation (21) \(c = dx\), \(b_m = l - x\), \(a_n = x - dx\), we integrate within the assigned limits, neglecting infinitesimals of the 2nd order and higher.

We have:

\[ \sum_{0}^{x} T_{f\max} = \frac{1}{2z} \int_{0}^{x} \left[ q\,(l-x)^2 - g\,x^2 \right] \frac{dx}{l} = \frac{x}{6zl} \left[ 3ql\,(l-x) + p\,x^2 \right] \ldots \tag{24} \]

From this equation one can construct a curve which will characterize the increase of \(\sum T_{f\max}\) along the span. For convenience of construction, we transform equation (24) into the following form:

\[ \left.\sum_{0}^{x} T_{f\max} = \frac{q l^{2}}{8z}\cdot \left[ 3\left(1-\frac{x}{l}\right) + \frac{4(m-1)}{3m}\cdot\left(\frac{x}{l}\right)^{2} \right]\frac{x}{l} \right., \tag{25} \]

where \(m\) has its former meaning, i.e.

\[ m=\frac{q}{g}=\frac{p+g}{g}. \]

According to equation (25), for \(m=4\) and the constant multiplier \(\dfrac{q l^{2}}{8z}\), the curve in Fig. (6-b) has been constructed.

The absolute greatest value of \(\sum T_f\) in the span will be obtained, according to formula (22), when

\[ x=x_0=\frac{l\sqrt{m}}{\sqrt{m}+1}\ldots\tag{25'} \]

and will be equal to

\[ \max \sum T_{f\max} = \frac{q l^{2}}{6z}\cdot \frac{3m+(m+2)\sqrt{m}}{(1+\sqrt{m})^{3}} = \frac{q l^{2}}{6z}\cdot \frac{3m+(m+2)\sqrt{m}}{1+3m+(3+m)\sqrt{m}} \ldots\tag{26} \]

For comparison, let us find the same value, but for \(c>dx\), for example, as the limiting case of variation, when

\[ c=\frac{l}{2}. \]

In this case the span will be divided into two parts, and for the first of them, after substitution in formula (21),

\[ b_n=\frac{l}{2},\quad a_n=0 \]

we obtain

\[ \sum_{0}^{1/2} T_{\max} = \frac{q l^{2}}{8z} = \frac{M_{\max}}{z}, \]

as was to be expected.

For the second, with \(b_n=0,\ a_n=\dfrac{l}{2}\), we similarly find

\[ \sum_{1/2}^{1} T_{\max} = -\frac{g l^{2}}{8z} = -\frac{q l^{2}}{8z}\cdot\frac{1}{m}. \]

Summing the values, we obtain:

\[ \sum_{0}^{1/2} T_{f\max} = \frac{q l^{2}}{8z}; \]

\[ \sum_{0}^{1} T_{f\max} = \frac{q l^{2}}{8z} - \frac{q l^{2}}{8z}\cdot\frac{1}{m} = \frac{q l^{2}}{8z}\left(\frac{m-1}{m}\right). \]

In Fig. (6-b) the corresponding curve has been constructed through three points.

Obviously, in this case

\[ \max \sum T_{f\max} = \frac{q l^{2}}{8z} = \frac{M_{\max}}{z} \ldots\ldots\ldots\tag{27} \]

From a comparison of formulas (26) and (27) we conclude that the true value for a simple beam under a uniformly distributed load is

\[ \boxed{\max \Sigma T_{f\max}}=\mu\cdot \frac{q l^{2}}{8z} = \boxed{\mu\cdot \frac{M_{\max}}{z}}, \tag{28} \]

where

\[ \mu= \boxed{ \frac{4[3m+(m+2)\sqrt{m}]} {3[1+3m+(3+m)\sqrt{m}]} }. \tag{29} \]

The coefficient \(\mu\), which depends on the ratio of the loads, gives the transition from the usual value \(\Sigma T_{\max}\), obtained for \(c=\dfrac{l}{2}\), i.e. for complete loading of the span, to the true, absolutely greatest value, at the same time characterizing the influence of the length of the segment at the extreme limits of its variation. This latter circumstance is especially important for us.

The variation of \(\mu\) is presented in the following table, calculated by formula (29):

\(m=\dfrac{g+p}{g}\)	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6	6.5	\(\infty\)
\(\mu\)	1.000	1.030	1.105	1.131	1.155	1.172	1.185	1.197	1.206	1.215	1.221	1.228	1.333

or graphically in Fig. (7).

From the table or the graph it is seen that the coefficient \(\mu\) varies within rather narrow limits and, with an infinite increase of the ratio \(m=\dfrac{p+g}{g}\), approaches asymptotically the value \(1.333\).

Consequently, the greatest magnitude of the error that can be made in computing \(\Sigma T_{f\max}\) for one position of the load and \(c=\dfrac{l}{2}\) will be equal to

\[ \frac{1.333-1.000}{1.333}\cdot 100=25\% \]

relative to the true value.

In reinforced-concrete floor beams usually \(m=2\), i.e. the possible error may reach

\[ \frac{0.105\cdot 100}{1.05}=10\%. \]

In reinforced-concrete bridges with a span, for example, of about 6 m, \(m\simeq 5\), i.e. in this case we shall underestimate the result by

\[ \frac{1.206-1.000}{1.206}\cdot 100=17.1\%, \]

if we compute \(\Sigma T\) for one placement of the load at \(\max M\) at the middle of the span, i.e. for \(c=\dfrac{l}{2}\); in fact, the error will be still greater, since the equivalent load toward the supports increases.

For comparison with the results obtained, in Fig. (6-b) a curve has been plotted, found for the same \(m=4\), but with \(c=0.1\cdot l\).

The procedure for constructing it presented no particular difficulties: the separate values of \(\sum T_{\max}\) by segments were obtained from formula (21), by substituting \(c=0.1\cdot l\) and the corresponding \(b_m\) and \(a_n\); by summing \(\sum T_{\max}\) from the left, the ordinates of the curve \(\sum T_{f\max}\) were found and plotted on the drawing. The greatest value is obtained at \(x_0=0.7\cdot l\) and is equal to

\[ \max \sum T_{f(c=0.1\cdot l)} = 1.144\cdot \frac{M_{\max}}{z}. \]

For the same \(m\), but with \(c=dx\), from the table we find \(\mu=1.185\), i.e.,

\[ \max \sum T_f = 1.185\cdot \frac{M_{\max}}{z}. \]

The difference is only

\[ \frac{1.185-1.144}{1.185}\cdot 100 = 3.5\%. \]

From a comparison of the results obtained it may be concluded that, if the live load is transmitted directly to the beam, then in important structures (for example, bridges, etc.) with a heavy live load it is quite sufficient to take \(c=0.1l\); in other cases one may limit oneself to \(c=0.25l\), or even \(c=l/2\).

