ARC DIAGRAMS FOR COMPRESSED-BENT RODS LOADED BY SEVERAL LONGITUDINAL FORCES

I. D. Kisenko

(Presented by the scientific seminar of the Department of Strength of Materials)

The problem of calculating compressed-bent rods is complicated to a certain degree when several longitudinal forces or a distributed longitudinal load are present. Exact solutions obtained for some such problems are very laborious and in a number of cases require the use of special functions. These circumstances hinder the broad use in engineering practice of the solutions that exist. In connection with this, approximate methods of solution have become widespread.

Since the total potential energy of a compressed-bent beam, as a system in a state of stable equilibrium, has a minimum value, the determination of the form of the elastic line is a problem of the calculus of variations. The application of the energy method to the calculation of a bent rod whose deflections under the action of a transverse load and several longitudinal forces are regarded as sums of ordinates of sinusoids with periods multiple of the length of the rod made it possible to obtain an approximate solution of the problem [1]. An estimate of the accuracy of the approximate solution permits the conclusion that, in simple cases of transverse loading, the energy method displays rapid convergence and gives results satisfactory for practical purposes already in the first approximation [2]. However, under a complex transverse load, especially under a load causing the appearance of inflection points of the elastic line, the energy method loses rapid convergence, and to obtain a solution of the required accuracy it is necessary to use several approximations. This thereby significantly narrows the practically expedient range of application of the energy method.

In those cases in which analytical solutions, exact and approximate, lead to excessively cumbersome equations, require the obtaining of several approximations or the use of special functions, the advantages of the graphical method stand out especially clearly. The graphical method for solving problems of longitudinal-transverse bending, based on the geometric interpretation of the solution in the form of vector diagrams, was first proposed in 1919 by N. G. Chentsov and further developed in the works of Soviet authors. It should be noted that in foreign technical literature the graphical method of vector diagrams is often credited to Howard, despite the fact that the latter’s proposals were made 9 years later.

N. G. Chentsov’s method can be used not only in a polar but also in a rectangular coordinate system [3]. In this case the vector (polar) diagrams are replaced by arc diagrams.

This article presents the application of the graphical method of arc diagrams to the calculation of compressed-bent rods loaded by several concentrated longitudinal forces and by a mixed transverse load. The indicated method is applicable both for stiffness constant over the entire length of the rod and for stiffness varying from segment to segment.

Method of arc diagrams

The differential equation of a compressed-bent rod in the coordinate system adopted in Fig. 1,

\[ \frac{d^2 M}{d x^2}+\alpha^2 M=q \tag{1} \]

by introducing the dimensionless variable \(\varphi=\alpha x\), which is the angular equivalent of length, is reduced to the form

\[ \frac{d^2 M}{d\varphi^2}+M=\mu, \tag{2} \]

where \(\mu=\dfrac{q}{\alpha^2}\) and \(\alpha^2=\dfrac{S}{EJ}\). Since \(\dfrac{dM}{dx}=Q\), then \(\dfrac{dM}{d\varphi}=\dfrac{Q}{\alpha}=T\).

The quantity \(T\) will hereafter be called the reduced transverse force.

Equation (2) can now be represented in the form of a relation between the bending moment and the reduced transverse force

\[ T\frac{dT}{dM}+M=\mu. \tag{3} \]

Integration gives the result

\[ (M-\mu)^2+T^2=\rho^2, \]

which makes it possible to conclude that the solution of equation (3) in the coordinate-axis system \(M\) and \(T\) is a circle of radius \(\rho\), having its center at the point \(O\) \((T=0,\ M=\mu)\). The value of the constant \(\rho\) becomes known directly when the diagram is constructed.

