Chart for Accelerating the Calculation of Moments at the Supports of a Loaded Bar under Symmetric Loading Using the Method of Foci

Unknown
Submitted 1928 | SovietRxiv: ru-192801.07811 | Translated from Russian

Abstract Generated abstract

This article presents a graphical aid for simplifying the calculation of support moments in beams or frame members with constant moment of inertia under symmetric loading, using the method of foci expressed through focal ratios. It derives fixity focal coefficients from the focal ratios and rewrites the end moment formulas as products of these coefficients and the load dependent quantity \(3\Omega/l\), values of which are available in handbooks for common symmetric load cases. A numerical example shows how the proposed chart yields the coefficients for elastically restrained ends and thereby reduces repeated arithmetic in frame calculations.

Full Text

Eng. D. E. Romanov.

Chart for Accelerating the Calculation of Moments at the Supports of a Loaded Beam under Symmetric Loading on Frames in Calculations by the Method of Foci

In calculating frames and beams by the method of foci, the device of focal distances is very widely used1. But, in essence, in an analytical calculation the focal distances are not at all needed, since in all the derivations use is made of the ratios

\[ \frac{a}{l-a} \quad \text{and} \quad \frac{b}{l-b}, \]

where \(a\) and \(b\) are the focal distances, and \(l\) is the length of the member.

These ratios are called, as is known, focal ratios and are often denoted by \(K_a\) and \(K_b\)—left and right2. They are computed directly from the angles of rotation of the elastic line, and the formulas for the computation are obtained even more simply than for focal distances.

Moreover, the extreme limits of this kind of focal ratios are rather narrow, since with a hinge \(K=0\), and with full fixity \(K=0.5\). This circumstance permits, in approximate calculations, by way of “estimates,” easily assigning the degree of fixity, estimating “by eye” the possible focal ratio within such narrow limits as 0 and 0.5.

Using the device of focal ratios, the moments at the supports of a loaded member are expressed by the following formulas2:

\[ M_a = -\left( \frac{\varphi_a}{\beta} - \frac{\varphi_b K_b}{\beta} \right) \frac{K_a}{1-K_aK_b} \tag{1} \]

\[ M_b = -\left( \frac{\varphi_b}{\beta} - \frac{\varphi_a K_a}{\beta} \right) \frac{K_b}{1-K_aK_b}, \]

where \(\beta\) is the angle of rotation of the elastic line at the support of a freely resting member under the action of a moment \(M=1\) at its opposite end, and \(\varphi_a\) and \(\varphi_b\) are the angles of rotation at the supports of this member due to the external load on it.

In the formulas cited, certain simplifications may be made for the case of a member with constant moment of inertia. This case is most frequently encountered in practice3.

For \(I=\text{const.}\), the angles, as is known from the theory of structures, are equal to:

\[ \varphi_a=\frac{A}{EI}; \quad \varphi_b=\frac{B}{EI}, \tag{2} \]

where \(A\) and \(B\) are the support reactions from a load equal to the area of the diagram of moments from the external load, as in a simple beam. In addition,

\[ \beta=\frac{l}{6EI}\ldots\ldots\ldots\ldots\ldots\ldots\ldots (3) \]

Substituting into formula (1) and carrying out the reductions, we have:

\[ \begin{aligned} M_a &=-\left(\frac{6A}{l}-\frac{6B}{l}K_b\right) \frac{K_a}{1-K_a\cdot K_b} =-\frac{6}{l}(A-B\cdot K_b)\, \frac{K_a}{1-K_a\cdot K_b}, \\[6pt] M_b &=-\left(\frac{6B}{l}-\frac{6A}{l}K_a\right) \frac{K_b}{1-K_a\cdot K_b} =-\frac{6}{l}(B-A\cdot K_a)\, \frac{K_b}{1-K_a\cdot K_b}. \end{aligned} \tag{4} \]

For various loads the quantities

\[ A,\ B \quad \text{or} \quad \frac{6A}{l},\ \frac{6B}{l} \]

are given in handbooks.