In the first case, with \(c=0.1\cdot l\), formulas (21) for a uniformly distributed load may be transformed into the following form:

\[ \boxed{ \begin{aligned} \sum_{n}^{m} T_{\max} &= \frac{1}{18z}\cdot\left(q\cdot b_m^{2}-g\cdot a_n^{2}\right)\\ \sum_{n}^{m} T_{\min} &= \frac{1}{18z}\cdot\left(g\cdot b_m^{2}-q\cdot a_n^{2}\right) \end{aligned} } \qquad \ldots \quad (30) \]

In those cases where the load is transmitted to the beam at the panel points—for example, from auxiliary transverse beams—it is, of course, necessary to take the length of the segment equal to the distance between panel points (the panel length), or equal to half the span in less important structures.

Influence lines of \(\sum T\) in continuous beams.

Everything stated above concerning the construction of influence lines, their loading, and the selection of the length of the segment for calculation is analogously applicable also to continuous beams.

Having the influence lines \(M\)* and the lever arms of the internal forces \(z\), it is not difficult to construct, from the basic equation (3), the influence line \(\sum T\) for any segment. As an example, Fig. 8 gives several typical influence lines \(\sum T\) in a two-span continuous beam of constant section. For clarity, each line is constructed in two forms: from the direct superposition, left on right, of the two moment lines for the sections forming the boundaries of the segment, and reduced to a horizontal closing line.

* The ordinates of the influence lines \(M\) for continuous beams may be taken directly from tables placed in various reference handbooks, for example \(^{9)}\,^{10)}\) and others.

The ordinates are plotted precisely, with the scale observed, and, for greater generality, the constant factor \(\dfrac{l}{z}\) has been taken equal to unity. In passing to the actual values of the influence lines \(\sum T\), the ordinates must be multiplied by the actual value of the quantity \(\dfrac{l}{z}\). It is easy to see that these lines may also be influence lines for increments of moments, if the constant factor is taken to be the quantity \(l\), owing to the fact that \(\sum T \cdot z = \Delta M\).

Of some interest is the influence line \(\stackrel{10}{\underset{5}{\sum}} T\) (Fig. 5-b), constructed for a portion equal to the right half of the left span. From its form one may immediately conclude that, in order to obtain the numerically greatest value of \(\stackrel{10}{\underset{5}{\sum}} T\), i.e. the greatest number of bent-up bars, both spans must be loaded with the full load; here we have a loading analogous to that for obtaining the maximum*) support moment.

Let us trace how the form of the influence lines \(\sum T\) may change when the beam has haunches at the supports. If the moment of inertia of such a beam is taken as constant for the calculation of deformations, it is clear that only the influence lines for those portions in which one of the boundaries is the support section or a section within the haunch will be changed; the remaining influence lines will remain unchanged.

In Fig. 8, the changed influence lines are plotted with dashed lines for the case when the lever arm of the internal forces in the span is equal to \(0.75\) of its value at the support, i.e. when

\[ z = 0.75 \cdot z_{\mathrm{в}} . \]

The order of construction is clear without special explanation.

Let us examine certain features inherent, in general, in all end spans of any continuous systems with a free end support.

The bending moment in a section of the end span is

\[ M_x = M_{ox} + \frac{M_{\mathrm{в}} \cdot x}{l}, \]

where \(M_{ox}\) is the moment as in a simple beam, and \(M_{\mathrm{в}}\) is the support moment (Fig. 9).

For the portion of this span from \(x\) to \(l\), according to equation (3),

\[ \stackrel{l}{\underset{x}{\sum}} T = \frac{M_{\mathrm{в}}}{z_{\mathrm{в}}} - \frac{M_x}{z} = \frac{1}{z} \cdot \left( M_{\mathrm{в}} \cdot \frac{z}{z_{\mathrm{в}}} - M_x \right) = \frac{1}{z} \cdot \left( \xi \cdot M_{\mathrm{в}} - M_x \right). \]

Substituting its value for \(M_x\), we obtain:

\[ \stackrel{l}{\underset{x}{\sum}} T = \frac{1}{z} \cdot \left( \xi \cdot M_{\mathrm{в}} - M_{ox} - \frac{M_{\mathrm{в}} \cdot x}{l} \right) \quad \text{or} \]

\[ \stackrel{l}{\underset{x}{\sum}} T = \frac{1}{z} \cdot \left[ M_{\mathrm{в}} \left( \xi - \frac{x}{l} \right) - M_{ox} \right] \ldots\ldots\ldots (31) \]

By this equation it is very convenient to construct influence lines \(\sum T\) for portions with a boundary at the support.

*) By “max.” is meant the greatest numerical value.

Here one may encounter rather curious special cases:

1. If $\xi=\dfrac{x}{l}$, then

\[ \sum_x^l T=-\frac{M_{0x}}{z}, \ldots \ldots \ldots \ldots \ldots \ldots \ldots (32) \]

i.e., in this case the sum of the shearing forces is equal to the analogous value of them in a simple beam, and the corresponding influence line $\sum T$ turns into a triangle (the entire right branch disappears) (Fig. 9-e).

1. If $\xi=1$, and $\dfrac{x}{l}=\dfrac{1}{2}$, then (Fig. 8-a)

\[ \sum_{1/2}^l T=\frac{1}{z}\left(\frac{M_{\mathrm{в}}}{2}-M_{0x}\right)\ldots \ldots \ldots \ldots \ldots \ldots (33) \]

Consequently, $\sum T$ is equal to the difference between one-half of the support moment and the total moment at midspan, as in a simple beam, divided by $z$.

1. For $\xi<\dfrac{x}{l}$, from equation (31) we see that the entire right branch of the influence line $\sum T$ will even change sign (Fig. 9-f), i.e., in this case, in order to obtain the greatest numerical value of $\sum_x^l T$, it will be necessary to load the first span and all the others alternately (analogously to the loading when calculating $\max M$ in the first span).

In practical examples,

\[ z=\sim 0.75\cdot z_{\mathrm{в}}, \quad \text{i.e.} \]

\[ \xi=\sim 0.75, \]

and, consequently, only for the section adjacent to the support, of length

\[ c=0.25\,l \]

and less, can the right branch of the influence line $\sum T$ turn into zero or change sign (Fig. 9-g).

1. To this same category belongs the case of determining $\sum T$ within the limits of a bent-up bar formed, for example, by an inclination of $1/3$.

