ON A METHOD FOR DETERMINING THE RADIUS OF CURVATURE OF THE NEUTRAL LAYER OF PLANE CURVED BEAMS OF ARBITRARY CROSS-SECTION

M. G. PINSKY

The normal stresses in the cross-section of a curved beam are determined by the approximate formula

\[ \sigma = \frac{P}{F} \pm \frac{M}{S}\,\frac{z}{r_0+z}, \tag{1} \]

where \(P\) is the tensile force,
\(M\) is the bending moment,
\(F\) is the area of the cross-section of the beam,
\(S\) is the static moment of the area of the cross-section with respect to the neutral line,
\(z\) is the distance of the point of the cross-section under consideration, at which the stress is determined, measured, when positive, in the direction of the outer arc from the neutral line,
\(r_0\) is the radius of curvature of the neutral layer.

The quantity \(r_0\), as is known, is determined in each particular case from the relation:

\[ r_0=\frac{F}{\displaystyle \int_F \frac{dF}{r_0+z}} . \tag{2} \]

If we put \(u=r_0+z\), then

\[ r_0=\frac{F}{\displaystyle \int_F \frac{dF}{u}} . \tag{3} \]

From (3) one can determine \(r_0\) for various forms of section \([1]\) analytically. There exist a number of approximate methods for determining \(r_0\) \([1], [2]\). In this note a device is proposed for calculating the denominator in expression (3)

\[ J=\int_F \frac{dF}{u}. \tag{4} \]

This device consists in the following. Suppose that we have an arbitrary cross-section of a plane beam (Fig. 1). Draw the line \(aa_1\) at a distance \(u\) from the axis of curvature of the beam. At a distance \(du\), draw a line parallel to \(aa_1\); then the elementary area

\[ dF=2\overline{ab}\cdot du. \tag{5} \]

To calculate the integral (4), it is necessary to reduce the elementary area \(dF\) (5) by \(u\). To do this, while preserving the width of the elementary strip, we reduce the height of the strip, i.e. \(\overline{ab}\),

\[ \frac{dF}{u}=2\,\frac{\overline{ab}}{u}\,du . \tag{6} \]

Connect point \(a\) with point \(o\), and at a distance of \(1\ \text{cm}\) from the axis of curvature draw a line parallel to the sought axis. This straight line intersects \(\overline{ao}\) at point \(f\). Considering the similarity of the triangles \(aob\) and \(foe\), we write the relation

Fig. 1

\[ \frac{\overline{ab}}{bo}=\frac{\overline{fe}}{eo}, \]

whence

\[ \overline{fe}=\frac{\overline{ab}}{bo}=\frac{\overline{ab}}{u}. \tag{7} \]

Thus, from (6) on the basis of (7),

\[ \frac{dF}{u}=2\,\frac{\overline{ab}\,du}{u}=2\overline{fe}\,du . \]

Transfer the segment \(\overline{fe}\) onto the straight line \(ab\); in this case \(\overline{fe}=\overline{a_2b}\), and then

\[ \frac{dF}{u}=2\overline{a_2b}\,du . \tag{8} \]

To calculate the integral (4), the cross section should be divided into several portions. Carrying out such a construction for each portion, we obtain a series of points. By connecting these points with a smooth line and summing the area of the figure thus transformed, we thereby calculate graphically the integral (4),

\[ J=\int_{\rho}\frac{dF}{u}=F_1, \]

where \(F_1\) is the transformed area (shaded, Fig. 2).

Having determined \(F_1\) by ordinary summation, one can compute from (3)

\[ r_0=\frac{F}{F_1}. \tag{9} \]

The calculation of these areas is performed with a planimeter.

To increase the accuracy of calculating \(F_1\), it may simply be increased; to do this, the straight line should be set off by \(n\) cm. Then formula (9) will take the form:

\[ r_0 = n - \frac{F}{F_1}. \]

Fig. 2

REFERENCES

Belyaev N. M. Strength of Materials, 7th ed., pp. 590–593, 1951.
Timoshenko S. P. Strength of Materials, Part II, p. 73, 1946.