The boundary conditions of the problem always determine the position of two boundary straight lines to which belong the end points of the arc corresponding to the end sections of the rod. For example, for the force scheme of Fig. 1 the boundary straight lines \((a)\) and \((c)\) coincide with the \(T\)-axis, since \(M=0\) at \(x=0\) and \(x=l\). The arc of the circle must be bounded by the central angle \(\varphi=\alpha l\). Consequently, rotation of the straight line \((a)\) through the angle \(\varphi\) determines

Fig. 1

such a straight line \((c_1)\), on which lies the right-hand point of the arc corresponding to \(x=l\). But this same point must belong to the boundary straight line \((c)\). Thus the intersection of the straight lines \((c)\) and \((c_1)\) gives the point \(C\) of the arc, whose radius \(\rho\) is equal to \(QC\).

In the arc diagram, each section of the rod corresponds to a point whose coordinates, in the chosen scale, represent the reduced transverse force and the bending moment in the specified section.

In the general case, by transverse or longitudinal loads, and also by a stepwise change in stiffness, the span is divided into intervals, each of which is represented in the diagram by an arc of a circle. The position of the center of the arc of any interval is always determined by the values of the intensity of the distributed transverse load, the longitudinal force, and the stiffness of the interval. The displacement of the ends of the arcs at the boundary of adjacent intervals is determined by the matching conditions, which in each particular case take into account the presence of a discontinuity in one or several functions representing the change in bending moment, transverse force, intensity of distributed load, stiffness, and longitudinal force. In Fig. 2 an arc

Fig. 2

diagram is constructed for the span \(AD\), which has three intervals. Denoting the value of a function on the left by the sign \(-\), and the value of a function on the right by the sign \(+\), the matching conditions at the boundaries of the intervals may be represented in the following form:

\[ \begin{aligned} M^+ &= M^- + M_1,\\ T^+ &= T^-, \end{aligned} \qquad \text{for } x=l_1, \]

\[ \begin{aligned} M^+ &= M^-,\\ T^+ &= T^- + \frac{P}{\alpha} \end{aligned} \qquad \text{for } x=l_1+l_2. \]

The point \(B^-\) of the arc of the first interval, corresponding to \(x=l_1\), must lie on the straight line \((b^-)\), obtained by rotating the left boundary straight line \((a)\) about the center \(O\) through the angle \(\alpha l_1\). From the matching conditions it follows that the point \(B^+\) of the arc

of the second segment corresponding to this same abscissa must lie on the straight line \((b^+)\), shifted along the \(M\)-axis by the amount \(M_1\) relative to the straight line \((b^-)\). In this case the points \(B^-\) and \(B^+\) must have equal abscissas on the diagram. The point \(C^-\) of the diagram, corresponding to \(x=l_1+l_2\), will belong to the straight line \((c^-)\), obtained by rotating the straight line \((b^+)\) about the center \(O\) through the angle \(\alpha l_2\). The conjugacy conditions at \(x=l_1+l_2\) show that the point \(C^+\), having the same ordinate as the point \(C^-\), must lie on the straight line \((c^+)\), shifted along the \(T\)-axis by the amount

\[ \frac{P}{\alpha} \]

relative to the straight line \((c^-)\). The straight line \((d_1)\), corresponding to \(x=l\), is obtained by rotating the straight line \((c^+)\) about the center \(O\) through the angle \(\alpha l_3\). But the same abscissa \(x=l\) also corresponds to the straight line \((d)\). Consequently, the point \(D\) is determined by the intersection of the straight lines \((d_1)\) and \((d)\). The further operations of drawing arcs between the straight lines \((d)\) and \((c^+)\), \((c^-)\) and \((b^+)\), \((b^-)\) and \((a)\) require no explanation.

A change in the intensity of the distributed load does not affect the relative position of the boundary straight lines, but the center of the arc of a segment with another value of \(q\) will have a different ordinate \(\mu\).

Calculation of rods loaded by several axial forces

The method of arc diagrams makes it possible to obtain a visual picture of the distribution of forces in a compressed-bent rod loaded by several longitudinal forces (Fig. 3).

Fig. 3

The conjugacy conditions at the boundary of the \((k-1)\)-st and \(k\)-th segments have the form:

\[ M^+ = M^- \tag{4} \]

\[ Q^+ = Q^- - S_k \frac{dv}{dx}, \tag{5} \]

where \(\dfrac{dv}{dx}\) is the angular displacement of the boundary section.