If the length of the bent-up bar is denoted by $t$, then

\[ \frac{x}{l}=\frac{l-t}{l}. \]

Further, without great error, one may put

\[ \xi=\frac{z}{z_{\mathrm{в}}}=\frac{h}{h_{\mathrm{в}}}=\frac{3h}{3h+t}. \]

In the present case the difference is always

\[ \xi-\frac{x}{l}=\frac{3h}{3h+t}-\frac{l-t}{l} =\frac{t\cdot(3h+t-l)}{l\cdot(3h+t)} \]

a negative quantity, owing to the fact that usually

\[ h<\frac{l-t}{3}. \]

Consequently, almost always the influence line of \(\Sigma T\) within the limits of a bent-up bar will have the form shown in Fig. (9-g), and loading along it will be analogous to that which is used to obtain \(\max M\) in a span, i.e., through the span.

Moreover, owing to the fact that the adjacent portions of the influence line of the support moment—for example, in a beam with equal spans—are almost equal to one another, the influence of the support moment on the magnitude of \(\Sigma T\) will be reflected only in the amount of one half of its greatest value*).

Practical application of influence lines.

On the basis of all that has been set forth above, the following procedure may be established for calculating bent-up bars by means of the influence lines of \(\Sigma T\).

a) In simple beams,

When calculating beams of responsible structures with heavy temporary loading—for example, railway bridges or similar structures in civil engineering (crane girders, etc.)—it is necessary to carry out the calculation fully by influence lines for \(\Sigma T\), taking the lengths of the portions equal to \(0.1 \cdot l\) (see above). If, in doing so, the load is taken into account in the form of an equivalent uniformly distributed load (for each portion it may be different), then \(\max \Sigma T\) and \(\min \Sigma T\) may be computed directly from ready-made formulas derived from the influence line, without constructing the influence lines themselves (by formulas 21 and 30).

Otherwise, having constructed the influence lines, we load them in the usual manner for max and min.

In any case, for each portion, \(c = 0.1 \cdot l\), we find the greatest and the least value of \(\Sigma T\).

The total sum of the individual \(\Sigma T\) of one sign, after substitution in formula (6), will give the total number of bent-up bars of one direction throughout the entire span.

It must be borne in mind that, with the accepted sign rule for \(M\) (bending downward—positive \(M\)), a positive value of \(\Sigma T\) corresponds to the inclination of the bent-up reinforcement to the left, and a negative value—to the contrary.

To find the places of bending, special curves may be constructed. We successively sum the found \(\Sigma T_{\max}\) for the portions and, laying off these quantities as ordinates from the horizontal, connect them by a smooth curve (Fig. 10). We obtain a diagram which we shall call the envelope of the greatest increments of \(\Sigma T\)**).

Any ordinate of such envelopes is equal to the greatest increment—possible for different positions of the load—of the shearing forces to the left of the section against which the ordinate is taken.

Above, in examining the question of the influence of the length of the portion on the accuracy of the results, we already dealt with similar curves, calling them the curves \(\Sigma T_f\). Consequently, the envelope of the greatest increments may be called the envelope \(\Sigma T_f\).

In a simple beam of constant section under a uniformly distributed load identical for all sections of the span, this envelope may be constructed directly from the exact equation (25) for \(\Sigma T_f\).

Having constructed the envelope, we assign the stirrup section \(f_e\), the distance between stirrups \(e\), and draw the inclined line \(ab\) at \(\angle \beta\) to the horizontal (Fig. 10), with the stirrups at equal distances.

*) Here we have a partial confirmation of the well-known rule of Prof. Mörsch, according to which, in calculations for shear within the limits of a bent-up bar, the greatest support moment from temporary loading is taken into account in the amount of one half... etc. See ¹) ²).

**) Similarly for the least values as well.

The angle \(\beta\) characterizes the increment \(\Sigma T_b\) taken up by the stirrups, and is determined by the condition:

\[ \tan \beta=\frac{\Sigma T_b\cdot m_1}{l\cdot m_2} =\frac{f_e\cdot\sigma_{e_1}\cdot m_1}{e\cdot m_2}; \qquad \ldots \qquad (33) \]

here \(m_1\) and \(m_2\) are the scale numbers of the epure drawing.

We draw, at the angle \(\beta\), i.e. parallel to the inclined line \(ab\), a tangent to the curve; the segment \(\dfrac{L}{ks}\) between the tangent and the inclined line, taken on the force scale, gives the absolute maximum value of \(\Sigma T_s\), according to which

\[ F_s=\frac{\Sigma T_s\cdot \cos 45^\circ}{\sigma_e} =\frac{\Sigma U_s}{\sigma_e}. \]

The further construction for finding the locations of the bends is entirely analogous to the method proposed by Prof. Löser\(^4\) in calculating bends from epures of bending moments, and is clear from the drawing (see drawing 10).

In beams of constant section, instead of the epure of the largest increment \(\Sigma T\), one may construct the epure of the largest increments of moments \((\triangle M)\), which is proportional to the epure \(\Sigma T_f\) (see below, example No. 4).

If the beam is completely symmetrical in outline, then the epure of the smallest increments need not be constructed, since it will be inversely symmetrical to the epure of the largest increments; that is, the reinforcement in the right half will be exactly the same as in the left, but inclined in the opposite direction, which is obvious.

It was noted above that in beams with a comparatively lighter live load (for example, in most civil structures, in bridges carrying an ordinary road, etc.), it is quite sufficient to take the length of the segment equal to one-half the span. In this case the influence line \(\Sigma T\) becomes a triangle identical with the moment line at midspan, and the greatest value of the magnitude \(\Sigma T\) of one direction over the whole span will be proportional to the greatest moment, since

\[ \Sigma T_{\max}=\frac{M_{\max}}{z}. \]

The moment \(M_{\max}\) is usually found in advance for selecting the beam section; consequently, it need not be found again.

But, as became clear in studying the influence of the length of the segment, in the present case, under the action of a moving load, the result found will always be somewhat smaller than the true, absolute maximum value \(\Sigma T_{f\max}\). And, not wishing deliberately to reduce the margin of strength in the bent-up bars, it is useful to increase the value found for \(\Sigma T_{\max}\), according to formula 28, by multiplying it by the coefficient \(\mu\), which, for a uniformly distributed load, may be found from formula 29 or else taken from the corresponding table or graph (drawing 7).

For other loads, the coefficient \(\mu\) may be determined approximately by comparison with a uniformly distributed load equivalent to the given case; approximately, \(\mu=1.07—1.18\).

It is quite obvious that, under the action of such moving loads, for which the greatest moment in the span occurs not at midspan but in other sections, it is precisely its value that must be introduced into the calculation; that is, the length of the segment should be taken equal to the distance from the support to the section with the greatest moment.

In such calculations, the points of bend-up are assigned for constructive reasons. If desired, they may also be found from the envelope of moments corresponding to the given load position for which the design \(M_{\max}\) was found, since here the corresponding branch of the envelope of the greatest increments of moments turns into the branch of the envelope \(M\) (Fig. 6).

c) In continuous beams.