Under a symmetrical load on the member,

\[ A=B=\frac{\Omega}{2};\quad \text{i.e.}\quad \frac{6A}{l}=\frac{6B}{l}=\frac{3\Omega}{l}. \]

Substituting, we obtain:

\[ \left| \begin{aligned} M_a&=-\frac{3\Omega}{l}(1-K_b)\, \frac{K_a}{1-K_a\cdot K_b}, \\[6pt] M_b&=-\frac{3\Omega}{l}(1-K_a)\, \frac{K_b}{1-K_a\cdot K_b}. \end{aligned} \right. \ldots\ldots\ldots (5) \]

Let us denote:

\[ \left| \begin{aligned} C_a&=(1-K_b)\, \frac{K_a}{1-K_a\cdot K_b}, \\[6pt] C_b&=(1-K_a)\, \frac{K_b}{1-K_a\cdot K_b}, \end{aligned} \right. \ldots\ldots\ldots (6) \]

and call them the fixity focal coefficients, respectively the left \(C_a\) and the right \(C_b\).

In this case formulas (4) take the form:

\[ \left| \begin{aligned} M_a&=-\frac{3\Omega}{l}\cdot C_a, \\[6pt] M_b&=-\frac{3\Omega}{l}\cdot C_b. \end{aligned} \right. \ldots\ldots\ldots (7) \]

As noted above, the focal ratios vary within the limits from 0 to 0.5.

Thus the fixity coefficients will also vary within these same limits, since for \(K_a = 0\), from formula (6),

\[ C_a = 0, \quad\text{and for } K_a = 0.5 \text{ and } K_b = 0, \]

we obtain \(C_a = 0.5\).

The limits, as can be seen, are narrow; moreover, most often in reinforced-concrete frames these fixity coefficients fluctuate from 0.200 to 0.300. This makes it possible in some cases of approximate calculation (e.g., preliminary calculation) to assign directly the values \(C_a\) and \(C_b\), which, with a certain amount of practice, can be done fairly accurately.

To accelerate calculations, the quantity

\[ \frac{3\Omega}{l}, \]

which enters formula (7), is given in handbooks\(^1\), for each type of symmetrical load, in the form

\[ \frac{3\Omega}{l} = \frac{6A}{l} = \frac{6B}{l}, \]

For example, we give several cases most often encountered in practice (taken from Löser, except for case 3).

No. of drawing Type of symmetrical load \(\dfrac{3\Omega}{l}\)
1 Uniformly distributed load \(q\) over the entire beam. \(\dfrac{3\Omega}{l} = \dfrac{1}{4}\cdot ql^2\)
2 Triangular load with height \(q\). \(\dfrac{3\Omega}{l} = \dfrac{15}{32}\cdot ql^2\)
3 Trapezoidal load with height \(q\) and small side \(e = l - 2a\). \(\dfrac{3\Omega}{l} = \dfrac{q}{4l}\left(l^3 - 2a^2l + a^3\right)\)
4 One concentrated force \(P\) at the middle of the span. \(\dfrac{3\Omega}{l} = \dfrac{3}{8}\cdot Pl\)
5 Two forces \(P\), located at a distance \(l/3\) from the supports. \(\dfrac{3\Omega}{l} = \dfrac{2}{3}\cdot Pl\)
6 Three forces \(P\) at a distance \(l/4\) from one another. \(\dfrac{3\Omega}{l} = \dfrac{15}{16}\cdot Pl\)

The fixity coefficients \(C_a\) and \(C_b\), however, are much more conveniently calculated in advance from formula (6), after calculating the focal ratios, of course for those members where a symmetrical load acts, and entered in the little table.

\(^1\) B. Löser. Bemessungsverfahren... 1925. S. 15–18 and other reference manuals.

To avoid these computations, the present article proposes a graph for determining these coefficients. Results from this graph are obtained quickly. Its accuracy is not inferior to that of a logarithmic slide rule, and its use is very simple, as we shall see from a numerical example.

Example. Let there be given a bar of span \(l = 7.0\ \mathrm{m}\), of constant cross-section, elastically fixed at the ends; the magnitude of the fixity is characterized by the focal ratios

\[ K_a = 0.235, \]

\[ K_b = 0.350. \]

The bar is loaded by two forces symmetrically, \(P = 5\ \mathrm{t}\) each, at distances \(e = 2.5\ \mathrm{m}\) from the supports (Fig. 7).