The point \(K^-\), belonging to the arc of the \((k-1)\)-st segment and corresponding to the end of this segment, and the point \(K^+\), belonging to the arc of the \(k\)-th segment and corresponding to the beginning of this segment, must belong to one straight line parallel to the \(T\)-axis. This follows from condition (4) of continuity of the bending moment at the boundary of the segments. On the other hand, the distances of these points from the axis of ordinates, determined by the quantities \(T^-\) and \(T^+\), must satisfy condition (5). This condition should be put in the form of a relation between \(T^+\) and \(T^-\).

For the boundary section, the function \(Q\) on the right and on the left has the values:

\[ Q^+ = Q_0 - (S_1 + S_2 + S_3 + \ldots + S_k)\frac{dv}{dx}, \]

\[ Q^- = Q_0 - (S_1 + S_2 + S_3 + \ldots + S_{k-1})\frac{dv}{dx}. \]

Here \(Q_0\) is the transverse force in the boundary section in the absence of longitudinal forces. After eliminating the unknown angular displacement \(\dfrac{dv}{dx}\), the conjugacy condition (5) assumes the form most convenient for geometrical interpretation:

\[ \frac{Q^{+}-Q_0}{Q^{-}-Q_0} = \frac{S_1+S_2+S_3+\ldots+S_k}{S_1+S_2+S_3+\ldots+S_{k-1}} \]

or

\[ \frac{ T^{+}-\dfrac{Q_0}{\alpha_k} }{ T^{-}-\dfrac{Q_0}{\alpha_{k-1}} } = \frac{\alpha_{k-1}}{\alpha_k}\, \frac{S_1+S_2+S_3+\ldots+S_k}{S_1+S_2+S_3+\ldots+S_{k-1}} . \tag{5a} \]

Let us imagine that the point \(K^{-}\) of the arc of the \((k-1)\)-th segment, corresponding to the boundary section and therefore having the abscissa \(T^{-}\), must belong to some straight line \((k-)\) (Fig. 4). In this case the point with abscissa

\[ \left(T^{-}-\frac{Q_0}{\alpha_{k-1}}\right) \]

must lie on the straight line \((k_1)\), obtained as a result of translating the straight line \((k-)\) parallel to itself by the amount

\[ -\frac{Q_0}{\alpha_{k-1}} \]

along the \(T\)-axis. It follows from condition (5a) that the point with abscissa

\[ \left(T^{+}-\frac{Q_0}{\alpha_k}\right) \]

will belong to such a straight line \((k_2)\), the abscissae of whose points are related to the abscissae of the corresponding (i.e., having the same ordinates) points of the straight line \((k_1)\) as

\[ \frac{\alpha_{k-1}(S_1+S_2+S_3+\ldots+S_k)} {\alpha_k(S_1+S_2+S_3+\ldots+S_{k-1})}. \]

Fig. 4

To construct the straight line \((k_2)\), it is sufficient to lay off the segment

\[ OB = OA \cdot \frac{\alpha_{k-1}(S_1+S_2+S_3+\ldots+S_k)} {\alpha_k(S_1+S_2+S_3+\ldots+S_{k-1})} \]

and connect point \(B\) with the point of intersection of the straight line \((k_1)\) and the axis of ordinates. On the straight line \((k+)\), obtained by translating the straight line \((k_2)\) parallel to itself along the \(T\)-axis by the amount \(\dfrac{Q_0}{\alpha_k}\), there must lie the point with abscissa \(T^{+}\), i.e., the initial point \(K^{+}\) of the arc of the \(k\)-th segment. Thus, if the straight line \((k-)\) contains the final point of the arc of the \((k-1)\)-th segment, then the initial point of the arc of the \(k\)-th segment will lie on the straight line \((k+)\) and will have the same ordinate.

The sequence of graphical operations in constructing an arc diagram is conveniently considered on a particular example (Fig. 5).

