Prof. G. V. Trapeznikov.

The Influent Method for Solving Beams

In the present article the author proposes a new method for solving beams, i.e., for finding for them reactions, fixed-end moments, angles of rotation of sections, and deflections at selected points on the axis of the beam.

All methods and procedures proposed for solving beams, generally speaking, pursue three principal aims:

1. To give the solution greater clarity, since this facilitates the checking of the solution.

2. To accelerate the often very lengthy process of determining the redundant unknowns in statically indeterminate beams and the deformations in all kinds of beams having varied and complicated loading.

3. To obtain the required quantities while avoiding, or reducing to a minimum, the determination of intermediate quantities not needed for the beam calculation itself, such as, for example, arbitrary constants.

The greatest clarity of solution is, of course, possessed by the historically very first method: integration of the differential equation of the elastic curve traced by the bent axis of the beam. But, in terms of speed of solution, this method almost always leaves the user of it in last place in comparison with those applying other methods. An acceleration effect can be achieved only if, when setting up the differential equations and subsequently integrating them, devices simplifying the solution are used. Such devices include:

1. A successful choice of the origin of coordinates, one or several, with the calculation arranged so that one or both arbitrary constants of integration, for one or several spans, become zero.

2. Taking the bending moment to the left or to the right of an arbitrary section, on the basis of considerations of obtaining a simpler expression for the moment, or on the basis of the considerations indicated in item 1.

3. The device of applying an additional load in the case of distributed loads.

4. Application of Clebsch’s rule, where it can be applied directly.

5. Artificial complication of the argument, with the calculation arranged so that Clebsch’s rule can be applied where it cannot be applied directly.

6. Use of the mechanical symmetry of the system.

But with a large number of spans with varied loading, or a large number of supports, all these devices do little to simplify the solution. The literal solution will be extraordinarily cumbersome, while the numerical one, in the operations of integration and substitution, will give every chance of making arithmetic errors. The checking of the solution can be carried out by solving the same problem anew.

The graph-analytical method has less clarity, but the simplicity of analogy in the transition from \(q_x\) to \(M_x\), from \(Q_x\) to \(\tg \varphi_x\), and from \(M_x\) to \(f_x\) makes it easy to accept the apperception of the deflection as a conditional (ficti—

of the bending moment in a beam loaded by a distributed load, in the form of a moment area, and of the angle of rotation of the beam cross-section as the conditional bending moment from the load of its area of shearing forces.

Again, with a variety of loads and an increased number of supports, difficulties arise, especially noticeable when individual spans of the beam are loaded by a distributed load whose law of variation of ordinates is expressed by some analytic curve. Devices in the form of the temporary removal of supports and the introduction and removal of temporary distributed loads, of course, facilitate the solution of the problem, but only for beams of the simplest type.

Castigliano’s method, almost entirely indispensable in solving curved beams and rigid frames, gives, for straight beams, in many cases, a solution far more complicated than that given by simple integration.

Methods of approximate solution, very simple and clear in representation, considerably narrow their field of application because, in applying these methods, we must first outline the approximate form of the elastic line and adopt, for each particular case, a curve parameterized in a definite manner. The latter requires great skill, does not always make it possible to estimate the magnitude of the error, and, in practice, these methods cannot be applied even with a moderate number of supports.

The method, which appeared not so long ago, of applying moments of higher degrees and which gives a fairly rapid solution for beams, has not yet won full acceptance, owing to the considerable abstractness of the formulas used.

Finally, Clapeyron’s method, while it admittedly lacks the advantage of visual clarity if one speaks of the formal application of his equation, surpasses all other methods in the speed with which the calculated quantities—support moments and reactions—are determined. As for the determination of deformations, they must, all the same, be found by some other method.

Having listed only the principal methods for solving beams, the author proposes for consideration his own method, while refraining, for the present, from evaluating it. The basis of the influent method is very simple: it consists in the fact that, for a simple cantilever beam, the magnitude of an action (deflection, angle of rotation of the cross-section) at an arbitrary point is found from two concentrated loads (forces, bending moments) situated to the left and to the right of the point under consideration.

When a distributed load is present on the beam, each element of this load is regarded as an elementary concentrated force, which leads to the solution of a certain integral. The development of the method, in applying it to beams of every kind, consists in transforming the given beam into a conditional cantilever one and makes it possible to apply the basic formulas obtained for a simple cantilever. The influent method gives, at the same time, the possibility of finding both the reactions in statically indeterminate beams and the magnitudes of deformations at an arbitrary point of any beam.

Proceeding to the exposition of the method, the author must say that, within the limits of a journal article, he cannot exhaust the solutions of various kinds of typical beams and confines himself to the minimum of material necessary in solving heterogeneous beams. Likewise, the conditions imposed by the size of the article compel him to refrain from a detailed description of the approximate solution of beams by the tabular method, as well as from including in the text the tables for accelerated solution. Those interested in a more detailed presentation of the applications of the influent method are referred by the author to his book bearing the same title as the present article.

I. Deformations of a Simple Cantilever.

Let us take once and for all the origin of coordinates at the fixed end, and direct the \(Y\)-axis downward. The abscissa of that point of the beam axis at which we need to determine the action of the load will be denoted by \(x_0\), the angle of rotation of the beam cross-section at this point by \(\varphi_0\), and the deflection by \(f_0\).

Suppose (Fig. 1) that, to the right and to the left of the chosen point, we have a concentrated force and likewise a concentrated bending moment.

Fig. 1.

