Prof. S. K. Konyukhov.

Static and Dynamic Analysis of the Cage.

1. Static analysis of the cage.

“Excess weight is just as meaningless in any object as a badge on a coachman’s coat—perhaps even more meaningless. A badge may, in the end, serve for identification, whereas excess weight signifies only a needless expenditure of force. For me it is a mystery what basis there can be for confusing gravity with force. Weight is very good in a pile driver for driving piles, but what is the point of setting excess weight in motion when nothing is thereby achieved? Why burden a machine intended for transport with special weight? Why not transfer the excess weight to the load that is being transported by the machine? Fat people cannot run as fast as thin people, yet we give a large part of our transport machines such heaviness, as though dead weight and volume increased speed? Poverty is to a considerable degree the result of dragging dead loads.”

So writes H. Ford in his book My Life and Work.

These words seem to have been written especially as a stinging criticism of the irrational use of weight in mine cages. It is difficult to say what grounds were used in constructing the overwhelming majority of mine cages; but, judging by the fact that the literature on cage construction is exhausted by only two works—N. Lessar’s Hoisting Cages (in French) and S. A. Zaborovsky’s Dynamic Analysis of a Two-Deck Mine Cage (in Russian)—and that most cages do not satisfy the economical use of material, one may, without hesitation, note here a dark and little-explored area in the art of construction. They are built, evidently, by eye, according to the principle: “as appropriate and proper.” Meanwhile, with increasing mine depths and the growing need to increase the extraction of minerals, the question of the weight of the cage acquires great significance.

To begin with, a light cage can be equipped with more reliable parachutes, and this is closely connected with the safety of work in enterprises. Kaleidoscopic views on parachutes—beginning with the complete denial of their usefulness, passing to an indifferent attitude, and finally ending with the prescription that parachutes be installed on cages—all this must give way to a single view: the advantage and necessity of supplying cages with rationally designed parachutes, whose action must be checked as often as the tested rope. It must not be forgotten for a single minute that the matter here concerns human life, and therefore penny-pinching calculations have no place here.

When it is necessary to speak of work under normal conditions, it should be noted here that reducing the dead load, which the weight of the cage is, will inevitably lead to easier operation of the machine and, consequently, to savings in driving power and to the use of lighter engines. If a cage of reduced weight is intended to replace an old heavy cage, then there are advantages here as well. Without changing the old rope, the same machine can raise one and a half times more people in a lightened cage on each hoist, and yet the load on the rope will be less than before. And this leads to a great reduction of time in mining, where all work is carried out urgently, while the lowering and raising of the labor force only takes away working time.

Therefore, reducing the time required for moving the labor force, the lighter cage makes it possible each time to lift a larger useful load on the same rope, which likewise saves time and increases the productivity of the enterprise. The time for lowering the fastening is reduced. Finally, the lighter cage makes it possible, with the same machine and the same rope, to serve lower horizons.

Lightening the cage and the possibility of raising more people in a shorter time is of special importance for such extraordinary events in the life of a mine as fire, explosion, flooding of the mine, etc.

By what means can the reduction in the weight of cages be achieved? Several can be outlined, but the greatest effect should be obtained by combining all methods into one—a comprehensive one.

1) Establishment of a normal type of hoisting cages, standardization of cages, which should provide both light and cheap cages with the necessary number of rationally selected and arranged members.

2) Transition from riveted joints to welding, which should give a saving in weight of approximately 31%.

3) Use, instead of iron, of the newest alloys, possessing low specific weight and high mechanical-technological properties.

Among these are aluminum alloys.

Table 1.

Material	Specific weight	Coefficient of expansion
German alloy	2.90–2.95	0.0000 255
American alloy	2.85–2.91	0.0000 246
Silumin	2.5–2.65	0.0000 222
Duralumin	2.8	0.0000 226

Table 2.

Material	Strength, hard	Strength, soft	Elongation %, hard	Elongation %, soft	Hardness, hard	Hardness, soft
German alloy	20	30	20	10	55	90
American alloy	20	30	20	10	60	90
Silumin	18	30	30	5	60	—
Duralumin	38	42	20	15	60	150

Strength is taken in kg/mm².
Elongation in % is calculated by the formula \(L = 11.3\sqrt{q}\).
Hardness (\(P = 500\) kg, ball \(\phi = 10\) mm) in kg/mm².

Ford did not hesitate to use vanadium steel for his automobiles on the broadest scale and achieved the most striking results, while the French before him had used this steel only for engine valves.

Here, of course, the deeply rooted routine also made itself felt: the French did not find in themselves enough common sense to use the vanadium steel known only to them, whereas Ford, from the breakage of a valve spindle (a collision occurred in a race between a French automobile and a Ford one), immediately determined what valuable specimen of material had fallen into his hands, and made enormous efforts to set up the metallurgy of vanadium steel in America. Indeed, vanadium steel has a tensile strength almost three times greater than ordinary carbon steel, and deserves great attention in construction.

It is therefore possible boldly to repeat Ford’s words: “vanadium must be used wherever strength and lightness are required.”

Alongside vanadium steel, aluminum alloys should also find wide application; they have already conquered for themselves an extensive field of use in the most varied branches of industrial technology, science, and everyday life, but are still only knocking at the door of mining engineering.

Coming closer to the question of calculating hoisting cages, let us ask: what can be expected from this innovation?

It is best, in the sense of clarity, to consider an example.

Let us imagine a mine 1000 meters deep, served by a rope with a specific weight of 0.00094 kg/cu. cm. Let the rope tension at the lugs be 100 kg, and at the cage 80 kg. Further denote by \(S\) the weight of the rope, by \(R\) the maximum radius of winding, and by \(r\) the minimum. Let the lower cage be sitting on the lugs, while the upper loaded cage has not reached the lugs; this case represents the usual phenomenon during hoisting, owing to the unequal stretching of both ropes. Let \(M\) denote the maximum moment of resistance on the drum shaft.

Now let us see what will happen if loads of 6000—5500 and 5000 kilograms are suspended from the rope.

\(Q\)	\(Q_1 = 6000\)	\(Q_2 = 5500\)	\(Q_3 = 5000\)
\(S\) kg.	10225	9405	8550
\(R\) cm.	395	381	371
\(r\) cm.	128	122	116
\(M\) kg. mt.	10574	9481	8432

The second moment of resistance is less than the first by 10.6%, and the third by 20.6%. In reality the economy in hoisting operation will amount to 4—6%, and only during maneuvering will it rise to the norms given above.

If the shaft is deepened to 1050 meters and the cage is suspended from a rope weighing 5000 kilograms, then

\[ S = 9255 \text{ kg}; \quad \text{and } M \max = 8700 \text{ kg mt.} \]

Assuming the shaft depth to be 1000 meters and \(Q = 6000\) kg, we obtain

\[ M \max = 10574 \text{ kg mtr.} \]

The difference in the magnitude of the maximum moment of resistance on the drum shaft will be approximately 17.7%.

