Prof. G. V. Trapeznikov.

A Grapho-Analytical Proof of the Theorems of Castigliano and Betti

As the basis for the proof of Castigliano’s theorem we shall take the following two propositions:

1. In its elastic deformation, a system obeys Hooke’s law, which analytically may be expressed as follows:

\[ f=\alpha P \]

i.e., the displacement at any point of the system, in any direction, due to the action of a load likewise applied at an arbitrary point, is proportional to the magnitude of this load. Here, by displacement we shall in general understand displacement, both linear and angular, and by load—a force or a moment causing such displacement.

1. To the system we shall apply the principle of independence of the action of forces, which for the present case may be formulated as follows:

The final form and magnitude of the displacement at any point of the system do not depend on the sequence in which the loads applied to the system follow one another.

Let there be a system with a series of loads: \(P_0\ P_1\ P_2,\ldots P_n\). We shall denote by \(f_{ik}\) the displacement produced by the load \(P_k\), at the point of application of a load \(P_i\) already present or to be present, in the direction of the latter. This displacement must correspond to the type of load \(P_i\); for a concentrated force it will be a linear displacement, and for a concentrated moment—an angular one.

Let us begin to deform the system, introducing one load after another, beginning with \(P_0\). For each of the loads we shall construct a displacement diagram. (Fig. 1).

For brevity, we shall denote potential energy by the letters PE.

By virtue of the work performed by the load \(P_0\), the system will acquire a stock of PE: (Fig. 1-a)

\[ U=U_{00}=\frac{P_0 f_{00}}{2}=\alpha_0\frac{P_0^2}{2}. \]

Upon application of the next load \(P_1\), the PE will receive an increment:

1. From the work expended by \(P_1\) in its own displacement: (Fig. 1-b)

\[ U_{11}=\frac{P_1 f_{11}}{2}=\alpha_1\frac{P_1^2}{2}, \]

1. From the work expended by \(P_0\), which has traversed the path \(f_{01}\) without change of its magnitude, due to the deformation of the system by the load \(P_1\): (Fig. 1-a)

\[ U_{01}=P_0 f_{01} \]

The total PE of the whole system will now be:

\[ U=U_{00}+U_{01}+U_{11} \]

Applying the remaining loads to the system, in the order of their numerical indices, we shall obtain in the end, when the last load \(P_n\) has been applied:

\[ \begin{aligned} U={}&\alpha_0\frac{P_0^2}{2}+\alpha_1\frac{P_1^2}{2}+\alpha_2\frac{P_2^2}{2}+\cdots+\alpha_{n-1}\frac{P_{n-1}^2}{2}+\alpha_n\frac{P_n^2}{2}+\\ &+P_0 f_{01}+P_1 f_{12}+P_2 f_{23}+\cdots+P_{n-1} f_{(n-1)n}+\\ &+P_0 f_{02}+P_1 f_{13}+\cdots\\ &\quad\cdot \qquad \cdot\\ &\quad\cdot \qquad \cdot\\ &\quad\cdot \qquad \cdot\\ &+P_0 f_{0(n-1)}+P_1 f_{1n}\\ &+P_0 f_{0n} \end{aligned} \]

The expression just obtained for the total PE of the system consists of two different parts:

1. Of the terms of the first column, whose sum gives the value of the PE obtained from the displacement \(P_0\) under the action of all loads, including its own action.

Fig. 1.

1. The PE terms of all the remaining columns, each of which gives the PE from the displacement of some load acting after the application of \(P_0\).

In view of this, the sum of the terms of all columns except the first will not depend on \(P_0\). Denoting the sum of the terms independent of \(P_0\) by \(\Phi\), we write:

\[ U=\alpha_0 \frac{P_0^2}{2}+P_0 \sum_{i=1}^{i=n} f_{0i}+\Phi . \]

Taking the partial derivative of \(U\) with respect to \(P_0\), we obtain:

\[ \frac{\partial U}{\partial P_0}=\alpha_0 P_0+\sum_{i=1}^{i=n} f_{0i}. \]

But we had:

\[ \alpha_0 P_0=f_{00} \]

and therefore:

\[ \frac{\partial U}{\partial P_0}=\sum_{i=0}^{i=n} f_{0i}=f_{P0}, \]

i.e. the partial derivative of the total potential energy of the entire system with respect to one of the external loads is equal to the displacement of this load in its own direction.