The angle of rotation and the deflection of the elastic line at the chosen point will be:

\[ EJ\varphi_0=\frac{P_1x_1^2}{2}+M_1x_1+P_2x_0(x_2-x_0)+\frac{P_2}{2}x_0^2+M_2x_0 \]

\[ EJf_0=\frac{P_1x_1^3}{3}+\frac{P_1x_1^2}{2}(x_0-x_1)+\frac{M_2}{2}x_1(2x_0-x_1)+\frac{P_2}{2}(x_2-x_0)x_0^2+\frac{M_2}{2}x_0^2 \]

In the presence of a number of forces and moments, the equations obtained become the following:

\[ EJ\varphi_0=\frac{1}{2}\sum P_1x_1^2+\frac{1}{2}x_0\sum P_2(2x_2-x_0)+\sum M_1x_1+x_0\sum M_1 \qquad\ldots\ldots\ldots (1) \]

\[ EJf_0=\frac{1}{6}\sum P_1x_1^2(3x_0-x_1)+\frac{1}{6}x_0^2\sum P_2(3x_2-x_0)+ \]

\[ +\frac{1}{2}\sum M_1x_1(2x_0-x_1)+\frac{1}{2}x_0^2\sum M_2 \qquad\ldots\ldots\ldots (2) \]

Example 1. Determine (Fig. 2) the deformation at the point of application of the force \(2P\). \(x_0=2l\)

Fig. 2.

\[ EJ\varphi_0=\frac{1}{2}Pl^2+\frac{1}{2}\,2P\cdot2l\,(2\cdot2l-2l)+2lM =\frac{9}{2}Pl^2+2Ml \]

\[ EJf_0=\frac{1}{3}P\cdot l^2(3\cdot2l-l)+\frac{1}{6}\,2P\cdot4l^2(3\cdot2l-2l)+\frac{1}{2}M\cdot4l^2 =\frac{37}{6}Pl^3+2Ml^2 \]

We have (Fig. 3) a distributed load whose law of distribution has an analytic expression. Let us find the deformations at the point with abscissa \(x_0\). Here the sign \(\Sigma\) will become \(\int\).

Fig. 3.

\[ EJ\varphi_0=\frac{1}{2}\left[ \int_a^{x_0} q_x x_1^2\,dx_1 + x_0\int_{x_0}^{b} q_x(2x_2-x_0)\,dx_2 \right] \]

\[ EJ f_0=\frac{1}{6}\left[ \int_a^{x_0} q_x x_1^2(3x_0-x_1)\,dx_1 + x_0^2\int_{x_0}^{b} q_x(3x_2-x_0)\,dx_2 \right] \]

There is no need to regard the last equations as special, since in composing them we proceed, all the same, from equations (1) and (2).

Example 2. Fig. 4. Determine the deflection at the end of the cantilever.

The equation of the load curve:

\[ q_x=\frac{q}{l^2}x^2 \]

Fig. 4.

\[ EJ f_l=\frac{q}{6l^2}\int_0^l x^2x^2(3l-x)\,dx = \frac{q}{6l^2}\left|\frac{3}{5}lx^5-\frac{1}{6}x^6\right|_0^l = \frac{13}{180}ql^4 \]

In determining the deformations at an arbitrary point of some portion of a beam, we thereby write the equation of the angles of rotation of the axis, or the equation of the elastic line in this portion.

Fig. 5.

Example 3. Drawing 5. Write the equation of the elastic line for the second portion.

\[ (2)\qquad 6EJy_2=P_1a^2(3x_2-a)+ \]

\[ +P_2x_2^2(3b-x_2)+P_3x_2^2(3c-x_2) \]

II. General principles in solving beams.

The ease of obtaining equations for the deformations of a simple cantilever beam, when the influence method is applied, compels one to seek, for beams of all kinds and types, the possibility of considering any beam as a purely cantilever beam. If at least one of the end supports of the beam has a fixing, then the transformation of the beam into a simple cantilever will consist in the mental removal of all its supports except one—the fixed one. This latter will also be the support of the cantilever beam; all the others will simply be points of application of concentrated loads—reactions, added to the given external load. The difference between such a conditional cantilever beam and an actual cantilever (under the same external load) will consist in the fact that,

that in the former, before it is solved, certain deformations at the points of application of the load-reactions will already be known. For supports lying on one straight line, at the support points the fictitious cantilever will have deflection ordinates equal to zero; for supports situated at different levels, the fictitious cantilever will have, at the support points, deflection ordinates equal to the ordinates of these supports measured from the fixed support. Finally, if there is a second extreme fixed support, the angle of rotation of the beam cross-section at the point corresponding to the indicated support must be equal to zero.

Thus any beam having at least one extreme fixed support may be considered as a cantilever beam in which, at certain places (at the points where the supports of the actual beam are located), unknown concentrated loads act (reaction-forces, reaction-moments), while at these same points certain deformations are known (deflection ordinates, angle of rotation of the cross-section).

In beams having no fixed supports, we may regard any cross-section as fictitiously fixed. Drawing 6 shows that, by adopting a fictitious fixing at some arbitrary section \(c\), we thereby divide the whole beam into two independent fictitious cantilevers, with a common place of fixing at \(c\). It is clear that the deformations will now be measured not from the \(X\)-axis, coinciding with the initial axis of the beam, but from the line \(t_c\), tangent to the elastic curve at the place of the fictitious fixing of the fictitious cantilevers. The tangent \(t_c\) will be the initial axis of each of the fictitious cantilevers. The point of fictitious fixing may be taken anywhere, and the possibility is not excluded of placing it at one of the extreme supports, as occurs for point \(d\); in this case the initial axis of the fictitious cantilever will be the tangent \(t_d\).

Drawing 6.

III. Two-support beam with one fixed end.

Drawing 7. The transformation of such a beam into a fictitious cantilever may be carried out by discarding the free support. In this case the deflection ordinate of the fictitious cantilever at point \(B\) will be equal to:

\[ f_B \]

When the support is restored, however, the reaction \(B\) will give an equal deflection ordinate:

\[ \frac{B l^3}{3EJ} \]

From the equality of these ordinates, we obtain the value of \(B\):

\[ B = \frac{3EJ}{l^3}\, f_B \ . \ . \ . \ . \ (3) \]

Drawing 7.