These considerations are quite sufficient to draw the attention of designers to the rational choice of the cage weight.

We shall carry out the calculation of the cage under two assumptions:

1. The cage is suspended from the rope.
2. The cage is seated on the lugs.

Each cage consists of the following principal parts: a) horizontal frames, b) vertical posts with grids, c) connections, and d) accessories or appurtenances of the cages, chiefly of a safety nature.

The form of the cage and its dimensions are determined to a considerable degree by the ground conditions in which the mine shaft is driven and by the intensity of the mine’s operation. Thus there are single-deck, two-deck, three-deck, and four-deck cages. Cages with a larger number of decks are almost never built. Each deck accommodates either one mine car or two; in the latter case the cars are arranged either side by side, i.e., on two tracks, or one behind the other, i.e., on one track.

The section of the cage by a horizontal plane is either a rectangle or a rectangle with beveled corners. This depends entirely on the section of the mine shaft.

As for the designs of cages, they are quite varied, but it would be advisable to establish, for cage construction, the same schemes as exist for roof trusses. Then, to a considerable extent, the unnecessary work of considering which design to prefer and why would be made easier. This problem awaits its solution, and its execution will not be without benefit.

When beginning the design of a cage, it is necessary to know in advance, at least, the cross-section of the shaft, the number of mine cars in the cage and their weight with load, as well as the distance between the supporting points of the wheels along the rail track and perpendicular to it.

After this it is necessary to settle on some design, preferring cages of known factories and paying attention to the fact that the suspension of the cage from the chains should be effected not by the horizontal members, but by the posts. The frames situated horizontally and the sides of the cage facing the guides are subjected to calculation.

I. The cage is suspended by chains from the posts.

a. Calculation of the frame.

The upper frame, from which the ends of the chains run and converge into a knot near the lower end of the rope, is subjected to the greatest action when the cage is suspended. Depending on how the lower ends of the chains are attached—either to the posts or to the horizontal members—the approach to solving the problem will be different.

Let the ends of the chains embrace the upper parts of the posts (vertical members).

In this case the upper frame is subjected to compression, and it is possible without difficulty to determine the forces compressing the individual members of the frame.

The cage is suspended either on two, on four, or on six chains. It is always endeavored to adjust the length of the chains so that there is equal tension in each.

Most often cages are suspended on four chains.

For this case a simple construction was proposed by Engineer Dessart.

Construct a right angle \(AOB\) (Fig. 1), in which \(OA\) and \(OB\) represent, to scale, half the sides of the rectangle formed by the four points of suspension of the chains. The line \(AB\) will represent half the diagonal of that same rectangle. Lay off the line \(AB\) on the side \(OB\), so that \(AB = OC\).

From point \(C\), as from a center, make an intercept with the length of the chain, on the scale adopted by us for the rectangle \(AOB\). In this way we obtain, on the side \(OA\), point \(D\). The straight line \(DC\) will represent, to scale, the length of each of the four suspension chains.

From point D downward, lay off the segment \(DG\) on the line \(OD\), and take it equal to the total load of the lower end of the rope, \(Q\) kg. The scale for \(DG\) is chosen in accordance with the drawing, \(a \text{ mm} = 100\) kg. Now draw through point \(G\) a horizontal line until it meets the straight line \(DC\) at point \(I\). In this way we obtain the segment \(GI\), representing the horizontal component of the tension in one chain, but on the scale \(4a \text{ mm} = 100\) kg. Or the straight line \(GI\) may be regarded as the sum of four horizontal components of the tension in the chain, but on the scale \(a \text{ mm} = 100\) kg.

Now transfer \(GI\) to \(AB\) in the position \(AX\), and draw \(XY\) parallel to \(CO\).

In that case \(XY = H_1\) will be the acting force along the side \(OB\) of the frame, and \(AY\) the acting force along the side \(OA\), equal to \(H_2\). The scale for \(H_1\) and \(H_2\) will be \(4a \text{ mm} = 100\) kg. To clarify how this strut arises, let us imagine the upper frame suspended symmetrically at 4 points \(A—B—C—D\). (Fig. 2). Let the point of application of the force be \(E\). The tension in the lower end of the rope will be \(Q\), and this tension, or resistance of the rope, is directed upward. Connect point \(A\) with \(C\) and point \(B\) with \(D\). Denote the intersection of the lines \(AC\) and \(BD\) by \(O\). It is not difficult to see from triangle \(OED\) that

\[ OE = ED \operatorname{Sin} \alpha \]

or

\[ Q/4 = S_1 \operatorname{Sin} \alpha, \]

where \(S_1\) will be the tension in each chain.

The direction of the tensions in the chains will run from the points \(A—B—C—D\) to point \(E\). Since the entire system is in equilibrium, in triangle \(OED\) the forces close, proceeding all the time in the direction opposite to the motion of the hands of a clock. Thus, the tension—that is, the resistance in member \(EO\)—will run from point \(E\) to point \(O\), and in member \(OD\) from point \(O\) to \(D\).

The resistance in member \(OD\) will be equal to

\[ Q/4 \cdot \operatorname{Cotang} \alpha. \]

Resolve the resistance of member \(OD\) into two tensions in the directions \(OM\) and \(ON\). Since in this case also the entire system is in equilibrium, in triangle \(OMD\) the forces close, proceeding in the direction of the motion of the hands of a clock. Thus, along member \(DA\) the force runs from point \(D\) to point \(A\), and along member \(DC\) from point \(D\) to point \(C\).

Exactly the same picture is obtained when considering the joints \(A\), \(B\), and \(C\). Hence it is clear that the members of the frame will be compressed.

The triangles \(OMD\) and \(OED\) in (Fig. 2) correspond to the triangles \(AOB\) and \(ODC\) in (Fig. 1).

To calculate the section of the frame members it is necessary to use Euler’s formula

\[ I = \frac{k \cdot H \cdot L^2}{n \pi^2 E}, \]

where \(I\) represents the minimum moment of inertia of the cross-section of the member, in kg, mm.

\[ k \text{— factor of safety } (k = 20) \]

\[ H \text{— force compressing the member, in kg.} \]

\[ L \text{— length of the member, in mm.} \]

\[ E \text{— Young’s modulus, in } \mathrm{kg/mm^2}. \]

Since with a pinned connection it is difficult to expect a firm fastening of the ends of the rods, \(n\) is taken equal to unity, i.e. the ends of the rods are assumed simply supported.

Example. A cage is suspended on four chains, the ends of which form a rectangle \(1200 \times 806\) millimeters. The length of the chain is 1200 mm.

The total load \(Q\) consists of: 1) the weight of the cage, 1800 kg; 2) the weight of 4 tubs filled with barren rock; the capacity of each tub is 6 hectoliters, and the weight with load is 1100 kg.