To determine \(C_a\) from the graph, one should find on the abscissa the point with \(K_b = 0.350\) and, going vertically to the intersection with the curve \(K_a = 0.235\), read the answer on the ordinate.

We obtain

\[ C_a = 0.166. \]

To determine \(C_b\) we proceed in the opposite way, i.e., on the abscissa we lay off the value \(K_a = 0.235\), note its intersection with the curve \(K_b = 0.350\), and obtain

\[ C_b = 0.292. \]

For greater mechanical ease in reading from the graph it is useful to apply a drawing triangle, with its right angle at the point of intersection.

And, since it is recommended that all computations of the fixity coefficients be carried out simultaneously for all loaded bars, the entire determination reduces to placing the triangle once on the graph (preferably a transparent one) and, moving it, entering all \(C_a\) and \(C_b\) in a small table.

Returning to our example, we find from a handbook, for the case of two forces, the value:

\[ \frac{3\Omega}{l} = \frac{3 \cdot P \cdot e(l-e)}{l} = \frac{3 \cdot 5 \cdot 2.5(7-2.5)}{7} = 24.1\ \mathrm{t \cdot m}. \]

Next, by formula (7), we obtain

\[ M_A = -\frac{3\Omega}{l}\cdot C_a = -24.1 \cdot 0.166 = -4.0\ \mathrm{t \cdot m}, \]

\[ M_B = -\frac{3\Omega}{l}\cdot C_b = -24.1 \cdot 0.292 = -7.05\ \mathrm{t \cdot m}. \]

It may also be advised, when calculating a system with a large number of loaded bars, to compute all values

\[ \frac{3\Omega}{l} \]

(separately for the permanent and live loads) for all bars simultaneously and to enter the results in a small table. Having it and also the small table of fixity coefficients (see above), there will naturally appear a table of all \(M\) at the supports, according to formula (7).

I hope that the brief indications given and the proposed graph will partly serve the purpose of simplifying the computational work of the structural engineer in determining force effects, and, at least somewhat, will shorten the thankless labor of arithmetical calculations.

For the article by Eng. D. E. Romanyova

Diagram page with load schemes and moment diagram.

Fig. 1.

Fig. 4.

Fig. 2.

Fig. 5.

Fig. 3.

Fig. 6.

Fig. 7.

For the article by Eng. D. E. Romanov

Dependence between the coefficient values \(C_a\) and \(C_b\) and the focal ratios \(K_a\) and \(K_b\), for a symmetric load on a member.

\[ M_a=-\frac{3\Omega}{\ell}\cdot C_a \]

\[ M_b=-\frac{3\Omega}{\ell}\cdot C_b \]

Vertical-axis label:

\[ C_a\,(C_b) \]

Horizontal-axis label:

\[ K_b\,(K_a) \]

Annotation on the upper boundary:

\[ K_a=0.50\,(K_b) \]

Diagonal annotation:

Line of equal \(C_a\) and \(C_b\).

Visible curve labels:

\[ 0.45,\quad 0.40,\quad 0.35,\quad 0.30,\quad 0.25,\quad 0.20,\quad 0.15,\quad 0.10,\quad 0.05 \]

Visible scale markings:

\[ 0.0,\ 0.05,\ 0.10,\ 0.15,\ 0.20,\ 0.25,\ 0.30,\ 0.35,\ 0.40,\ 0.45,\ 0.50 \]

  1. Dr. Ing. E. Suter, Die Methode der Festpunkte. 1923.
    Ing. A. Strassner, Neuere Methoden. 1925.
    Prof. I. V. Ulyaninsky, Calculation of Frame Structures. 1927.
    Eng. L. A. Samoilov, Calculation of Frames by the Method of Foci. Proceedings of MIIT, issue III, 1926. ↩

  2. Prof. V. I. Rudnev, Structural Mechanics, vol. II, 1926. (Prof. Rudnev uses reverse ratios.)
    Eng. K. A. Kartashov, Calculation of Frames by the Method of Focal Ratios. 1927. ↩↩

  3. The insignificant influence of haunches, if present, is often neglected without great detriment to accuracy. ↩

Submission history

Chart for Accelerating the Calculation of Moments at the Supports of a Loaded Bar under Symmetric Loading Using the Method of Foci