The proof proposed by the author of the article convincingly emphasizes the following points, which are very necessary for the assimilation of Castigliano’s theorem:

1. Castigliano’s theorem is applicable only to such systems which, in their deformation, follow Hooke’s law.

2. In applying Castigliano’s theorem, coordinate axes play no role, since the load itself (in the drawing: a straight arrow—a force, or a circular arrow—a moment) plays the role of a positive axis for determining the direction of displacement.

A fine and, at the same time, simple example illustrating the latter conclusion is the problem given below of a bar deformed by two oppositely directed axial forces. (Drawing 2).

Drawing 2.

The PE of the lower part of the bar will be:

\[ U_1=\frac{P_1^2(l-a)}{2EF}, \]

the PE of the upper part:

\[ U_2=\frac{(P_1-P_2)^2 a}{2EF}. \]

The total PE is expressed as:

\[ U=U_1+U_2=\frac{P_1^2(l-a)+(P_1-P_2)^2 a}{2EF}. \]

The partial derivatives of \(U\) with respect to \(P_1\) and \(P_2\) give:

\[ \frac{\partial U}{\partial P_1}=\frac{P_1l}{EF}-\frac{P_2a}{EF}=f_{P_1}\ldots\ldots\ldots (1) \]

\[ \frac{\partial U}{\partial P_2}=-\frac{P_1a}{EF}+\frac{P_2a}{EF}=f_{P_2}\ldots\ldots\ldots (2) \]

Considering equation (1), we see that the quantity \(\frac{P_1l}{EF}\) is the tensile deformation of the bar by the force \(P_1\) over the length \(l\) and, at the same time, it is the displacement of the point of application of \(P_1\) in its own direction and from its own action. The quantity \(\frac{P_2a}{EF}\) is the compressive deformation by the force \(P_2\) over the length \(a\) and, at the same time, the displacement of the point of application of \(P_1\) from the action of the load \(P_2\). Since the latter has a direction opposite to \(P_1\), this displacement is obtained with a minus sign.

The same reasoning is easily applicable also to the quantities entering equation (2).

We shall now give a second proof of Castigliano’s theorem, which will also be suitable for the reciprocal theorem, giving, as is known:

\[ \frac{\partial U}{\partial f_{P_0}}=P. \]

In drawing 1 there were displacement diagrams for each of the loads—

Fig. 3.

loads, and we followed the displacements only of those points to which loads had already been applied. Now we shall also trace the displacement of these points, but only independently of whether loads are present at them or not. In other words, we shall consider all points simultaneously upon the application of each of the loads.

Fig. 3. We load the system, beginning with the load \(P_0\). Upon applying it, we obtain:

1. In Fig. 3-a, a triangle with base \(f_{00}\) and height \(P_0\).
2. In Fig. 3-b—c— . . . — . . . —d, only the displacements of the corresponding points of future application of loads will be recorded:

\[ f_{10},\ f_{20},\ \ldots,\ f_{n0}. \]

Applying further the load \(P_1\), we obtain:

1. In Fig. 3-b, a triangle with base \(f_{11}\) and height \(P_1\). Its inclined straight line will now proceed not from the origin of coordinates, but from the end of the segment \(f_{10}\).
2. In Fig. 3-a, a rectangle with base \(f_{01}\) and height \(P_0\).
3. In Fig. 3-c— . . . —d, the displacements of the still unloaded points \(2, 3, \ldots, n\) will be recorded, in the form of segments:

\[ f_{21},\ f_{31},\ \ldots,\ f_{n1}. \]

When, finally, the last load \(P_n\) has been applied, we shall obtain the diagrams shown below in Fig. 4 by solid lines.

Fig. 4.

In each of these diagrams the height of the trapezoids (for the last diagram, of the triangle) is equal to the corresponding load, and the abscissa of the extreme right-hand point is equal to the total displacement of the point of its application.

Let us now load the system differently. Let us begin to apply all the loads simultaneously and let them increase to their limiting values in such a way that this increase proceeds in proportion to their magnitudes. With this method of loading the system, we obtain diagrams establishing the relation between load and displacement in a different form. All the diagrams will appear as triangles, as is shown in drawing 4 by the dashed line. The heights of these triangles will be equal to the heights of the corresponding trapezoids, and the bases—to the total displacements of the given loads.