Thus,

\[ Q = 1800 + 4 \cdot 1100 = 6200\ \text{kg}. \]

In Fig. 3 the Désar epure is given. From this epure it can be seen that the stress \(XY\) along the long side of the frame is \(H_1 = \sim 1000\) kg, and the stress \(AY\) along the short side is \(H_2 = \sim 650.\sim 700\).

If these quantities are substituted into Euler’s formula, we shall find the moment of inertia of the cross-section, and from this moment we shall select the dimensions and shape of the section.

Thus:

\[ I_1 = \frac{k \cdot H_1 \cdot L_1^2}{n \cdot \pi^2 \cdot E} = \frac{20 \cdot 1000 \cdot 1200^2}{1 \cdot 10 \cdot 20000} = 143300. \]

If we settle on a U-shaped profile, then the suitable profile will be \(65 \times 43.5 \times 7\), for which

\[ I_{\min} = 178250 \]

In exactly the same way \(I_2\) and the corresponding sectional profile will be found.

II. The cage is suspended on chains from the frame.

This method of suspension is not as favorable for the design of a light cage as the first, and such suspension of the cage is resorted to only in cases where, for some reason, it is impossible to attach the suspension to the posts. In this case the frame will be subjected to bending and compression with bending, or to complex compression.

The calculation begins by determining the section for bending. Very often it turns out that the computed section will also satisfy complex resistance. Otherwise it is necessary to reinforce the section. The rod or beam of the frame being calculated is regarded as a beam on two supports, the two nearest posts being taken as the supports.

Let the bending moment in the dangerous section be \(M\) kg·mtr. Assuming the stress \(R = 4\) kg/mm\(^2\), we obtain the basic equation

\[ \frac{I}{Z} = \frac{M}{R}. \]

Let, further, \(S\) sq. mm be the section of the beam. We find graphically, from Désar’s epure, the component \(H\).

Knowing this component, we determine the additional stress

\[ R^1 = \frac{H}{S}\ \text{kg/mm}^2. \]

The total maximum stress will be the sum \(R + R^1\), and it must not exceed 6 kg/mm\(^2\).

Example. The cage has two stories, and in each there are three tubs arranged in a row. The cage frame has been given the form of a rectangle with cut-off

with angles of \(2540 \times 1400\) mm, which is caused by the round form of the shaft section. Along the long walls of the cage there are three parallel uprights. The guides are placed on the short sides of the cage. The cage is suspended on six chains, of which the two inner, shorter ones are auxiliary, so that the frame is calculated for suspension on four chains. As suspension points we shall take the vertices of a rectangle \(1750 \times 1460\) mm. This means that the suspension points project somewhat beyond the limits of the main contour of the cage. At the beveled corners of the cage there are four uprights forming a quadrilateral with a section of \(2200 \times 1200\) mm. These uprights are the supporting ones (Fig. 4). The chain length is 1500 mm.

Let us assume the weight of the cage to be 2000 kg; the weight of six cars with a load of rock we shall take as 6600 kg. Thus, the total weight of the cage will be 8600 kg. With a knee width of 450 mm, we obtain the greatest bending moment of the frame in section \(P_2P_2^1\), equal to

\[ \frac{8600}{4}\cdot \frac{450}{2}=483750\ \text{kg. mm}. \]

Taking the allowable stress \(R = 4\ \text{kg/mm.}^2\), we obtain

\[ \frac{I}{Z}=\frac{483750}{4}=120938. \]

In exceptional cases, when the cage will hang only on two spare chains, the dangerous moment will be

\[ M_1 = 2150 \times 1100 = \frac{8600}{4}\cdot \frac{2200}{2} = 2{,}365000. \]

For such exceptional cases of a temporary character, one may allow a stress \(R \le 20\ \text{kg./mm.}^2\).

If box iron is taken for the frame, then it is easy to select dimensions knowing the section modulus \(W_x = 120938\). According to the German standard assortment, one may settle on a trough-iron profile No. 18 with dimensions \(180 \times 70 \times 7\) mm, for which \(W_x = 150\) cm.

But one may take two strips of the same box iron with a section of \(200 \times 8\) mm.

The section modulus for such a composite section is calculated by the formula

\[ W_x=\frac{BH^3-bh^3}{6H}+\frac{b\cdot h_1^3}{6h_1} =\frac{60\cdot 200^3-44\cdot 183^3}{6\cdot 200} +\frac{44\cdot 16^3}{6\cdot 16} =153765\ \text{mm}^3. \]

Thus, under ordinary conditions the stress per 1 sq. mm will be 3.15 kg, and in the extreme case \(\sim 15.4\ \text{kg/mm.}^2\).

To calculate the additional stress caused by compression of the frame, it is necessary to use the Dessar diagram (Fig. 5). Constructing this diagram for the case under consideration, we obtain \(H_1 = \sim 1900\) kg. This stress falls on the sectional area of the frame, equal to \(200 \times 8 + 200 \times 8 + 44 \times 4 \times 8 = 4608\ \text{sq. mm.}\)

Per one sq. mm this gives

\[ R_1=\frac{1900}{4608}=\sim 0.41\ \text{kg}. \]

In order to offer proper resistance to this action, the beam must possess a minimum moment of inertia equal, according to Euler’s formula, to

\[ I_{\min}=\frac{k\cdot H L^{2}}{n\cdot \pi^{2}E} =\frac{20\cdot 1500\cdot 2200^{2}}{1\cdot \pi^{2}\cdot 20000} =\sim 920.000. \]

The section adopted for the frame has a moment of inertia equal to \(\sim 8.500.000\). The total stress of the frame beam will be equal to \(3.15+0.32=3.47\ \mathrm{kg/mm^{2}}\), and in the extreme case \(=15.4+0.32=15.72\ \mathrm{kg/mm^{2}}\).

III. The cage stands on cams.

Let us consider the question from the point of view of statics, i.e., assuming the absence of shocks, shaking, and impacts.

Example. Suppose there is a four-story cage accommodating one mine car on each story. The three lower stories have a height of 1200 mm, and the uppermost one 1750 mm. The length of the cage is 1200 mm, and the width is 760 mm. On the long walls there are three posts. There are no diagonals (Fig. No. 6). Let us cut the cage along the plane of symmetry perpendicular to the long sides of the frame (Fig. 7). We shall consider one half of the frame \(B A D D^{1} A^{1} B^{1}\) as a beam supported at the points \(D\) and \(D^{1}\), where the cams are located, the distance between them being 640 millimeters.

What load does this beam carry? First of all let us turn our attention to the floor of the cage. Its area is approximately 1 sq. meter. Practice shows that one square meter of cage floor amounts to about 150 kg. Thus 75 kg will fall on the half of the frame under consideration. Let this weight be uniformly distributed along the beam. Since the beam consists of the link \(D D^{1}=640\) millimeters and two side members \(B D\) and \(B^{1}D^{1}\), each of which has a length of 750 mm, the total length of the beam will be \(=2140\) mm. Dividing the load of 75 kg proportionally to the lengths of the individual parts of the beam, we obtain a load on each side member of 26 kg, and on the middle part 23 kg. We shall assume the loads of 26 kg to be applied at the points \(A\) and \(A^{1}\) (Fig. 8).