Here we must proceed in a manner entirely analogous to the preceding case. In continuous beams with a heavy load it will be necessary to carry out the whole calculation completely by means of the influence lines \(\Sigma T\), dividing each span into ten parts, and, in the final result, to construct separately for each span the envelopes of the greatest and least increments of \(\Sigma T\) or \(\Delta M\), if the beam is of constant cross-section.

The procedure for computing ordinates and constructing envelopes remains the same as in simple beams; the entire difference will be, first, in the form of the influence lines \(\Sigma T\) for the individual portions, and therefore also within the limits of loading, and, second, in the very form of the envelope of increments. It goes without saying that if the spans of the continuous beam are wholly symmetrical, then the envelope of the least increments need not be constructed, again owing to its inverse symmetry to the envelopes of the greatest increments.

When considering the question of the influence lines \(\Sigma T\) for continuous beams, it has already been noted that their construction can easily be carried out by using ready-made ordinates of influence lines of bending moments, taken from existing tables. In this case it is more convenient to construct them with the constant multiplier

\[ \frac{l}{z}, \]

where \(z\) is the span arm of the internal forces.

For the haunched portions it is better to retain the same multiplier \(l/z\), allowing for variability by introducing the ratio

\[ \xi = \frac{z}{z_{\mathrm{v}}} \quad \text{(see above).} \]

To speed up such calculations for continuous beams, both equal-span and unequal-span but with certain different ratios of span lengths, ready-made tables of ordinates of the influence lines \(\Sigma T\) may be compiled for successive portions equal to \(0.1 \cdot l\), or even \(\frac{1}{12} \cdot l\), with the constant multiplier

\[ \frac{l}{z}. \]

For haunched portions such tables may have branches for various \(\xi\), for example from \(\xi = 1\) (a beam without haunches) to \(\xi = 0.5\), with gradations of \(0.1\); intermediate values will be taken in calculations by interpolation.

In developing and compiling such tables for beams with haunches, it is useful to take into account the effect of the variability of the moment of inertia on the magnitude of the ordinates. If such tables are compiled, then the computational work in constructing the influence lines \(\Sigma T\), already insignificant without them, will almost disappear entirely.

The indicated calculation of bent-up bars in continuous beams by influence lines \(\Sigma T\) and by envelopes of the greatest increments, although it gives us complete accuracy and definiteness of results, is nevertheless rather cumbersome and therefore will be justified only in special cases noted above (railway bridges, etc.).

In beams with a relatively lighter live load, by analogy with simple beams, one may, without great error, divide each span for calculation into only two portions, with the boundary at a section usually close to the middle of the span, where the bending moment is greatest.

When obtaining \(\Sigma T_{\max}\) of one direction in each span, there is absolutely no need to draw the influence line \(\Sigma T\): it is only necessary to have an idea of its form in order to know exactly which spans must be loaded with the live load. Knowing the loading, we find the moments at the boundaries of the segment and, by the fundamental equation (3), compute the design \(\Sigma T\), from which, after specifying the number of stirrups, the required area of bent-up bars of one direction is determined.

Similar influence lines have already been constructed by us (Figs. 8 and 9). Their form showed that loading according to them is completely identical with the loading used in obtaining the numerically greatest value of the adjoining support moment.

But the latter, in continuous beams, is usually already found in advance for selecting the section; consequently, for the calculation of bent-up reinforcement there will always be lacking only the corresponding span moment (see example No. 3).

Even the calculation of these span moments can be avoided by replacing them, quite correctly, by the greatest moments in the spans, for the following reasons.

First, as the investigation of the influence of the length of the segment has shown us, by dividing each span into only two segments we shall always obtain an overestimated result, which, in order to approximate the true, absolutely greatest value in simple beams, is multiplied by a coefficient \(\mu\) greater than unity; a similar coefficient \(\mu>1\) will also occur in continuous beams.

Second, in continuous beams with haunches, for a segment within the limits of the haunch, as we saw above, the loading is completely identical with the loading used in obtaining the maximum span moment, but not the support moment (see Fig. 9-g).

Thus, we shall come closer to the true value of \(\Sigma T_f\), and, moreover, it may be asserted that we shall not in the least exaggerate the result, if we introduce into the fundamental equation (3) simultaneously the numerically greatest values of both the support moment and the moment in the span.

Both bending moments will always have been found in advance for selecting sections; consequently, the determination of the total required number of stirrups and bent-up bars in each span, with the aid of the fundamental equation and formula (6), is reduced to very simple operations—so simple that there is absolutely no need for any special auxiliary tables for calculating \(\Sigma T\).

Of course, it is entirely possible, on the basis of the method set forth, with the aid of the influence lines \(\Sigma T\) and envelopes of increments, to prepare for a number of continuous beams, with and without haunches, ready-made tables giving directly the absolutely greatest values of \(\Sigma T_f\) in each span, in both directions, for certain constant multipliers. For example, under a uniformly distributed load the constant multiplier will be the quantity

\[ \frac{q l^2}{z}, \]

and under concentrated loads \(P\), fixed at their points of application—

\[ \frac{P \cdot l}{z} \]

and so forth.

However, the author sees no particular need for such tables, since in important structures with heavy moving loads, for quite understandable reasons, one cannot in any case avoid constructing envelopes of the greatest increments of shearing forces; in other cases, the results, very simply found by direct substitution of the already available bending moments into the fundamental equation, will differ negligibly from those absolutely greatest values that such tables could give.

Examples

1. Determination of the total quantity of bent-up bars in a simple beam under a uniformly distributed load.

Problem statement: A single-span, simply supported reinforced-concrete floor beam, having a design span \(l = 8.25\) m and an internal-force lever arm in the span \(z = 0.68\) m, is subjected to a permanent uniformly distributed load \(g = 2.45\) t/m and a live load \(p = 2.85\) t/m. Determine for it the total required area of bent-up bars.

Solution:

The load ratio is

\[ m=\frac{g+p}{g}=\frac{2.45+2.85}{2.45}=2.165. \]

From the table compiled according to formula (29), we find

\[ \mu=1.114. \]

Further,

\[ \Sigma T_{f\max} = \mu\cdot\frac{M_{\max}}{z} = \mu\cdot\frac{(g+p)\cdot l^2}{8\cdot z} = 1.114\cdot\frac{(2.45+2.85)8.25^2}{8\cdot0.68} = 73.8\text{ t}. \]

The length of the part of the span having a positive increase of \(\Sigma T\), according to formula (25), is

\[ x_0= \frac{l\cdot\sqrt{m}}{\sqrt{m}+1} = \frac{8.25\cdot\sqrt{2.165}}{\sqrt{2.165}+1} = 4.92\text{ m}. \]

With two-legged stirrups of RE 8 mm, with \(f_e=1.01\text{ cm}^2\), placed every 0.25 m, their area over the length \(x_0 = 4.92\) m is

\[ F_b=1.01\cdot\frac{4.92}{0.25}=19.85\text{ cm}^2. \]

Further, by formula (6), with \(\psi=1\), we obtain the total area of bent-up bars in one direction:

\[ F_s=0.707\cdot\left(\frac{73860}{1200}-19.85\right)=29.4\text{ cm}^2. \]

2. Example of checking stresses in bent-up bars.

Problem statement. *) In the main girder of a railway beam bridge with span \(l = 6.80\) m (Fig. 11), check the stress in the third bent-up bars from the left, consisting of 2 RE 36 mm. All dimensions are indicated in Drawing (11-a). The permanent load is \(g = 3.52\) t/m, and three concentrated loads of \(G = 0.13\) t each, located at midspan and over the supports. The live load is the standard train of 1907.