The centers of the arcs of the first and second segments have ordinates

\[ \mu_1=-\frac{q}{\alpha_1^2}=-\frac{qEJ}{S_1}, \]

\[ \mu_2=-\frac{q}{\alpha_2^2}=-\frac{qEJ}{S_1+S_2}. \]

Fig. 5

The boundary conditions \((M=0\) for \(x=0\) and \(T=-\dfrac{ql}{\alpha_2}\) for \(x=l)\) determine, respectively, the left \((a)\) and right \((c)\) boundary straight lines.

The construction of the arc diagram is carried out in the following order.

1. Rotate the straight line \((a)\), which in this case coincides with the \(T\)-axis, through the angle \(\alpha_1 l_1\) about the center \(O_1\) to the position \((b^-)\). The point \(B^-\) must lie on this straight line.

2. Translate the straight line \((b^-)\) parallel to itself by the amount

\[ -\frac{Q_0}{\alpha_1}=-\frac{ql_1}{\alpha_1} \]

along the \(T\)-axis into the position \((b_1)\).

1. Draw the straight line \((b_2)\), the abscissas of whose points are related to the abscissas of points with the same ordinates on the straight line \((b_1)\) as

\[ \frac{\alpha_1(S_1+S_2)}{\alpha_2 S_1}. \]

1. Translate the straight line \((b_2)\) parallel to itself by the amount

\[ \frac{Q_0}{\alpha_2}=\frac{ql_1}{\alpha_2} \]

in the direction of the \(T\)-axis to the position \((b^+)\). The point \(B^+\) must lie on this straight line.

1. Rotate the straight line \((b^+)\) through the angle \(\alpha_2 l_2\) about the center \(O_2\) to the positio-

…line \((c_1)\). The point of intersection of the line \((c_1)\) with the right boundary line \((c)\) is the terminal point of the arc of the right segment (point \(C\)).

1. Draw an arc of radius \(O_2C\) between the lines \((c)\) and \((b^+)\). The intersection of the arc with the line \((b^+)\) determines the point \(B^+\).

2. Draw through the point \(B^+\) a straight line parallel to the \(T\)-axis until it intersects the line \((b^-)\) at the point \(B^-\).

3. Draw an arc of radius \(O_1B^-\) between the lines \((b^-)\) and \((a)\). The intersection of the line \((a)\) and the arc determines the point \(A\), corresponding to the abscissa \(x = 0\).

When using the diagram, it is necessary to bear in mind that the scale along the \(T\)-axis changes from segment to segment depending on \(\alpha\).

The presence of a mixed transverse load, including force factors of various types, is accounted for in the diagram, as was shown above, by mutual displacement of the boundary lines, the character and magnitude of which are determined by the type and magnitude of the force factor. A change in stiffness, while leaving the functions \(M\) and \(Q\) continuous, causes a discontinuity in the function \(T\):

\[ \frac{T^+}{T^-}=\sqrt{\frac{EJ^-}{EJ^+}}, \tag{6} \]

where \(EJ^-\) is the stiffness of the left segment,
\(EJ^+\) is the stiffness of the right segment.

Consequently, the boundary lines must receive a mutual angular displacement such that the abscissas of any two points belonging to these lines and having identical ordinates will be related by relation (6).

The linear and angular displacements may be calculated from the formulas:

\[ v=\frac{M_0-M}{S}, \]

\[ \Theta=\frac{Q_0-\alpha T}{S}, \]

where \(M\) and \(T\) are the bending moment and the reduced transverse force, determined for the selected section from the diagram,
\(M_0\) and \(Q_0\) are the bending moment and transverse force determined for the same section without taking into account the influence of the longitudinal forces.

LITERATURE

1. Rozanov O. N. An approximate method for calculating beams of variable stiffness, “Tekhnika vozdushnogo flota,” No. 10, 1928.

2. Makushin V. M. On the calculation of compressed-bent beams in the case of application of longitudinal forces within the span, “Trudy Kr. MMMI im. Baumana,” issue 2/1, 1936.

3. Rojansky V., Beth R. Camptograms for Beams in Compression, “Journal of Applied Mechanics,” vol. 14, No. 3, 1947.