The deformation in an arbitrary section with abscissa \(x_0\) will be equal to the difference of the deformations due to the external load and the reaction \(B\). Denoting (Fig. 7) the deformations of a simply supported beam respectively by \(y_0'\) and \(y_0\), we obtain:

\[ EJ y_0' = EJ \varphi_0 - \frac{1}{2} B x_0 (2l - x_0) \ldots \ldots \ldots \ldots \ldots \ldots (4)' \]

\[ EJ y_0 = EJ f_0 - \frac{1}{6} B x_0^2 (3l - x_0) \ldots \ldots \ldots \ldots \ldots \ldots (5)' \]

\[ y_0' = \varphi_0 - \frac{3 f_B}{2l^3} x_0 (2l - x_0) \ldots \ldots \ldots \ldots \ldots \ldots (4) \]

\[ y_0 = f_0 - \frac{f_B}{2l^3} x_0^2 (3l - x_0) \ldots \ldots \ldots \ldots \ldots \ldots (5) \]

To find the maximum deflection ordinate it is necessary to solve the equation: \(y_0' = 0\), i.e.:

\[ \varphi_0 - \frac{3 f_B}{2l^3} x_0 (2l - x_0) = 0, \]

and substitute the obtained value of \(x_0\) into equation (5).

Example 4. Fig. 8. Determine the reaction \(B\) and the position of the maximum deflection ordinate.

Fig. 8.

\[ EJ f_B = \frac{1}{6} \left[ 2 \cdot 12^2 (3 \cdot 14 - 12) + \int_0^{12} x^2 (3 \cdot 12 - x)\, dx + 12^2 \int_{12}^{14} (3x - 12)\, dx \right] = \]

\[ = 5328\, tm^3. \]

\[ B = \frac{3 \cdot 5328}{12} = 7{,}514\, t. \]

We apply equation \((4)'\).

\[ 0 = \frac{1}{2} \left[ \int_0^{x_0} x^2\, dx + x_0 \int_{x_0}^{14} (12x - x_0)\, dx + x_0 2(2 \cdot 14 - x_0) \right] - \]

\[ - \frac{7{,}514}{2} x_0 (2 \cdot 12 - x_0), \]

whence:

\[ x_0^2 - 44{,}243 x_0 + 215 = 0 \]

and

\[ x_0 = 5{,}57\, m \]

IV. A Two-Span Beam with Two Fixed Ends.

Fig. 9. As in the preceding case, the temporary removal of the right support transforms the beam under consideration into a simple cantilever with a deflection arrow equal to \(f_B\) at point \(B\). Restoring the support is associated with applying the force \(B\) and the moment \(m_B\), which must reduce the deflection arrow and the angle of rotation of the section at point \(B\) to zero. These conditions will be written:

\[ EJ\varphi_B-\frac{Bl^2}{2}+m_Bl=0; \]

\[ EJf_B-\frac{Bl^3}{3}+\frac{m_Bl^2}{2}=0. \]

Fig. 9.

Hence the reactions of the right support are determined:

\[ B=\frac{6EJ}{l^3}\left[2f_B-l\varphi_B\right]; \ldots \ldots \ldots \ldots \ldots (6)' \]

\[ m_B=\frac{2EJ}{l^2}\left[3f_B-2l\varphi_B\right]; \ldots \ldots \ldots \ldots \ldots (7)' \]

For the given beam the reactions can be expressed in general form through the external load:

\[ B=\frac{1}{l^3}\left[\Sigma Px^2(3l-2x)+6\Sigma Mx(l-x)\right]; \ldots \ldots (6) \]

\[ m_B=\frac{1}{l^2}\left[\Sigma Px^2(l-x)+\Sigma Mx(2l-3x)\right]; \ldots \ldots (7) \]

Example 5. Fig. 10. Determine the reactions of the right support. We apply equations (6) and (7).

Fig. 10.

\[ B=\frac{1}{64l^3}\left[3Pl^2(12l-2l)+P\cdot4l^2(12l-4l)+q\int_0^{4l}x^2(12l-2x)\,dx\right]=\frac{31}{32}P+2ql; \]

\[ m_B=\frac{1}{16l^2}\left[3P\cdot l^2(4l-l)+P\cdot4l^2(4l-2l)+q\int_0^{4l}x^2(4l-x)\,dx\right]= \]

\[ =\frac{17}{16}Pl+\frac{64}{3}ql^2; \]

The deformation, at an arbitrary point with abscissa \(x_0\), is found as the difference between the deformations due to the external load and due to the reactions of the right support (Fig. 9).

\[ EJ y_0' = EJ\varphi_0-\frac{B}{2}x_0(2l-x_0)+m_B x_0 \ldots \ldots \ldots (8)' \]

\[ EJ y_0 = EJ f_0-\frac{B}{6}x_0^2(3l-x_0)+\frac{m_B}{2}x_0^2 \ldots \ldots \ldots (9)' \]

Replacing \(B\) and \(m_B\) by their values (6) and (7), we obtain:

\[ EJ y_0' = EJ\varphi_0-\frac{x_0}{2l^3} \left[ l(4l-3x_0)(\sum Px^2+2\sum Mx)- \right. \]

\[ \left. -2(l-x_0)(\sum Px^3+3\sum Mx^2) \right] \ldots \ldots \ldots (8) \]

\[ EJ y_0 = EJ f_0-\frac{x_0^2}{6l^3} \left[ 3l(2l-x_0)(\sum Px^2+2\sum Mx)- \right. \]

\[ \left. -(3l-2x_0)(\sum Px^3+3\sum Mx^2) \right] \ldots \ldots \ldots (9) \]

Example 6. Fig. 11. Find the deflection arrow at the point: \(x_0=12\,m\).

Fig. 11.