Solution:

The length of the segment occupied by the bars under consideration is \(c = 0.89\) m. The distances from the boundaries of the segment to the supports are:

\[ a_1=0.65+\frac{0.89}{2}=1.09\text{ m}, \]

\[ b_2=6.80-1.09-0.89=4.82\text{ m}. \]

) The problem statement is taken from Perederii, G. P.—Course on Reinforced-Concrete Bridges*, 1923, p. 307.

Arm of the internal forces, according to the condition,

\[ z = 0.84\ \text{m}. \]

Further,

\[ \frac{c}{z}=\frac{0.89}{0.84}=1.06. \]

In Fig. (11-b) the corresponding influence line \(\Sigma T\) has been constructed. Placing the live load on it in the unfavorable position, we find:

\[ \Sigma T^{p}_{\max}=\left(180+80\cdot 0.32\right)\frac{1.06}{6.8\cdot 2}=16\ \text{t}. \]

From the dead load, by formula (21), and from the influence line, we have:

\[ \Sigma T_g= \frac{1.06\cdot 3.52}{2(6.8-0.89)} \cdot \left(4.82^2-1.09^2\right) + \frac{0.13\cdot 1.06}{2} =7.64\ \text{t}. \]

The total is

\[ \Sigma T_{\max}=16.0+7.64=23.64\ \text{t}. \]

From the drawing it is seen that the stirrups—double, of RE 8 mm (four-cut)—are placed every 25 cm, i.e., their area in the section under consideration is

\[ F_b=\frac{0.89}{0.25}\cdot 2.01=7.15\ \text{cm}^2. \]

Next we find the actual stress:

\[ \sigma_e= \frac{\Sigma T}{\psi\cdot F_b+\sqrt{2}\cdot F_s} = \frac{23640}{\,^{7}/_{9}\cdot 7.15+\sqrt{2}\cdot 20.36} = 695\ \text{k}/\text{cm}^2. \]

3) Example of determining the total amount \(F_s\) in a continuous beam.

Condition: For a two-span continuous beam with spans \(l=l_1=l_2=7.35\) m, find the total amount of required bent-up reinforcement. The dimensions of the beam are indicated in Fig. (12). The dead load consists of a uniformly distributed load \(g=0.6\ \text{t}/\text{m}\) and a concentrated force \(G=11.05\ \text{t}\) at midspan. The live load is \(P=8.55\ \text{t}\), likewise at midspan.

For these loads it was previously found that the greatest moment at midspan is \(M_{\max}=+27.30\ \text{t m}\), and the greatest negative moment at the intermediate support is \(M_b=-31.1\ \text{t m}\).

Solution: We take the lengths of the sections equal to the distance between the concentrated forces (between the nodes),*) i.e., we shall have four equal sections:

\[ c_1=c_{II}=c_{III}=c_{IV}=l/2. \]

Mentally imagining the form of the corresponding influence lines, we conclude that, in order to obtain \(\Sigma T^I_{\max}\), it is necessary to load the first span with the live load, and for \(\Sigma T^{II}_{\max}\)**)—both spans.

Thus, for calculating \(\Sigma T\), we lack only the moment \(M\) at midspan from full loading.

From the tables in Beton-Kalender we find:

\[ M=\frac{g l^2}{16}+0.156\cdot (P+G)\cdot l = \frac{0.6\cdot 7.35^2}{16} + 0.156\cdot (8.55+11.05)\cdot 7.35 = \]

\[ =+24.42\ \text{t m}. \]

) The boundary between sections is taken to be that section in the span in which the greatest span bending moment is obtained.
*) We find the greatest value numerically.

Further, from equation (3), we shall have: *)

\[ \Sigma T^I_{\max}=\frac{M_{\max}}{z_0}=\frac{27.3}{0.9\cdot0.69}=+44.0\ \text{t}. \]

\[ \Sigma T^{II}_{\max}=\frac{M_{\mathrm{в}}}{z_{\mathrm{в}}}-\frac{M}{z_0} =-\frac{31.1}{\tfrac{7}{8}\cdot0.94}-\frac{24.42}{0.9\cdot0.69} =-77.2\ \text{t}. \]

The right span is symmetrical; consequently the values of \(\Sigma T\) for it will be equal in magnitude to the analogous values for the left span, but opposite in sign, i.e. inversely symmetrical.

On the basis of what has been said,

\[ \Sigma T^{III}_{\max}=-\Sigma T^{II}_{\max}=+77.2\ \text{t}., \]

\[ \Sigma T^{IV}_{\max}=-\Sigma T^{I}_{\max}=-44.0\ \text{t}. \]

We install single two-leg stirrups of \(RE\ 1/4''\) at \(e=25\) cm, with \(f_e=0.63\ \text{cm}^2\). Their total area in each section is

\[ F_b=\frac{c\cdot f_e}{e}=\frac{735\cdot0.63}{2\cdot25}=15.55\ \text{cm}^2 \]

The required area of bent-up bars according to form. (6), with \(\psi=1\), bearing in mind that, according to the standards, not more than \(1/3\,\Sigma T\) is transferred to the stirrups:

\[ F_s^{IV}=F_s^I=\frac{2\Sigma T}{3\sigma_e}\cdot\cos45^\circ =\frac{2\cdot44000}{3\cdot1200}\cdot0.707 =17.3\ \text{cm}^2 \]

\[ F_s^{II}=F_s^{III}= \left(\frac{\Sigma T}{\sigma_e}-\psi\cdot F_b\right)\cos45^\circ = \left(\frac{77200}{1200}-15.55\right)0.707 =34.4\ \text{cm}^2 \]

We adopt in the I-st and IV-th sections

\[ 2\,RE\ 1\tfrac{1}{8}''+1\,RE\ 1'' \quad \text{with } F_s=17.89\ \text{cm}^2, \]

and in the II-nd and III-rd sections, \(4\,RE\ 1\tfrac{1}{8}''+2\,RE\ 1''\) with \(F_s=35.78\ \text{cm}^2\).

These bars may be distributed among the sections constructively, taking into account that, according to the signs obtained before \(\Sigma T\), the bars in the I-st and III-rd sections must be inclined to the left, and in the II-nd and IV-th—to the right.