The weight of the floor can be calculated more accurately when the finished construction of the frame, cross members, decking, rail, etc., is already available. In addition to the actual weight of the floor, two vertical forces from the weight of the car with empty rock are transmitted to the frame. Since the capacity of the car is taken as 6 hectoliters, the weight of the loaded car will be 1100 kilograms. This weight is transmitted to the frame by means of the car wheels at a track gauge of 450 millimeters. Thus each force is equal to 275 kg. Finally, four equal forces will act on the frame, composed of the weight of the cage, equal to 1800 kg, and the weight of three cars, 3300 kg, excluding the already considered weight of the floor of the lower story. Thus each force will be equal to

\[ \frac{1800+3300-150}{4}=1238\ \mathrm{kg}. \]

Two such forces will act on the beam representing half of the cage frame. These forces are applied at the points \(A\) and \(A^{1}\), the distance between which is equal to 760 mm.

Here we have a cantilever beam loaded symmetrically with concentrated loads and, between the supports, carrying a uniformly distributed load (Fig. 8). The bending moment from the concentrated load will be, in the dangerous section,

\[ M_{x}=1264\times 155-1550.5\times 95=48622.5. \]

And from the uniformly distributed loads

\[ M_{x}=\frac{23\times 95^{2}}{640\cdot 2}=\sim +162.2. \]

Thus, the total bending moment will be about 48,800 kg·mm. If one assumes a load per 1 sq. mm equal to 3 kg, then

\[ \frac{1}{Z_i}=\frac{48800}{3}=16266 \]

If one takes a steel strip of channel-shaped section \(80 \times 47 \times 8.5\), then for it

\[ \frac{1}{Z_i}=28834. \]

In that case the load per 1 sq. mm will be

\[ \frac{48800}{28834}=1.69\ \text{kg/mm}^2 \]

If we assume that the frame is cut by the plane of symmetry \(P_2 P_2^1\), then we obtain the beam \(E D A B C E G\), situated on two supports \(D\) and \(E\), the distance between which will be equal to 1380 mm.

Let us consider the loads that this beam carries.

Here again it is necessary to take into account the weight of the lower-floor decking as a uniformly distributed load.

Dividing half the weight of the floor (75 kg) in proportion to the lengths 1380 and 320, we obtain a load on the length \(D E\) approximately equal to

\[ \frac{75 \times 1380}{1380+640}=51.2\ \text{kg}=\sim 52\ \text{kg}. \]

On the share of the beams \(D E\) and \(E G\) there will then fall equally 11.5 kg. We assume these latter forces to be applied as concentrated forces at points \(D\) and \(E\) (Fig. 9).

Then it is necessary to take half the weight of the loaded paneling (550 kg) and assume that this weight acts as a concentrated load at the same points.

Finally, it is necessary to take half the load transmitted to the lower frame through the posts from the upper floors. This will still amount to 2475 kg.

The forces transmitted by the posts of the lower frame are not equal to one another. These forces essentially consist of the following parts:

1). The self-weight of the cage, reduced by the weight of the lower floor. This is \(1800 - 150 = 1650\) kg. The cage doors with their accessories are attached to the corner posts. If the weight of these doors is taken as 140 kg, then each post will carry a load of \((1650 - 140):6 = 251.6\) kg. Thus, an intermediate post will transmit a force of 251.6 kg, and the corner posts \(251.6 + 35 = 286.6\) kg.

2). The weight of the paneling located on the three upper floors, which amounts to 3300 kg.

Let the cage be cut by the plane of symmetry \(P_2 P_2^1\), and let the distance between adjacent posts be 600 mm. In that case each half of the frame, 1380 mm long, may be considered as a beam on three supports. The corner points of the posts will carry a load equal to one quarter of the weight of the paneling, i.e. 275 kg. The reactions at the support points will be equal in magnitude and opposite in sign to the required components.

In order not to complicate the derivations, we shall use the determination of the reactions at points A, B, and C by means of the formulas of strength of materials (see Kirpichev, Strength of Materials, Part II, 1923, pp. 218–219).

\[ Y_1 + 2Y = 2P \]

\[ Y = -\frac{3}{2}\,\frac{P l_1^2\left(1-\frac{1}{3}l_1\right)}{l^3} \]

Substituting the numerical values into these formulas, we obtain

\[ Y_1 + 2Y = 573\ \text{kg}. \]

\[ Y = -\frac{3}{2}\cdot \frac{236.5\cdot 690^2\cdot(600-230)}{600^3} = \sim 350 . \]

In that case \(Y_1=-127\). The reactions at points A, B, and C will be

\[ R_A = R_C = -350 \quad \text{and} \quad R_B = +127 \]

A small difference will result if, in general, the own weight of the frame is neglected. Then in the formula for \(Y\), instead of 286.5, one would have to enter the weight of the pipes as 275 kg.

In that case the reactions will be

\[ R_A = R_C = -337\ \text{kg}, \quad \text{and} \quad R_B = 124\ \text{kg}. \]

Taking into account that three cars located on the three upper floors must be introduced into the calculations, we obtain the loads

\[ -3R_A = -3R_C = +1011 \quad \text{and} \quad -3R_B = -372 \]

Consequently, the loads transmitted to the individual posts will be

\[ 286.6 + 1011 = 1297.6 \]

\[ 251.6 - 372 = -120.4 \]

The bending moment in the dangerous section of the beam will be equal to \(\sim 85000\).

If, as before, the load per \(1\ \text{sq. mm}\) is taken as \(4\ \text{kg}\), then

\[ \frac{1}{Z}=\frac{85000}{4}=\sim 21250 \]

Since channel iron with a section of \(80 \times 47 \times 8\) was taken for the beam, the actual load will be

\[ \frac{85000}{28834}=\sim 2.95\ \text{kg/mm}^2. \]

If the cage side had not three posts, but two, then the bending moment at the dangerous point would be about 120,000, i.e. approximately 41% greater. Suppose that the posts are not located in one plane (Fig. 10). Such a case is represented by the construction of the cage at the Wiebe—Marie mine.

Let us cut the cage frame by the plane \(P_1P_1^1\) and replace the curved beam (frame) by a straight beam representing the projection of the frame onto the plane of the greatest bending moment, which for the given case is equal to \(\sim 634{,}000\ \text{kg mm}\).

In this case it is possible to take as the composite beam channel iron with a section of \(100 \times 52 \times 8\). The two of our beams should be tied with strip iron \(200 \times 8\). To increase the strength of the beam, we shall reinforce the lower flanges of the channel iron with four plates \(200 \times 160 \times 8\), placed under the lower rails.

If the same frame is cut along the plane \(P_2P_2^1\), then the bending moment in the dangerous section will be \(=\sim -150{,}000\ \text{kg mm}\). In this case one may limit oneself to a beam with a section of \(100 \times 52 \times 8\).