First we find the value of \(EJ f_{12}\)

\[ EJ f_{12} = \frac{1}{6} \left[ 10.8^2(36-8)+7.10^2(36-10)+ \right. \]

\[ \left. +5.12^2(36-12)-2.6^2(36-6) \right] -\frac{3}{2}\,6(24-6) =8378\,tm^3; \]

\[ EJ y_0 = 8378-\frac{12^2}{6.16^3} \left[ 3.16(32-12)(10.8^2+7.10^2+5.12^2-2.6^2-2.3.6)- \right. \]

\[ \left. -(48-24)(10.8^3+7.10^3+5.12^3-2.6^3-3.3.6^2) \right] \simeq 211\,tm^3 \]

To find the maximum deflection arrow, one must first find its abscissa by solving the equation:

\[ y_0'=0 \]

and substitute the obtained value of the abscissa into equation (9).

V. Two-span beams with hinged supports.

Transforming a beam with hinged supports into a fictitious cantilever presents no difficulties. Indeed (Fig. 12), drawing a tangent to the elastic line through a point with some abscissa \(x_0\), we may consider the transverse section coinciding with the point of tangency as fictitiously fixed. In this case the tangent will be the initial axis of two cantilever beams: one to the right and the other to the left of the fixed section. In doing so, the reactions \(A\) and \(B\) must of course be included among the external loads, and the deflections of the fictitious cantilevers must be reckoned from the tangent as from the initial axis. The left fictitious cantilever will have, at the support point \(A\), the deflection \(f_A\), and the right cantilever, at the support point \(B\), the deflection \(f_B\). These deflections are very easily determined by regarding each part of the divided beam as an independent cantilever.

Fig. 12.

We determine the deformations at the point with abscissa \(x_0\):

\[ y_0'=\frac{f_B-f_A}{l}\ . \tag{10} \]

\[ y_0=f_A+\frac{f_B-f_A}{l}x_0 =\frac{1}{l}\left[f_A(l-x_0)+f_Bx_0\right]\ . \tag{11} \]

There is no need to express in general form the fictitious deflections at the supports in terms of the external load.

Example 7. Fig. 13. Determine the deflection at midspan, between the supports.

Fig. 13.

The reactions will be found:

\[ \sum M_A=0=-2P\cdot l+P\cdot 3l+M-B\cdot 4l \]

\[ A=\frac{7}{4}P-\frac{M}{4l},\qquad B=\frac{5}{4}P+\frac{M}{4l}. \]

Deflections of the fictitious cantilevers at \(A\) and \(B\): (for \(x_0=2l\))

\[ EJf_A=\frac{A}{3}\,8l^3-\frac{2P}{6}\,l^2(6l-l) =3Pl^3-\frac{2}{3}Ml^2, \]

\[ EJf_B=\frac{B}{3}\,8l^3-\frac{P}{6}\,l^2(6l-l)-\frac{M}{2}\,4l^2 =\frac{5}{2}Pl^3-\frac{4}{3}Ml^2. \]

We substitute the obtained quantities into (11):

\[ EJy_{2l}=\frac{1}{4l}\left[\left(3Pl^3-\frac{2}{3}Ml^2\right)2l+ \left(\frac{5}{2}Pl^3-\frac{4}{3}Ml^2\right)2l\right] =\frac{11}{4}Pl^3-Ml^2. \]

VI. A Three-Span Beam with One Fixed End

Fig. 14. Let us agree, in beams having more than two supports, to denote all supports and the reactions at them by the letter \(A\) with an index indicating the ordinal number of the support. The extreme left support will have reaction \(A_0\), the following ones respectively: \(A_1\), \(A_2\), and so on. We shall denote the span lengths by the letter \(l\) with a numerical index of the span: \(l_1\), \(l_2\), and so on.

Fig. 14.

Application of the former method of temporarily removing supports gives the following conditions for the support points of the beam axis:

\[ EJ y_{A_1}=0=EJ f_{A_1}-\frac{A_1 l_1^3}{3} -\frac{A_2}{6}l_1^2\left[3(l_1+l_2)-l_1\right]= \]

\[ =f_{A_1}-\frac{A_1 l_1^3}{3} -\frac{A_2}{6}l_1^2(2l_1+3l_2) \]

\[ AJ y_{A_2}=0=EJ f_{A_2} -\frac{A_1}{6}\left[3(l_1+l_2)-l_1\right] -\frac{A_2}{3}(l_1+l_2)^3= \]

\[ =f_{A_2}-\frac{A_1}{6}l_1^2(l_1+3l_2) -\frac{A_2}{3}(l_1+l_2)^3 \]

The simultaneous solution of the last equations gives, for the reactions \(A_1\) and \(A_2\), the expressions:

\[ A_1=6EJ\, \frac{2(l_1+l_2)^3 f_{A_1}-l_1^2(2l_1+3l_2)f_{A_2}} {l_1^3 l_2^2(3l_1+4l_2)} \qquad . . . . . \ (12) \]

\[ A_2=6EJ\, \frac{2l_1 f_{A_2}-(2l_1+3l_2)f_{A_1}} {l_1 l_2^2(3l_1+4l_2)} \qquad . . . . . \ (13) \]

Example 8. Fig. 15. Determine the reactions of the free supports. \(q=\dfrac{4P}{l}\).