4) Example of calculating bent-up bars in a bridge beam by means of the influence lines of \(\Sigma T\).

It is required to calculate the stirrups and bent-up bars in the main rib of a reinforced-concrete girder bridge for a standard-gauge railway, with design span \(l=8.0\) m.
Load—“N”—1925.

Previously, in the calculation, it was found:

1. Permanent load, uniformly distributed, \(g=4\) t/m of beam.
2. The beam has a constant total depth \(d_0=1\) m, useful depth \(h=0.92\) m.

We divide the span into 8 sections, each of \(c=1\) m. We construct the influence lines \(\Delta M\) for each section, but, since the beam is symmetrical, it is sufficient to construct only four i.l.:

\[ \Delta M_{0-1},\ \Delta M_{1-2},\ \Delta M_{2-3},\ \Delta M_{3-4}\quad(\text{Fig. }13). \]

*) The values of \(z\) have been taken approximately: for a T-beam—\(0.9\,h\), for a rectangle—\(7/8\).

We load with a concentrated temporary load at \(\max \Delta M\) and find, using the tables of static moments of train “H,” from the temporary load:

\[ \Delta M_{0-1}=\frac{1}{2\cdot 8}\cdot(375+125\cdot 1)=36.25\ \text{t. m.} \]

\[ \Delta M_{1-2}=\frac{1}{2\cdot 8}\cdot 375=23.45\ \text{t. m.} \]

\[ \Delta M_{2-3}=\frac{1}{2\cdot 8}\cdot(225+100\cdot 0.5)=17.20\ \text{t. m.} \]

\[ \Delta M_{3-4}=\frac{1}{2\cdot 8}\cdot(112.5+75\cdot 1)=11.72\ \text{t. m.} \]

\[ *)\ \Delta M_{4-5}=\frac{30}{2\cdot 8}\cdot(3+1.5)=8.45\ \text{t. m.} \]

\[ \Delta M_{5-6}=\frac{30}{2\cdot 8}\cdot(2+0.5)=4.69\ \text{t. m.} \]

\[ \Delta M_{6-7}=\frac{30}{2\cdot 8}=1.875\ \text{t. m.} \]

\[ \Delta M_{7-8}=0\quad(\text{the entire i. l. is negative}). \]

Further, from the permanent load by formula (21), for example, for the segment \((1—2)\)

\[ \Delta M_{1-2}=\frac{c\cdot g}{2(l-c)}(b_2^2-a_1^2)=\frac{g}{14}\cdot(b_2^2-a_1^2). \]

By this formula, taking into account the reverse symmetry for the right-hand part, we find from the permanent load:

\[ \Delta M_{0-1}=-\Delta M_{7-8}=\frac{4\cdot 7^2}{14}=14\ \text{t. m.} \]

\[ \Delta M_{1-2}=-\Delta M_{6-7}=\frac{4}{14}\cdot(6^2-1^2)=10\ \text{t. m.} \]

\[ \Delta M_{2-3}=-\Delta M_{5-6}=\frac{4}{14}\cdot(5^2-2^2)=6\ \text{t. m.} \]

\[ \Delta M_{3-4}=-\Delta M_{4-5}=\frac{4}{14}\cdot(4^2-3^2)=2\ \text{t. m.} \]

Adding the quantities obtained, we find the values of the greatest increments of moments by segments:

\[ \begin{aligned} \Delta M_{0-1}&=36.25+14=+50.25\ \text{t. m.}\\ \Delta M_{1-2}&=23.45+10=+33.45\ \text{t. m.}\\ \Delta M_{2-3}&=17.20+6=+23.20\ \text{t. m.}\\ \Delta M_{3-4}&=11.72+2=+13.72\ \text{t. m.}\\ \Delta M_{4-5}&=8.45-2=+6.45\ \text{t. m.}\\ \Delta M_{5-6}&=4.69-6=-1.31\ \text{t. m.}\\ \Delta M_{6-7}&=1.875-10=-8.125\ \text{t. m.}\\ \Delta M_{7-8}&=-14\ \text{t. m.} \end{aligned} \]

*) For the segments of the right half of the span, loading by two 30-ton axles turned out to be less disadvantageous.

Ordinates of the diagram of the greatest increments of moments:

\[ \begin{aligned} y_0 &= 0.\\ y_1 &= 50.25\ \text{t·m}.\\ y_2 &= 50.25 + 33.45 = 83.70\ \text{t·m}.\\ y_3 &= 83.70 + 23.20 = 106.90\ \text{t·m}.\\ y_4 &= 106.90 + 13.72 = 120.62\ \text{t·m}.\\ y_5 &= 120.62 + 6.45 = 127.07\ \text{t·m}.\\ y_6 &= 127.07 - 1.31 = 126.76\ \text{t·m}.\\ y_7 &= 126.76 - 8.125 = 118.635\ \text{t·m}.\\ y_8 &= 118.635 - 14.00 = 104.635\ \text{t·m}. \end{aligned} \]

From the ordinates found, we construct the diagram of the greatest increments \(\Delta M\) (Fig. 13).

We propose to place double stirrups of RE 8 mm, with \(f_e = 2.01\ \text{cm}^2\) (area of 4 branches), every 25 cm, uniformly along the span.

We shall take the lever arm of the internal forces as

\[ z = 7/8\,h = 7/8 \cdot 0.92 = 0.81\ \text{m}. \]

Then the total increment of moments taken by the stirrups within the entire span is

\[ \Delta M^b_{\max} = \frac{z \cdot f_e \cdot l \cdot \sigma_{e_1}}{e} = \frac{81 \cdot 2.01 \cdot 800 \cdot 700}{25 \cdot 10^5} = 36\ \text{t·m}. \]

Further, on the basis of what was set forth above, we draw the inclined line \(ab\) and the tangent \(df\), parallel to it. In the drawing we obtain the segment \(ks\), which, when taken in the scale of the ordinates of the diagram, gives

\[ \max \Delta M_s = 105\ \text{t·m}, \]

i.e., the absolutely greatest increment of moments that must be transferred to the bent-up bars.

From it the required quantity of bent-up reinforcement is

\[ F_s = \frac{\Delta M_s \cdot 0.707}{z \cdot \sigma_e} = \frac{105 \cdot 10^5 \cdot 0.707}{81 \cdot 900} = 102\ \text{cm}^2. \]

We adopt

\[ 20\ \text{RE}\ 26\ \text{mm},\quad \text{with } F_s = 106.18\ \text{cm}^2. \]

The locations of the bends are found graphically from the diagram \(\max \Delta M\) (see Fig. 13), with the condition that in each plane two bars will be bent up.

Owing to the symmetry of the beam, we do not construct the diagram of the least increments of moments; instead, in the right half we take the same number of bent-up bars as in the left, and at the same distances.