Since the PE, under the new method of loading the system, must be the same in magnitude as under the former method, we must obviously have equality between the sum of the areas of the new triangles and the sum of the areas of the former trapezoids. The sum of the areas of the triangles will be written as follows: (Drawing 4)

\[ U=\frac{1}{2}\sum_{i=0}^{i=n} P_i f_{P_i}. \]

Previously we had:

\[ f_{ii}=\alpha_i P_i, \]

where the coefficient \(\alpha_i\) determined the magnitude of the steepness of the inclined straight line of the trapezoids, i.e. the stiffness of the system at the given point, under the action of a load at the same point and, moreover, of a unit load. Now, however, we have triangles with different steepnesses of the inclined straight lines, and consequently with different values of the coefficients \(\alpha\):

\[ f_{P_i}=\alpha_{P_i}P_i \]

When the system is deformed by loads applied simultaneously and increasing proportionally to their magnitude, we shall each time have the same picture for some \(P_i\). Therefore we may imagine that the system is deformed by only one load \(P_i\), but with a different degree of increase in deformation in comparison with the case when it is actually deformed by one \(P_i\). As for the other loads, they, as it were, do not directly influence the displacement \(P_i\), but only change the displacement coefficient (the stiffness for the given point).

Since this is so, having:

\[ U=\frac{1}{2}\sum_{i=0}^{i=n} P_i f_{P_i} =\frac{1}{2}\sum_{i=0}^{i=n} \alpha_{P_i}P_i^2,\ldots\ldots (3) \]

we shall always obtain:

\[ \frac{\partial U}{\partial P_k}=\alpha_{P_k}P_k=f_{P_k}. \]

Thus Castigliano’s theorem has been proved. For the derivation of the reciprocal theorem we shall use the equality:

\[ P_i=\frac{f_{P_i}}{\alpha_{P_i}} \]

Making the substitution of the values \(f_{Pi}\) into equation (3), we obtain:

\[ U=\frac{1}{2}\sum_{i=0}^{i=n}\frac{f_{Pi}^{2}}{\alpha_{Pi}}. \]

And, consequently:

\[ \frac{\partial U}{\partial f_{Pk}}=\frac{f_{Pk}}{\alpha_{Pk}}=P_k. \]

The graph of the second variant of the proof of Castigliano’s theorem (Fig. 4) leads to an easy proof of Betti’s theorem, in its most general form.

The condition of equality of the sum of the areas of the displacement diagrams, when the system is loaded in one way and in the other way, is expressed analytically as follows:

\[ \frac{1}{2}\sum_{0}^{n} P_i f_{ii} +P_0\sum_{1}^{n} f_{0i} +P_1\sum_{2}^{n} f_{1i} +P_2\sum_{3}^{n} f_{2i} +\ldots = \]

\[ =\frac{1}{2}\left[ P_0\sum_{0}^{n} f_{0i} +P_1\sum_{0}^{n} f_{1i} +P_2\sum_{0}^{n} f_{2i} +\ldots \right]. \]

Or, after simplification, by cancellation of individual terms of the equality:

\[ P_0\sum_{1}^{n} f_{0i} +P_1\sum_{2}^{n} f_{1i} +P_2\sum_{3}^{n} f_{2i} +\ldots +P_{n-1}f_{(n-1)n} = \]

\[ =P_n\sum_{0}^{n-1} f_{ni} +P_{n-1}\sum_{0}^{n-2} f_{(n-1)i} +P_{n-2}\sum_{0}^{n-3} f_{(n-2)i} +\ldots +P_1 f_{10}. \]

The last equality can be written already in quite compact form:

\[ \sum_{k=0}^{k=n-1} P_k \sum_{i=k+1}^{i=n} f_{ki} = \sum_{k=1}^{k=n} P_k \sum_{i=0}^{i=k-1} f_{ki}. \]

Betti’s theorem, obtained in such an analytic formulation, could be stated as follows:

If all the loads deforming a system are arranged in an arbitrary numerical sequence, then the sum of the products of the loads by their displacements due to the action of all loads lying from the beginning of the sequence up to them is equal to the sum of the products of the loads by their displacements due to the action of all loads lying from them to the end of the sequence.

Assigning the value

\[ k=0 \]

we obtain:

\[ P_0 f_{01}=P_1 f_{10}, \]

i.e. the special case first shown by Maxwell: every pair of loads in an elastically deformed system is connected in such a way that equality holds between the products of each load and its displacement due to the action of the other load.