Fig. 15.

\[ 6EJ f_{A_1} =\frac{4P}{l}\int_{0}^{3l} x^2(9l-x)\,dx +3P\cdot 9l^2(15l-3l) \]

\[ +P\cdot 9l^2(21l-3l) =\frac{4P}{l}\left|3lx^3-\frac{x^4}{4}\right|_{0}^{3l} +486Pl^3=729Pl^3. \]

\[ 6EJ f_{A_2}=\frac{4P}{l}\int_0^{3l} x^2(18l-x)\,dx+3P\cdot25l^2(18l-5l)+P\cdot36l^2(21l-6l)= \]

\[ =\frac{4P}{l}\left[6lx^3-\frac{x^4}{4}\right]_0^{3l}+1515Pl^3=2182Pl^3. \]

\[ A_1=\frac{2\cdot216l^3\cdot729-9l^2\cdot15l\cdot2182}{27l^3\cdot9l^2\cdot21l}\,Pl^3=3.99P. \]

\[ A_2=\frac{2\cdot3l\cdot2182-15l\cdot729}{3l\cdot9l^2\cdot21l}\,Pl^3=3.80P. \]

The equations of deformation for the first and second spans are written as follows:

\[ y_1'=\varphi_1-\frac{x_1}{2EJ}\left[A_1(2l_1-x_1)+A_2\{2(l_1+l_2)-x_1\}\right]\ .\ .\ .\ .\ (14) \]

\[ y_1=f_1-\frac{x_1^2}{6EJ}\left[A_1(3l_1-x_1)+A_2\{3(l_1+l_2)-x_1\}\right]\ .\ .\ .\ .\ (15) \]

\[ y_2'=\varphi_2-\frac{1}{2EJ}\left[A_1x_2^2+A_2x_2\{2(l_1+l_2)-x_2\}\right];\ .\ .\ .\ .\ (16) \]

\[ y_2=f_2-\frac{1}{6EJ}\left[A_1l_1^2(3x_2-l_1)+A_2x_2^2\{3(l_1+l_2)-x_2\}\right];\ .\ .\ (17) \]

VII. Three-span beam with two fixed ends.

Fig. 16. In this beam the following conditions must be satisfied:

Fig. 16.

\[ EJy_{A_1}=0=EJf_{A_1}-\frac{A_1l_1^3}{3}-\frac{A_2}{6}l_1^2(2l_1+3l_2)+\frac{m_2}{2}l_1^2; \]

\[ EJy_{A_2}=0=EJf_{A_2}-\frac{A_1}{6}l_1^2(2l_1+3l_2)-\frac{A_2}{3}(l_1+l_2)^3+\frac{m_2}{2}(l_1+l_2)^2; \]

\[ EJy'_{A_2}=0=EJ\varphi_{A_2}-\frac{A_1}{2}l_1^2-\frac{A_2}{2}(l_1+l_2)^2+m_2(l_1+l_2); \]

Whence the expressions for the reactions of the middle and right supports are obtained:

\[ A_1=\frac{3EJ}{l_1^3l_2^3}\left[(l_1+l_2)^3 f_{A_1}-l_1^2(l_1+3l_2)f_{A_2}+l_1^2l_2(l_1+l_2)\varphi_{A_2}\right]\ .\ .\ (18) \]

\[ A_2=\frac{3EJ}{l_1l_2^3(l_1+l_2)} \left[ l_1(l_1+4l_2)f_{A_2}-(l_1+l_2)(l_1+3l_2)f_{A_1} -l_1l_2(l_1+2l_2)\varphi_{A_2} \right]\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (19) \]

\[ m_2=\frac{EJ}{l_1l_2^2(l_1+l_2)} \left[ 3l_1(l_1+2l_2)f_{A_2}-3(l_1+l_2)^2f_{A_1} -l_1l_2(3l_1+4l_2)\varphi_{A_2} \right] \tag{20} \]

It is not difficult, for the given beam, to express all reactions in terms of the external load, but the equations obtained will be very cumbersome and inconvenient to use. It is much better to carry out the calculation by first determining the magnitudes of the deformations of the conditional cantilever at the supports, and only then substituting the values found into equations (18), (19), and (20). In this case the verification of the solution can be carried out in parts.

Fig. 17.

Example 9. Drawing 17. Determine the fixing moment at the right support: \(q=\dfrac{P}{l}\).

We first find the deformations of the conditional cantilever at the support points.

\[ 6EJ f_{A_1}=P\cdot 4l(12l-2l)+2P\cdot 16l^2(15l-4l) +\frac{P}{4l^2}\int_0^{4l}x^3(12l-x)\,dx= \]

\[ =392Pl^3+\frac{P}{4l}\left|3lx^4-\frac{x^5}{5}\right|_0^{4l} \simeq 533Pl^3. \]

\[ 6EJ f_{A_2}=P4l(21l-2l)+2P\cdot 25l^2(21l-5l) +\frac{P}{4l^2}\int_0^{4l}x^3(21l-x)\,dx= \]

\[ =876Pl^3+\frac{P}{4l}\left|\frac{21}{4}x^4-\frac{x^5}{5}\right|_0^{4l} \simeq 1161Pl^3. \]

\[ EJ\varphi_{A_2}=P\cdot 4l^2+2P\cdot 25l^2 +\frac{P}{4l^2}\int_0^{4l}x^3\,dx=70Pl^2. \]

Applying equation (20), we find:

\[ m_2= \frac{3}{4l\cdot 9l^2\cdot 7l} \left[ 12l\cdot 10l\cdot \frac{1161Pl^3}{6} -3\cdot 49l^2\cdot \frac{533Pl^3}{6} -\right. \]

\[ \left. -4l\cdot 3l\cdot 10l\cdot \frac{70Pl^2}{2} \right]\simeq 71Pl. \]

To determine the deformations (Fig. 18) in each of the spans, we discard the right support and thereby transform the beam into a conditionally cantilevered one. Upon restoration of the supports, the deformations at arbitrary points of the spans will be found as the difference between the deformations of the conditional cantilever: from the action of the external loads and from the action of the applied reactions at the centers and edges of the right supports.