Since in reality the frame constitutes a single whole, both moments will be smaller than those calculated.

For making the frame, soft rolled steel is used, with a tensile strength of 44 kg/mm² and 20% elongation.

For making the frame of a cage at the Flémalle mine, a strip of box iron with a section of 80 × 47 × 8 mm and a length of 4.32 meters was taken. This strip was given the shape of a rectangle with dimensions of 1380 × 760 mm. For this purpose the strip is heated and, while hot, subjected to hammer blows by means of swages. Then it is checked to see how far the frame has preserved flatness, whether its right angles are exact, and whether the dimensions of length and width are correct. The free ends of such a bent frame must be almost exactly butted together. These ends are cleaned of everything that might interfere with welding. Both ends of the frame are heated in a large furnace, while on a neighboring forge a wedge plate is prepared from iron of the best quality. When the ends of the beam and the wedge are brought to a white heat, the frame is quickly carried to the anvil and, with powerful hammer blows, the wedge plate is driven into the joint, first along the beam and then along the flanges. The entire welding is carried out in three operations. If the fists are located under the short side of the frame, then the weld is made on the long side, despite the fact that this side is subjected to the greatest bending. This is done for the reason that the principal loads for the beam are in the form of impacts and shocks, and not loads of a constant character.

At the present time welding of even a more complex kind is performed very conveniently and successfully by means of the flame of gas burners (so-called autogenous welding).

The cage walls serve to connect the frames with one another; their form or construction depends chiefly on three circumstances: 1) the arrangement of the guides (the guides may be located on one side or on two sides); 2) the method of suspension of the cage and the degree of inclination of the mine shaft. In approaching the question of suspension of the cage, it must be noted that the simplest and most desirable suspension is one in which the chain is directed along the axis of the cage. In this case the work of the frictional forces in the shaft is reduced to a minimum. This kind of suspension in the direction of the center of gravity of the cage—or, more precisely, passing through the center of gravity of the cage—may be called axial-central. It is sufficient to make use, for this purpose, of one chain passing through the center of the upper frame; when the cage is used for lowering and raising people, as a precaution, two symmetrically arranged chains must also be attached. This method of suspension is used in vertical shafts. In inclined shafts this method is not applicable. Here four, six, and even more chains are used. Besides this, it is sometimes necessary here to make the cage structure not rigid.

3). On the number of mine cars placed side by side, in other words, on the width of the wall. Let the walls of the cage consist, in addition to the sides of the frames, of two parallel posts, but without diagonals.

Such a simplified construction is naturally applicable in the case when the wall of the cage has a small width.

We shall choose the method of suspension on four chains, although other constructions may also be used. If the mine shaft is inclined, then there is no need for diagonals in the walls, since with diagonals angle pieces for connection with the frames and posts are unavoidable, and this in turn causes excessive weight of the cage and its rigidity. Even if one takes into account that the diagonals are subjected to insignificant forces and serve to maintain the sheathing of the cage walls, they may safely be replaced by box iron of small cross-section (p. 11). Such a system of walls is convenient for all kinds of guides, but is not so capable of withstanding large bending moments.

Let us apply these general considerations to a cage at the Flémalle mine.

Suppose the cage is in a suspended state. Whatever the system for suspending the cage from the ropes (by means of posts or by means of a frame), the posts are always subjected to one and the same tensile stress, amounting to approximately one quarter of the total load \(Q\). When the cage is placed on catches, the most unfavorable case, with regard to the resistance of the cage to the forces acting on it, will be the case in which the catches support the lower sill.

Under these conditions the posts are subjected to longitudinal bending under the action of a force \(DG = F \text{ kg}\), given by the Descares diagram.

In applying Euler’s formula to this case, each post between two adjacent sills may be regarded as a beam supported at both ends.

From the Descares diagram we readily obtain the acting force \(F = \sim 1600 \text{ kg}\).

And then

\[ I=\frac{kF \cdot L^{2}}{n \cdot \pi^{2}E} =\frac{20 \cdot 1600 \cdot 1200^{2}}{1 \cdot 10 \cdot 20000} =\sim 225,000 \]

Such a moment of inertia is fully satisfied by channel iron with a profile \(80 \times 47 \times 8\), the weight of which per linear meter is \(10.11 \text{ kg}\). The walls may be supplemented with diagonals of the same channel iron, with profile \(50 \times 25 \times 6\), weighing \(4.14 \text{ kg}\) per linear meter, or else with posts of channel iron with profile \(60 \times 30 \times 6\), weighing \(5.10 \text{ kg}\) per linear meter.

After this it is easy to find the approximate weight, disregarding the weight of the gusset connecting plates and the weight of the rivets.

Item	Weight
4 posts, \(5 \text{ m}\) long \(t \times 10.11 \text{ kg}\)	\(202 \text{ kg}\)
16 diagonals, \(1.7 \text{ m}\) long \(t \times 4.14 \text{ kg}\), or 4 posts, \(4.8 \text{ m}\) long \(t \times 5.10 \text{ kg}\)	\(103\)
Total	\(305 \text{ kg}\)

Sometimes the posts used are not vertical, but inclined or converging at one point. This type of cage wall is used when the cage is of small height, has one deck, and is intended, for example, for servicing shaft sinking and drainage. Here it is convenient to use a central suspension of the cage. The guides in this case may be placed either on the side or on the front. A cage of this kind was built in Belgium for sinking a shaft with a large inflow of water. The cage was single-deck and weighed only \(195 \text{ kg}\), despite the fact that it had been designed with a large margin of strength.

Turning to multi-deck cages, it must be noted that the use here of posts converging at one point has its disadvantages.

Multi-deck cages, like long trains, are not very flexible in motion; equipping them with auxiliary devices is usually complex and heavy, and all this produces additional stresses in certain parts of the cage, not so insignificant that they can be neglected.

It is particularly difficult to connect the lower platform with the upper roof of the cage into a single whole, so that the heavy roof of the cage does not cause significant bending in the upper frame.

Let us suppose that the cage frames are fastened to the joints of posts converging at one point. Thus these posts take upon themselves the greater part of the load of the cage, while the vertical posts are sufficiently unloaded.

Let the cage be suspended from the rope. The layout of the cage is given (in Fig. 12). The cage weighs \(1800 \text{ kg}\), and 4 mine cars with load weigh \(4400 \text{ kg}\). Thus, the total weight of the cage is \(6200 \text{ kg}\). We distribute this load at the points where the frames are fastened to the inclined posts. There will be sixteen such points, so that at each point there will be a concentrated load of \(6200 : 16 = 388 \text{ kg}\). If we find—

... a diagram of the cage drawn to scale, it is not difficult to do this, knowing the dimensions of the car, so that the direction of the inclined posts becomes known, and consequently also the angle at point A \((\angle H A J)\).

Projecting the force applied at each corner onto the direction of the inclined post, we obtain the force acting along the post.