For comparison, upon the diagram of the greatest increments of moments there is superposed the diagram of the greatest moments, computed for the given case.* From the drawing we see that the first diagram entirely covers the second, with a discrepancy in the greatest ordinates of

\[ \frac{127.5 - 100.75}{127.5}\cdot 100 = 21\%. \]

* The calculation itself is not given for the sake of brevity.

Conclusion.

Let us enumerate the principal propositions of the new method for calculating bent-up bars.

1. The basic equation (3), which relates the bending moments at the boundaries of a segment to the sum, or resultant, of the shearing forces on that segment, makes it possible to determine, for a beam of practically any form, the required result entirely without knowledge of the transverse forces, the need for which falls away, provided only that we are not interested in the magnitude of the shearing stresses in the sections*).

2. Influence lines of \(\Sigma T\), constructed from the basic equation, make it possible to determine the actually required number of bent-up bars and stirrups for a beam under the action of a moving load, or to find, under the same conditions, the true acting stress in the bent-up bars in check calculations. The latter circumstance will be valuable where the sufficiency of the provided bent-up reinforcement has raised some special, serious doubts.

3. Influence lines of \(\Sigma T\), even when represented schematically, make it possible to know the limits of loading; and, if we propose to assign the locations of the bends constructively, the subsequent calculation can be carried out without influence lines, by the basic equation (3). In this case part of the moments has been found beforehand in selecting the sections; consequently, the calculation is considerably simplified.

4. Diagrams of the greatest increments of \(\Sigma T\), as a result of calculation by influence lines of shearing forces, make it possible to find the truly absolutely greatest required number of bent-up bars in the span, and at the same time make it possible to distribute these bars properly along the span. On these diagrams of the growth of \(\Sigma T\) one may draw the stepped line of increase of the adopted quantity of bent-up reinforcement, which, in a rationally designed beam, must enclose the curve \(\Sigma T_f\), or, as an extreme limit, cut into it with some corners while protruding with others (Fig. 10). Thus, in our opinion, a pressing problem in the design practice of reinforced-concrete bridges is solved.

5. The established analogy between the influence line of \(\Sigma T\) and the influence line of the horizontal projection of the force in the outgoing diagonal of a truss will perhaps permit further generalizations and will help in investigations, for example, of the question of calculating bent-up bars in eccentrically compressed or tensioned members.

6. The method of calculating bent-up bars discussed in the present article is fully applicable only to beams in simple bending, but, with certain restrictions and assumptions, may also be applied to eccentrically compressed or eccentrically tensioned members.

The point is that the basic equation (3) gives us \(\Sigma T\) along the geometric axis, but owing to the fact that in beams the neutral axis coincides with the geometric one, we also determined from it the resultant of the principal tensile forces on the segment, etc.

When, however, \(M\) acts together with \(N\), the neutral axis, as is known, will not coincide with the geometric one; thus, although we do have the possibility, in accordance with the principle of independent action of forces, to determine from equation (3) the sum of the shearing forces along the geometric axis, this does not solve the question, because for calculating bent-up bars one needs \(\Sigma T\) along the neutral axis, along which alone it is possible to find the resultant of the principal tensile forces. Before proceeding to the investigation of this question, let us note that, since in crossbeams

*) In reinforced-concrete bars, determining the magnitude of the shearing stresses has no practical meaning, since according to the standards all shearing stresses must be transferred to stirrups and bent-up bars. Moreover, as was noted at the beginning, the permissible limit of \(15\ \mathrm{kg}/\mathrm{cm}^2\) is rarely exceeded.

bending prevails over the action of the normal force, the neutral axis will deviate only slightly from the geometric one, i.e., the cross-beam may be considered, without great error, as a beam, and the method set forth may be applied in full. The assumption will tend toward increasing the factor of safety in the bent-up bars.

For the posts of frames, however, in which the action of the compressive normal force predominates, the calculation of bent-up bars is almost unnecessary, since usually the entire section is in compression; in exceptional cases, when the eccentricities are large, the posts may be treated as beams.

Tomsk, March 1928.

Literary sources cited in the work

1. Mörsch, E. Der Eisenbetonbau, 5th ed., I.
2. German Concrete Union (E. V.). Design and Calculation of Reinforced-Concrete Structures (translation), 1928, ch. III, pp. 317–371.
3. Foerster, M. Die Grundzüge des Eisenbetonbaues, 1926, pp. 361–377.
4. Löser, B. Bemessungsverfahren..., 1925, pp. 65–94.
5. Perederii, G. Course on Reinforced-Concrete Bridges, 1923, pp. 51, 287, 307.
6. Gosplan of the USSR. Temporary Technical Conditions and Standards for the Design and Construction of Reinforced-Concrete Structures, 1926.
7. Scientific-Technical Committee of the NKPS. Temporary Technical Conditions and Standards for the Design and Construction of Railway Structures, 1926.
8. Timoshenko, S. Course in Structural Statics, Part I, 1926, p. 142.
9. Griot, G. Tables for Continuous Beams, 1928.
10. Handbook for Engineers of Construction Specialties, I, 1928.
11. Romanov, D. “Determination of the Sum of Shearing Forces in a Reinforced-Concrete Member of Variable Height from Bending Moments,” Proceedings of the Siberian Technological Institute, vol. 48 (II), issue 4, 1925.
12. Proceedings of the STC of the NKPS, issue No. 25, 1926.

Tomsk. OkrLit No. 919.
Print run: 500 copies.

8th Transpechat. Makarovsky Lane, No. 6.

Table 1.

For the article by Eng. D. E. Romanesvet.

a)

b)

Drawing 1.

c)

Distribution of compressive stresses.

With regard to concrete in tension.

Without regard to concrete in tension.

Drawing 2.

Plate II.
To the article by Eng. A. S. Romanov.

Fig. 3

Visible labels and inscriptions in the figure:

\(R\), \(A\), \(B\)

\(a\), \(b\)

\(1'\), \(1\), \(2'\), \(2\), \(F\), \(O_1\), \(O_2\)

\(z_1\), \(z_0\), \(z_2\)

\(K\)

\(u_1\), \(u_2\)

\(a_1\), \(a_2\), \(b_1\), \(b_2\)

\[ \frac{a_1}{z_1} \]

\[ \frac{a_2}{z_2} \]

\[ -\left(\frac{b_1}{z_1}-\frac{b_2}{z_2}\right) \]

\[ \frac{b_2}{z_2} \]

\[ \frac{b_1}{z_1} \]

\[ \left(\frac{a_2}{z_2}-\frac{a_1}{z_1}\right) \]

\(y_1\), \(y_2\)

\(\omega_a\), \(\omega_b\)

\[ \text{I.L. } \sum_{1}^{2} T \]

\[ \ell \]

\[ \lambda_a \]

\[ \lambda_b \]

Fig. 4

Visible labels and inscriptions in the figure:

\(R\), \(A\), \(B\)

\(F\), \(O_{1-2}\)

\(\beta\), \(\varphi\), \(\varphi_1\), \(\alpha\)

\(1\), \(2\)

\(h_1\), \(h_2\)

\(D_{1-2}\)

\(u_{1-2}\)

\(a_1\), \(a_2\), \(b_1\), \(b_2\)

\[ \frac{a_1}{h_1} \]

\[ \frac{a_2}{h_2} \]

\[ \frac{b_2}{h_2} \]

\[ \frac{b_1}{h_1} \]

\[ -\left(\frac{b_1}{h_1}-\frac{b_2}{h_2}\right) \]

\[ \left(\frac{a_2}{h_2}-\frac{a_1}{h_1}\right) \]

\[ \text{I.L. } D_{1-2}\cdot \cos\alpha \]

\[ \ell \]

Table III
for the article by Eng. D. E. Romanov.