For the first span we obtain:

Fig. 18.

\[ EJ y_1' = EJ \varphi_1 - \frac{A_1}{2} x_1 (2l_1 - x_1) - \frac{A_2}{2} x_1 \left[2(l_1 + l_2) - x_1\right] + m_2 x_1 \ldots \tag{21} \]

\[ EJ y_1 = EJ f_1 - \frac{A_1}{6} x_1^2 (3l_1 - x_1) - \frac{A_2}{6} x_1^2 \left[3(l_1 + l_2) - x_1\right] + \frac{m_2}{2} x_1^2 \ldots \tag{22} \]

and for the second:

\[ EJ y_2' = EJ \varphi_2 - \frac{A_1}{2} l_1^2 - \frac{A_2}{2} x_2 \left[2(l_1 + l_2) - x_2\right] + m_2 x_2 \ldots \tag{23} \]

\[ EJ y_2 = EJ f_2 - \frac{A_1}{6} l_1^2 (3x_2 - l_1) - \frac{A_2}{6} x_2^2 \left[3(l_1 + l_2) - x_2\right] + \frac{m_2}{2} x_2^2 \ldots \tag{24} \]

Fig. 19.

Example 10. Figure 19. Determine the angle of rotation of the cross-section of the beam at the point with abscissa: \(x_2 = 8\,m\).

We compute the following quantities:

\[ EJ f_{A_1} = \frac{1}{6}\left[5 \cdot 9(18 - 3) + 8 \cdot 36(24 - 6)\right] \cong 976\,tm^3, \]

\[ EJ f_{A_2} = \frac{1}{6}\left[5 \cdot 9(30 - 3) + 8 \cdot 64(30 - 8)\right] \cong 2080\,tm^3, \]

\[ EJ \varphi_{A_2} = \frac{1}{2}\left[5 \cdot 9 + 8 \cdot 64\right] \cong 278\,tm^2, \]

\[ EJ \varphi_{8m} = EJ \varphi_{A_2} \cong 278\,tm^2, \]

\[ A_1 = \frac{3}{6 \cdot 4^3}\left[1000 \cdot 976 - 36 \cdot 18 \cdot 2080 + 36 \cdot 10 \cdot 2 \cdot 278\right] = 6{,}30\,t, \]

\[ A_2 = \frac{3}{6 \cdot 4^3 \cdot 10}\left[6 \cdot 22 \cdot 2080 - 10 \cdot 18 \cdot 976 - 64 \cdot 14 \cdot 278\right] = 3{,}90\,t, \]

\[ m_2 = \frac{1}{6 \cdot 16 \cdot 10}\left[3 \cdot 6 \cdot 14 \cdot 2080 - 3 \cdot 100 \cdot 976 - 6 \cdot 4 \cdot 34 \cdot 278\right] = 4{,}58\,tm. \]

The angle of rotation of the selected section will be:

\[ EJ\,y^{1}_{8m}=278-\frac{6.30}{2}\cdot36-\frac{3.90}{2}\,8(20-8)+4.58\cdot8=14.14\,tm^{2}. \]

VIII. A Three-Span Beam with Free Supports

Fig. 20. Let us take, in the beam, the redundant unknown reaction \(A_2\). We shall replace the curved beam by a conventional fixing above the support \(A_0\). Drawing from the point of fixing a tangent to the curved axis of the beam, we shall measure the deformations from this straight line. From the drawing it is clear that in the present case the following geometrical relation holds:

Fig. 20.

\[ \frac{\left[f^{R}_{A_2}\right]}{\left[f^{R}_{A_1}\right]} = \frac{l_1+l_2}{l_1} \ldots\ldots (a) \]

where by \(\left[f^{R}_{A_1}\right]\) and \(\left[f^{R}_{A_2}\right]\) are meant the deflection ordinates of the conventional cantilever at the points of the corresponding supports. The superscripts \((R)\) indicate that, in calculating these ordinates, the action of the support reactions was also taken into account.

The absolute values of the conventional deflection ordinates will be:

\[ EJ\left[f^{R}_{A_1}\right] = \frac{A_1}{3}l_1^3 + \frac{A_2}{6}l_1^2(2l_1+3l_2) - EJ f_{A_1} \ldots\ldots (b) \]

\[ EJ\left[f^{R}_{A_2}\right] = \frac{A_1}{6}l_1^2(2l_1+3l_2) + \frac{A_2}{3}(l_1+l_2)^3 - EJ f_{A_2} \ldots\ldots (c) \]

In order to eliminate the reaction \(A_1\), which is unnecessary for us, we write the statics equation—the sum of moments about the support \(A_0\):

\[ \Sigma M_0-A_1l_1-A_2(l_1+l_2)=0 \ldots\ldots (d) \]

The simultaneous solution of equations \((a)\), \((b)\), \((c)\), and \((d)\) leads to finding the value \(A_2\):

\[ A_2= \frac{3EI}{l_1^2(l_1+l_2)} \left[ f_{A_2}l_1-f_{A_1}(l_1+l_2) \right] - \frac{l_1\Sigma M_0}{2l_2(l_1+l_2)} \ldots\ldots (25) \]

Example 11. Fig. 21. Determine the reaction \(A_2\)

\[ \Sigma M_0=M+3Pl \]

\[ EJ f_{A_1} = \frac{M}{2}l^2+\frac{P}{6}l^2(9l-l) = \]

\[ = \frac{1}{2}Ml^2+\frac{4}{3}Pl^3, \]

Fig. 21.

\[ EJ f_{A_2}=\frac{M}{2}\,2l(3l-2l)+\frac{P}{6}\,9l^2(12l-3l)=6Ml^2+\frac{27}{2}Pl^3, \]

\[ EJ\left[f_{A_2}l_1-f_{A_1}(l_1+l_2)\right] =\left(6Ml^2+\frac{27}{2}Pl^3\right)l -\left(\frac{1}{2}Ml^2+\frac{4}{3}Pl^3\right)4l +4Ml^3+\frac{49}{6}Pl^4. \]

\[ A_2=\frac{3}{l\cdot 9l\cdot 4l}\left(\frac{49}{6}Pl^4+4Ml^3\right) -\frac{l}{2\cdot 3l\cdot 4l}(M+3Pl) =\frac{5}{9}P+\frac{7}{24}\frac{M}{l}. \]

As for finding the deformations in general form, this presents no difficulty. However, another path may be recommended for finding them. After finding the reaction, the bending moment over the middle support is found. Then the separate spans are considered as two-support beams, only having, as a load, a bending moment over one of the supports, which is added to the externally prescribed load. With such a solution we can make use of the simpler formulas already available to us: (10) and (11).