The stresses in the horizontal members and in the inclined posts are shown (in Fig. 12) to the right and above each member.

Thus, the following table may be written.

Member	H F	F D	D B	B A	H J	F G	D E	B C
Stress . . .	+390	+790	+1180	+1570	−60	−60	−60	−60

Let the cage be set down on the catches and fully loaded. Then at points H and J reactions arise the same as for a cage without inclined posts. They were calculated earlier, and each reaction was equal to 1238 kg. At points B, C, D, and E there will be concentrated loads, each equal to 388 kg. As for points F and G, the same forces of 388 kg will act in them; in addition, the posts carry part of the load of the lower platform, namely that part which is not taken directly by the frame. Thus the frame carries the weight of the car, 1100 kg, plus its own weight, amounting to 150 kg, which makes 1250 kg. For the posts there remains

\[ \frac{6600}{4} - 1250 = 300 \text{ kg}, \]

and for each post 75 kg. These 75 kg are distributed along the length FG and GH. The most unfavorable case for calculation will be that in which the load of 75 kg is concentrated at the nodes F and G. Thus, the concentrated load at each of the nodes F and G will be equal to 463 kilograms.

Now it is possible to compile a table of the stresses in the bars. These stresses are written (in Fig. 12) to the left and below each member.

Member	H F	F D	D B	B A	H J	F G	D E	B C
Stress . . .	−1250	−795	−400	0	−185	−65	−60	−60

The members HF and JG are subjected to the greatest action. Choosing for these members U-shaped iron with a section of \(80 \times 45 \times 6\), we fully ensure resistance, since \(J_{\min} = 194000\), and the section is \(1100 \text{ mm}^2\). The weight of one running meter of this iron will be 8.6 kg.

It is necessary to say separately a few words on the question of the posts of the upper part of the cage.

For these posts the same rolled iron has been selected as for the posts H A and J A, i.e. \(80 \times 45 \times 6\), despite the fact that both in the suspended condition of the cage and when the cage is on the catches the vertical posts are, as it were, unloaded from forces. These posts have their own definite task: to take the shocks when being set down on the catches.

As for the diagonals, they are theoretically useless, but practically necessary for greater rigidity of the cage posts. Their transverse-section dimension is taken smaller than that of the inclined posts, but such that riveting work can be carried out. Accordingly, one may settle on a profile \(40 \times 40 \times 5\), with a weight per running meter of 3 kg.

For the horizontal bars the section has been calculated earlier, and they will be suitable both for the short sides and for the long sides of the frame, i.e. \(80 \times 47 \times 8\).

Thus, we can now calculate the weight of our cage:

4 posts, each \(5\ \mathrm{mtr}\) long \(\times 8.6\ \mathrm{kg}\)	. . . .	172 kg.
4 posts, each \(4.8 \times 8.6\ \mathrm{kg}\)	. . . .	165 »
Diagonals \(11.4\ \mathrm{mtr}\) long \(\times 3\ \mathrm{kg}\)	. . . .	35 »
Total	. . . .	372 kg.

The form or construction of the cage may be somewhat modified, keeping the posts inclined, but not allowing them to converge along the height of the cage into a single point. This will be only a variation of the previous construction. Suppose, for example, that the ends of the inclined posts form, at the level of point \(A\) (Fig. 12), a square. The cage can be suspended by the protruding ends of these inclined posts. A cage of this type has been built at the Daldubsh mine. It holds four mine cars, arranged two on one deck and one behind the other.

(See Die Entwickelung des Niederrheinisch Westfälischen Steinkohlen Bergbaues, Bd. V. S. 302 Fig 210). The displacement of the cage suspension points to the side from the line of symmetry (vertical) makes it necessary to strengthen the section of the inclined posts. If, for example, the suspension points form a square with a side of \(600\ \mathrm{mm}\), then, when drawing up the suspension sketch, it will not be difficult, from purely geometrical considerations, to calculate the new section of the inclined post. Let us choose a channel-iron section \(80 \times 47 \times 8\), with a weight per running metre of \(10.11\ \mathrm{kg}\).

As for the vertical posts, their section may be reduced, for the inclined posts will now take up a more considerable share of the load from shaking and jolts. For the vertical posts one may take a channel-iron section of dimensions \(50 \times 25 \times 6\), with a weight per running metre of \(4.14\ \mathrm{kg}\) (Fig. 13).

For greater rigidity of the cage body, diagonals may be installed if the guides are placed at the side; if, however, the guides are placed frontally, it is better to replace the diagonals by a solid axial bar, as is done in the sketch (Fig. 13). For this bar one may take channel iron with the same section as that of the vertical posts, i.e. \(50 \times 25 \times 6\), with a weight per running metre of \(4.14\ \mathrm{kg}\).

In that case the principal weight of the cage frame will be as follows:

4 inclined posts, each \(5\ \mathrm{mtr}\) long \(\times 10.11\ \mathrm{kg}\)	. . . .	202 kg.
6 vertical bars, each \(4.8\ \mathrm{mtr}\) long \(\times 4.14\)	. . . .	120 »
Total	. . . .	322 kg.

Let us consider another case, when the cage wall has an axial post and diagonals. This type of cage is very convenient for carrying out all systems of guides and for the arrangement of the central suspension of the cage. The cages prove sufficiently strong and elastic, since all the additional reinforcement of the cage is arranged along the axial post. The corner posts play only a secondary role and theoretically are superfluous.

Let the cage again be suspended on a rope, and let it weigh \(6200\ \mathrm{kg}\). The cage has four decks, as is shown in the diagram (Fig. 14). At point \(A\) there will act a force equal to half the weight of the loaded cage, i.e. \(3100\ \mathrm{kg}\). We distribute this force uniformly among the joints located not only along the outer contour (\(K, E, H, K, M, J, G\) and \(D\)), but also among the inner joints (\(C, F, J\) and \(L\)). Then in each of these twelve joints there will act a force

\[ 3100 : 12 = 258. \]

All these forces will act vertically downward, and since the whole system is in equilibrium, a force equal in magnitude to the sum of the nodal forces, but directed vertically upward, must be applied at point \(A\). This force will represent the resistance of the rope.

Denote the stress of member \(AB\) by \(a\), the stress of member \(AC\) by \(b\), that of member \(BC\) by \(d\), etc., as shown (in Fig. 14). Then somewhere to the side let us draw a vertical line (in the drawing to the right of Fig. 14) from point 1 and lay off on it, to some scale, the forces acting at the joints. We shall then obtain twelve segments, whose sum is \(3100\) kg. If through the end of the first segment we draw a horizontal line (parallel to \(BC\)), and through point 1 a line parallel to \(AB\), we obtain a closed triangle, the sides of which are marked in the drawing by \(a\), \(d\), and \(2\).