Drawing 5.

a)

b)

Scale of ordinates:
at \(3\ \mathrm{mm} - 0.1 \cdot \dfrac{q l^{2}}{8z}\)

Note: \(m = 4\).

Drawing 6. Curves \(T_{f\max}\) for various \(C\).

\(\mu\)

Drawing 7. Dependence between \(\mu\) and \(m\).

Table IV.
For the article by Eng. V. E. Romanov.

Drawing 8. Lines of influence of \(\sum T\) in a two-span beam.

Supports: \(A\), \(B\), \(C\).
Span lengths: \(l\), \(l\).
Panel points: \(0, 1, 2, \ldots, 20\).

a)

Labels on the diagrams:

• infl. \(M_{10}\)
• infl. \(M_{5}\)
• I.L. \(\displaystyle \sum_{5}^{10} T\)

b)

Labels on the diagrams:

• infl. \(M_{2}\)
• infl. \(M_{6}\)
• I.L. \(\displaystyle \sum_{2}^{5} T\)

c)

Labels on the diagrams:

• infl. \(M_{10}\)
• infl. \(M_{8}\)
• I.L. \(\displaystyle \sum_{8}^{10} T\)

Scale of ordinates:

\[ \text{in }1\text{ mm} \;-\; 0.01\,\frac{l}{z} \]

Note: The influence lines of \(\sum T\) in beams are drawn with dashed lines for

\[ \xi=\frac{z}{z_{0}}=0.75, \]

and with solid lines for a beam of constant cross-section.

Plate V.

Fig. 9. Influence lines \(\sum T\) in a three-span beam.

a)

\[ \xi=\frac{z}{z_B} \]

b)

Influence line

\[ \sum_{4}^{8} T \]

c)

Influence line

\[ \sum_{10}^{15} T \quad \text{at } \xi=0.75 \]

d)

Influence line

\[ \sum_{5}^{10} T \quad \text{at } \xi=0.75 \]

e)

Influence line

\[ \sum_{5}^{10} T \quad \text{at } \xi=0.5 \]

f)

Influence line

\[ \sum_{5}^{10} T \quad \text{at } \xi=0.25 \]

g)

Influence line

\[ \sum_{8}^{10} T \quad \text{at } \xi=0.75 \]

Scale of ordinates:

\[ \text{in }1\text{ mm } - 0.01\,\frac{\ell}{z_B} \]

Note:
1) \(\xi\) is the ratio of the lever arm of the internal forces in the span to the lever arm at the support, i.e.

\[ \xi=\frac{z}{z_B}. \]

2) The ordinates \(M_B\) are taken as for a beam with \(J=\text{const}\).

3) The influence lines are shaded as in a simple beam.

TABLE VI
to the article by Eng. D. E. Romanov

Fig. 10. Determination of the locations of bends according to the diagram of the greatest principal stresses, \(\sum T\).

Visible labels and notations in the figure:

• \(\sum T\)
• \(A\), \(B\)
• \(\ell\)
• \(\alpha\), \(\beta\)
• \(\sum T_{\max}^{b}\)
• \(b\)
• diagram of greatest principal stresses, \(\sum T\)
• \(C_s\)
• \(s\)
• \(\sum U_s\)
• \(15^\circ\)
• \(\sum T_s\)
• \(45^\circ\)
• \(90^\circ\)
• \(g\)
• \(k\)
• \(f\)
• \(d_f \parallel ab\)

Fig. 11.

Visible labels and notations in the figure:

• \(a)\)
• \(1.0\)
• \(0.25\), \(0.89\), \(0.89\), \(0.97\)
• double of \(\varnothing 8\) mm
• \(2 \varnothing 22\)
• \(2 \varnothing 22\)
• \(2 \varnothing 36\)
• \(2 \varnothing 34\)
• \(0.30\)
• \(1\), \(2\)
• \(c = 0.89\)
• \(0.25\)
• \(6.80\ \mathrm{m}\)

Fig. 11.

Visible labels and notations in the figure:

• \(1.09\)
• \(L \cdot \sum T\)
• \(1.5\), \(1.5\), \(1.5\)
• \(0.32\)
• \(1.06\)
• \(c/2 = 1.06\)
• \(b)\)
• \(\ell_2 = 4.82\)

Fig. 12.

Visible labels and notations in the figure:

• \(a)\)
• \(0.7\ \mathrm{m}\)
• Beam facade.
• \(I\) sec.
• \(II\) sec.
• \(III\) sec.
• \(IV\) sec.
• \(g \parallel p\)
• \(g \parallel p\)

Visible load-scheme labels:

• \(b)\)
• \(\ell/2\)
• \(g\), \(p\)
• Load scheme.
• \(g\), \(p\)
• \(\ell/2\)
• \(g\)
• \(7.35\)
• \(7.35\)

Fig. 12.

Table VIII.

to the article by Eng. D. E. Romanov.

[Diagram labels visible on the plate:]

• \(2 \varnothing 26\)
• \(4 \varnothing 8\)
• \(l = 8\ \mathrm{m}\)
• \(0.25\)
• \(0.5\)
• \(30\)

Scales:

• lengths: \(1.5\ \mathrm{cm} = 1\ \mathrm{m}\);
• ordinates: \(1\ \mathrm{cm} = 20\ \mathrm{t\,m}\).

Curve of bending moments.

Curve of variable moments.

\(\max \Delta M\)

\(20 \varnothing 26\ \mathrm{mm}\)

\(M_{\max} + \Delta M\)

\(df \parallel ab\)

\(1\ \mathrm{m}\)

L.B. \(\Delta M_{0-1}\)

L.B. \(\Delta M_{1-2}\)

L.B. \(\Delta M_{2-3}\)

L.B. \(\Delta M_{3-4}\)

\(30\ \mathrm{t}\)

\(0.5\)

Plan of the influence of the increments of moments.

Drawing 13.