XI. Multi-support beams.

The solution of beams with more than three supports, in the form in which we have done it up to now, presents no fundamental difficulties; however, the cumbersomeness of the formulas obtained sets a limit to their application in practice. Therefore the solution of such beams is better carried out using the theorem of three moments. Clapeyron’s equation, in its canonical form, as was indicated in the introduction, has its shortcomings, of which the chief one is the difficulty of determining the quantities entering into the right-hand side of this equation, when the loads are complex and varied. Here the influence method removes this difficulty, transforming the right-hand side of the equation so that its use becomes very simple.

Fig. 22.

Let us derive Clapeyron’s equation by applying the influence method. Drawing 22 represents two adjacent, arbitrarily chosen spans of a multi-support beam.

Here:

\(A_n\) is the total reaction of the middle support,

\(A''_{0(n-1)}\) and \(A'_{0(n+1)}\) are the shearing forces near the \(n-1\)-th and \(n+1\)-th supports, the first being in the transverse section of the beam lying to the right of support \(n-1\), while the second is in the section lying to the left of support \(n+1\). These shearing forces are obtained only at the expense of the external load prescribed on the beam.

\(\mathfrak{M}_{n-1}\) and \(\mathfrak{M}_{n+1}\) are the support moments, considered positive when they tend to load their span (to bend the beam with convexity downward).

\[ \frac{\mathfrak{M}_{n}-\mathfrak{M}_{n-2}}{l_n} \quad \text{and} \quad \frac{\mathfrak{M}_{n}-\mathfrak{M}_{n+1}}{l_{n+1}} \]
are the shearing forces due to the action of the support moments.

We denote the support moments in Gothic type in order not to confuse them with the bending moments belonging to the given load, which will be denoted in Roman type, with indices of the corresponding spans.

Having adopted a fictitious fixed end over the support \(A_n\) (Fig. 22), we obtain the following geometrical relation between the absolute values of the deflection ordinates of the fictitious cantilevers with overhangs \(l_n\) and \(l_{n+1}\), at the ends of the latter:

\[ \frac{\left[f^R_{n-1}\right]}{\left[f^R_{n+1}\right]} = \frac{l_n}{l_{n+1}}, \quad \text{or} \quad \left[f^R_{n-1}\right]l_{n+1} - \left[f^R_{n+1}\right]l_n = 0 \qquad \ldots (26) \]

Let us write the expressions for \(\left[f^R_{n-1}\right]\) and \(\left[f^R_{n+1}\right]\) in expanded form, adopting the following two rules:

1. We shall take as the origin of coordinates for the abscissae of concentrated loads the free end of the given span, i.e. for the left span—its left end, and for the right span—its right end.

2. We shall consider concentrated moments belonging to the external loads positive in the case when, being in the left span, they are directed clockwise, and in the right span—counterclockwise. Fig. 23 explains the latter rule.

Fig. 23.

Let us find the magnitudes of the end deflection ordinates of the fictitious cantilevers. Fig. 22.

\[ \begin{aligned} EJ\left[f^R_{n+1}\right] ={}& \frac{A''_{0(n+1)}}{3}l_n^3 + \frac{\mathfrak{M}_n-\mathfrak{M}_{n-1}}{3l_n}l_n^3 + \frac{1}{2}\sum M_n(l_n-x)\left[2l_n-(l_n-x)\right] \\ &- \frac{1}{6}\sum P_n(l_n-x)^2\left[3l_n-(l_n-x)\right] + \frac{\mathfrak{M}_{n-1}}{2}l_n^2 - \frac{1}{3}A''_{0(n-1)}l_n^3 + \frac{1}{6}\mathfrak{M}_{n-1}l_n^2 \\ &+ \frac{1}{3}\mathfrak{M}_n l_n^2 + \frac{1}{2}\sum M_n(l_n^2-x^2) - \frac{1}{6}\sum P_n(l_n-x)^2(2l_n+x) \qquad \ldots (27) \end{aligned} \]

\[ \begin{aligned} EJ\left[f^R_{n+1}\right] &= -\frac{A'_{0(n+1)}}{3}l_{n+1}^{3} -\frac{\mathfrak{M}_n-\mathfrak{M}_{n+1}}{3l_{n+1}}l_{n+1}^{3} -\frac{1}{2}\Sigma M_{n+1}(l_{n+1}-x)\left[2l_{n+1}\right. \\ &\left.\quad -(l_{n+1}-x)\right] +\frac{1}{6}\Sigma P_{n+1}(l_{n+1}-x)^2\left[3l_{n+1}-(l_{n+1}-x)\right] -\frac{\mathfrak{M}_{n+1}}{2}l_{n+1}^{2} \\ &= -\frac{1}{3}A'_{0(n+1)}l_{n+1}^{3} -\frac{1}{6}\mathfrak{M}_{n+1}l_{n+1}^{2} -\frac{1}{3}\mathfrak{M}_{n}l_{n+1}^{2} -\frac{1}{2}\Sigma M_{n+1}\left(l_{n+1}^{2}-x^{2}\right) \\ &\quad +\frac{1}{6}\Sigma P_{n+1}(l_{n+1}-x)^2(2l_{n+1}+x) \qquad \ldots\ (28) \end{aligned} \]

The magnitudes of the shear forces \(A''_{0(n-1)}\) and \(A'_{0(n+1)}\) will be found by considering each span separately, as a cut two-support beam loaded only by the specified load.