Through the point of intersection of \(a\) and \(d\) we draw a vertical line, and through point \(0\) a line parallel to member \(AD\). We then obtain an isosceles trapezoid, whose lower base is the line \(1—0\), and whose sides are the lines \(a\) and \(c\). Its upper, smaller base will represent, on the adopted scale, the stress of member \(AC\), denoted by \(b = 2584\), or, in round numbers, \(2600\) kg.

At point \(A\) four forces meet. In the diagram on the right these forces form a closed quadrilateral. At point \(B\) three forces meet. In the diagram on the right these forces form a closed triangle, etc. All the stresses are written above each member, with the corresponding sign indicating tension or compression of the bar.

Let the cage be placed on brackets located under the lower floor of the cage.

Again we distribute half the weight of the cage among the twelve joints, as shown (in Fig. 15). The diagonals, which are always in tension, and the horizontal beams undergo the same stresses that were found when the cage was suspended on the cable. As for the central post, it experiences a different stress.

At joint 1 there will act a force \(3100 - 258 = 2842\), or, in round numbers, \(2840\) kg. The member \(r\) is subjected to this force (a compressed member). On beam \(l\) there will act a compressive force \(2840 - 3 \cdot 260 = 2060\); on beam \(g\) there will act a force \(2060 - 258 \cdot 4 = 1028 = \sim 1030\) kg.

Beam \(b\) will have a compressive force of \(515\) kg.

Applying Ritter’s method, cutting the lattice system into parts, we obtain the same result. Since in the second case (the cage rests on brackets) the central post is subjected to a greater action, equal to \(2840\) kg, the calculation of its cross-section must be based on it.

By Euler’s formula we write

\[ I_{\min}=\frac{20 \cdot 2840 \times 1200^{2}}{1 \cdot \pi^{2} \cdot 20000}=414.370. \]

If channel iron with profile \(120 \times 55 \times 7\) is taken, its minimum moment of inertia will be

\[ I_{\min}=432.000 \]

The cross-section of this iron is \(1504\ \mathrm{mm}^{2}\). The weight per running meter is \(13.3\) kg.

If we now allow a load of \(3\) kg per square millimeter for the diagonals, then the cross-section of the diagonals will be

\[ 300 : 3 = 100\ \text{sq. mm.} \]

For purely structural and assembly considerations, it is necessary to take channel iron with profile \(50 \times 25 \times 6\), whose cross-sectional area is \(528\ \mathrm{mm}^{2}\), and whose weight per running meter is \(4.14\) kg.

For the frame we take channel iron with profile \(80 \times 47 \times 8\). The role of the outer posts is to support additional parts of the cage, such as doors, etc. For these posts the same channel iron is taken as for the diagonals, i.e. with profile \(50 \times 25 \times 6\).

Thus, walls without frames will give the following weight:

\[ \begin{array}{rllr} 2 & \text{axial uprights, each 5 meters long} \times 13.3 &= \ldots & 133\ \text{kg} \\ 4 & \text{corner uprights, each 4.8 long} \times 4.14 &= \ldots & 80\ \text{''} \\ 16 & \text{diagonals, each 1.3 meters long} \times 4.14 &= \ldots & 86\ \text{''} \\ \hline & \text{Total} && 299\ \text{kg} \end{array} \]

Of particular interest, on account of its simplicity of design, strength, and lightness, is the cage shown (in Figs. 6–7) (side, front, and plan views). This design has parallel uprights but has no diagonals. The cage may be suspended either on four chains, or on six, or on two. With this design any type of guides may be used. In deep vertical shafts, when the cage has large dimensions, diagonals are installed for greater rigidity of the structure. In inclined shafts diagonals are not installed. Diagonals are also not installed when on each deck only one mine car is placed, even though the cage is multideck.

The presence of an axial upright considerably reduces the deflection of the frame. Thus, when only one mine car is placed on each deck, there is no need to lay transverse beams under the rails.

If the deck of the cage is able to accommodate two mine cars arranged one behind the other, then the transverse beam under the rails is installed precisely in the plane of the middle uprights.

Let the cage be suspended on a rope by the ends of the four outer uprights. If the weight of the cage remains 6200 kg, then each upright is subjected to a force of 1550 kg. When the cage is being driven, the cage suspension system on four chains can be replaced by a central suspension by the ends of the two axial uprights, and the former system left as a reserve. The load on each central upright will be 3100 kg.

If the cage rests on catches, then for this case the following stresses were found: the outer uprights are compressed by a force of 1298 kg, while the middle upright is subjected to tension by a force of 120 kg.

The choice of profile for the uprights presents no difficulty.

Let us take the outer uprights. For them one may use Euler’s formula and thus determine the minimum moment of inertia.

\[ I_{\min}=\frac{20 \times 1298 \times 1200^2}{1 \cdot \pi^2 \cdot 20{,}000}=189389 \]

If channel iron with the profile \(80 \times 47 \times 8\) is taken, then for it \(I_{\min}=275079\), which is quite sufficient. The weight of one running meter of such iron is 10.11 kg.

For selecting the profile of the axial uprights, let us assume a load of 3 kg per sq. mm. Since the greatest action to which this upright is subjected is 3100 kg, the cross-sectional area must be \(3100:3=1033\) sq. mm. If channel iron with the profile \(80 \times 45 \times 6\) is taken, its cross-sectional area will be 1100 sq. mm, and the weight of one running meter is 8.6 kg.

Thus, the basic weight of the uprights consists of

\[ \begin{array}{rlr} 4\ \text{uprights, each }5\ \text{mtr} \times 10.11\ \text{kg} &=& 202\ \text{kg} \\ 2\ \text{uprights, each }5\ \text{mtr} \times 8.6\ \text{kg} &=& 86 \\ \hline \text{Total} && 288\ \text{kg} \end{array} \]

An extremely important aspect in the construction of cages is the installation of the uprights. In fact, these parts are subjected to large compressive forces not only of a static nature, but also of a dynamic one. In addition, the uprights are subjected to longitudinal bending.

Most of the cages at the Fleyemal mine were made of strip iron. And in the appearance of the uprights of these cages the traces of deformation were very clearly manifested. In the plane of least resistance of the posts to longitudinal bending, the sinusoidal form of bending could be seen quite distinctly.

In addition, during operation the connection between the posts and the frame was constantly weakening, so that the weakened connections had often to be tightened.

The connection of the cage posts with the frame is made either by a bolted joint or by a riveted one. In both cases it is necessary here to use fastening parts of such dimensions that their cross-section offers proper resistance to shearing.

The greatest shearing force in the present case is in the form of the reaction of the cage upon the post when the cage is set on the dogs. For a cage with vertical posts without diagonals this reaction was equal to 1297.6 kg mm, or 1298 kg. If a load of 2.4 kg is allowed per square millimeter, then the necessary section resisting the shearing forces will be

\[ 1298 : 2.4 = 541 \text{ sq. mm.} \]

The riveted joint is made as shown in Fig. 16. One rivet, placed in the center, has a diameter of 20 mm and directly connects the post with the frame. Four rivets with a diameter of 16 mm are intended for connecting the clamp, which embraces the frame, to the post. The shearing area will thus be equal to \(\sim 1110\) sq. mm.

\[ \Omega = \frac{\pi}{4}(20^2 + 4 \times 16^2) = \sim 1110 \text{ sq. mm.} \]

The outer posts are bent at the lower frame in such a way that they constitute a supporting surface for the frame.