Fig. 24.

The static moment of the forces acting on the left span is taken with respect to its right end. We obtain:

\[ \Sigma P_n(l_n-x)-\Sigma M_n-A''_{0(n-1)}l_n=0, \]

whence:

\[ A''_{0(n+1)}=\frac{1}{l_n}\left[\Sigma P_n(l_n-x)-\Sigma M_n\right]. \]

We write the same equation for the moments of the right span, taking them with respect to the left end of the span:

\[ \Sigma P_{n+1}(l_{n+1}-x)-\Sigma M_{n+1}-A'_{0(n+1)}l_{n+1}=0, \]

whence:

\[ A'_{0(n+1)}=\frac{1}{l_{n+1}}\left[\Sigma P_{n+1}(l_{n+1}-x)-\Sigma M_{n+1}\right]. \]

Substitution of the quantities \(A''_{0(n-1)}\) and \(A'_{0(n+1)}\) into equations (27) and (28), and of the quantities \(\left[f^R_{n-1}\right]\) and \(\left[f^R_{n+1}\right]\) into equation (26), gives:

\[ \mathfrak{M}_{n-1}l_n+2\mathfrak{M}_n(l_n+l_{n+1})+\mathfrak{M}_{n+1}l_{n+1} = -\frac{1}{l_n}\left[\Sigma P_n x\left(l_n^2-x^2\right)+\Sigma M_n\left(l_n^2-3x^2\right)\right] \]

\[ -\frac{1}{l_{n+1}}\left[\Sigma P_{n+1}x\left(l_{n+1}^2-x^2\right)+\Sigma M_{n+1}\left(l_{n+1}^2-3x^2\right)\right] \qquad \ldots\ (29) \]

The equation obtained is an analytic expression of the theorem of three moments, in which the right-hand side has taken a form very convenient for computation.

It should be remembered that the coordinates of the concentrated loads of each span \((x)\) are measured (when considering the spans in pairs) from the free ends of each span: for the left one—from the left, for the right one—from the right. This latter rule greatly simplifies the right-hand side of Clapeyron’s equation, giving it a symmetric form.

A similar equation, only in a less general form and with an asymmetric right-hand side, is found in Prof. Rivoš’s course on Strength of Materials.

The reactions are determined as follows. The reaction of the middle support in any pair of spans will consist of the sum of the shearing forces located in the sections nearest to the support, to the right and to the left of it.

Thus:

\[ A_n=A'_{0n}+A''_{0n}+\frac{\mathfrak{M}_n-\mathfrak{M}_{n-1}}{l_n} +\frac{\mathfrak{M}_n-\mathfrak{M}_{n+1}}{l_{n+1}}. \]

But:

\[ A'_{0n}=\Sigma P_n-A''_{0(n-1)} \]

and

\[ A''_{0n}=\Sigma P_{n+1}-A'_{0(n+1)} \]

The second terms of the right-hand sides of the equations have already been found above (see equation 28 and following).

Hence we obtain:

\[ A'_{0n}=\frac{1}{l_n}(\Sigma P_n x+\Sigma M_n), \]

\[ A''_{0n}=\frac{1}{l_{n+1}}(\Sigma P_{n+1}x+\Sigma M_{n+1}) \]

The total reaction of the \(n\)-th support \(A_n\) will be represented by the expression

\[ A_n=\frac{1}{l_n}(\Sigma P_n x+\Sigma M_n+\mathfrak{M}_{n-1}-\mathfrak{M}_n) +\frac{1}{l_{n+1}}(\Sigma P_{n+1}x+\Sigma M_{n+1}+\mathfrak{M}_{n+1}-\mathfrak{M}_n). \]

The determination of deformations presents no particular difficulty. Having determined the support moments and reactions, we solve the problem of finding the deformations in the selected span as a problem with a beam on two free supports, in which, in addition to the given load, there are also two additional loads in the form of concentrated moments at the ends (the support moments found).

Drawing 25.

Example 12. Drawing 25. Determine the support moments.

Using equation (29), we set up the following equations:

\[ 2\mathfrak{M}_1\,14+\mathfrak{M}_2\,8 = -\frac{1}{6}\int_0^6 2x(36-x^2)\,dx -\frac{1}{8}\,5.3(64-9); \]

\[ \mathfrak{M}_{1}\,8 + 2\mathfrak{M}_{2}\,14 + \mathfrak{M}_{3}\,6 = -\frac{1}{8}\,5.5(64-25) - \bigl[3.4(36-16) + 2.3(36-9) \]

\[ \mathfrak{M}_{2}\,6 + 2\mathfrak{M}_{3}\,14 = -\frac{1}{6}\bigl[2.3(36-9) + 3.2(36-4) - \]

\[ -\frac{1}{8}\int_{0}^{8}\frac{3}{8}(8-x)x(64-x^{2})\,dx \]

The subsequent simultaneous solution of the three equations obtained presents not the slightest difficulty; therefore, in solving the example, we shall stop at this point.

In my book, bearing the same title as the present article, tables are given, as well as a tabular method for the very rapid solution of beams by means of the influence method. The essence of the solution is that, by means of the influence method, such quantities of integrals have been found and entered in tables as:

\[ \int q_x x^2(3l - x)\,dx, \qquad \int q_x x(l^2 - x^2)\,dx \]

and so forth, computed for loads distributed according to a linear law, for various degrees of loading of the spans. In Fig. 26 some cases of span loading and types of load are given.

Having such tables, it is easy to calculate any beam. Using them, it is not difficult, with a high degree of accuracy, to calculate even a beam (Fig. 27) whose law of loading has no analytical expression, or whose expression is very complex. In such a case one may, by straightening by eye the arcs of the load curve, solve the beam tabularly, assuming that the separate portions of its spans carry a load distributed according to different, but linear, laws.

Fig. 27