As regards the upper frame, it does not experience sharp impacts when the frame is set on the dogs. Only when material of great length is lowered—wooden beams, rails—does this frame experience a comparatively small pressure from this load, placed on the roof of the cage and tied to the rope. For fastening the frame to the posts here, rivets of 18 mm diameter are used, 2 in number; the diameter of 20 millimeters is retained only at the central part of the fastening.

The shearing force at the central post is equal to 120 kilograms. Here it would be sufficient to place one rivet 20 millimeters in diameter, but the fastening here is made analogously with the neighboring fastenings.

Thus, we have

30 rivets of 20 millimeters diameter.
12 rivets of 18 millimeters diameter.
84 rivets of 16 millimeters diameter.

Weight of rivets . . . 15 kg.
30 straps weighing . . . . . . . . . 70 kg.
Total . . . 85 kg.

For good assembly of the cage it is necessary to take care to provide a press for good riveting. It is not recommended to punch holes for rivets, but to drill them; nor to use rivets of excessive length, otherwise a strong riveted joint will not be obtained. The holes are drilled one millimeter smaller in diameter than the rivet shank. When the parts of the cage are ready for assembly, accurately marked out and fitted, the preliminary assembly of the parts is carried out with the aid of bolts, placing the base of the cage on a firm platform set level. With the aid of a plumb line the positions of the vertical …

of the uprights, while a level is used to check the parallelism of the horizontal platforms attached to the uprights.

When all this has been done, the holes for the rivets are drilled through on site to the exact diameter; in this process inaccuracies in the centering of the rivets are corrected.

Since the uprights are located along the periphery of the frames, it is convenient to fasten them to the latter by means of forged braces \(80 \times 80 \times 8\), for which the very best cast steel is used. These braces are turned with their inner angle toward the interior of the cage; with their horizontal flange they stand on the upper flange of the frame, and with their vertical flange they adjoin the widest side of the upright, back to back. Having described the principal parts of the cage from the constructional point of view, and having shown how these parts may be approximately calculated from the point of view of statics, on the basis of the hypothesis of the distribution of static forces over the nodes, it is now necessary to show how to complete the construction of the cage and what additions this cage requires in order that it may be put into operation.

Among such additions, or accessories, the following parts must be included:

1. Sheet iron for the floors.
2. Rails for the mine cars.
3. Stops to prevent the cars from rolling inside the cage.
4. Rails.
5. Doors.
6. Finishing.
7. Bearing surfaces of the cage for setting it on the dogs.
8. Springs for suspending the cage.
9. Guide brackets.
10. Parachutes.

All these additional parts are necessary for completing the construction of the cage, and the absence of any one of them would affect the operation of the cage and the convenience of using it, as well as the durability of its service.

The question of parachutes is intended to be treated in a separate article; therefore we shall deal only with the remaining parts.

The floors of the cage are made of sheet, corrugated steel 5 mm thick and weighing about 42 kilograms per square meter. Such steel is quite sufficient to resist the load of the cage and lasts as long as the cage itself. The roof of the cage is usually made in the form of a vault with a small rise, and for rigidity thin and high angles are attached to the sheets of the vault, arranged along the edges of the roof. If it is intended to lower material on the roof, it is covered with plates of flat woven rope, which considerably protects the cage from shocks when loading materials. Such a woven covering is also convenient for accommodating working personnel when the shaft of the mine is inspected at low speed.

The floor sheets are connected to the cage frame either rigidly or by inserting them into slots. Such removable floors are necessary when lowering tall and long objects that do not fit within a single deck.

Rails for convenient rolling of cars into the cage may be of the mine-rail type, or angle or square iron may be used. If one car is placed on a deck, then the rail may be considered as a beam resting on two supports and with fixed ends. A quite sufficient section is, for example, \(75 \times 50 \times 14\) mm. In any case, calculating such a beam presents no difficulty.

If two cars are placed on a deck, arranged one behind the other when viewed from the loading point, then in the middle of the cage, perpendicular to the direction of the track, a transverse beam has to be laid. In such a

In this case each rail may be regarded as a beam on three supports, and in its calculation Clapeyron’s formula may be used.

The calculation of such beams is given in my article on the beam on three supports and its application to the solution of practical problems.

Turning to the question of the installation in the cage of devices for holding mine cars in position, it must be noted that the immobility of the cars in the cage during travel has a favorable effect both on the integrity of the cage and on the strength of the hoisting rope. A massive car, not held in place by anything, may easily roll sideways, knock out the comparatively lightly arranged door, and thereby jam the cage; if conductors are arranged at the front of the cage, they may suffer in such an impact. But all devices for holding cars in place must be so designed that, while performing their role distinctly and reliably, they in no case cause delay in the operation of the cage, nor serve as an obstacle when workers are being lowered. There are stops, manual, foot-operated, and automatic. According to the design of the stops they may be divided into two classes: stops with brackets and stops with sliding bolts (toggle system and latch system). An idea of these stops may be obtained from the schematic drawings placed in the collective work edited by H. Bansen and K. Teiwes, entitled Die Bergwerks-Maschinen (Springer edition, Berlin). In Drawing 17 the design of a stop of the latch type is given. Here there is a pivoting lever a, rotating about point O and with one end entering an opening pierced in rod b. Rod b is guided in its motion by bearings c—c. At the left end of rod b there is a head d, freely entering the slot of the angle iron that serves as the rail in the cage. The flat head d projects considerably above the vertical wall of angle n and thus serves as an obstacle to the rolling of the car wheels. By pressing with the foot on the long arm of the lever, the movement of head d and rod b to one side or the other is achieved. Angle c, with the curved outline of its vertical wall, serves as a stop against the movement of lever a clockwise.

In intensive work for issuing the mineral, more complex systems for rolling cars in and out, of a purely automatic character, are used, and the floors at the cages are made inclined. To accelerate the lowering and raising of workers, ready-made platforms could conveniently be used, by means of which a group of workers capable of fitting into the cage can be supplied at one time. The unloading of the cage could be carried out alternately, now from one side, now from the other. The boarding of the working personnel should be the same.

For the construction of the little doors either mesh or sheet steel 4 mm thick is used. The doors must in no case restrict the working floor area of the cage. When doors made of sheet steel are used, holes 20 mm in diameter are drilled in the sheets. In this way the cage is open to air.

Sheet steel 3 mm thick is also used for sheathing the remaining sides of the cage. Holes 20 mm in diameter are likewise drilled in this sheathing, for access of air and to lighten the weight of the cage. The spacing between the centers of the holes is made equal to 30 mm. The weight of such sheets is reduced by approximately one half.

To protect that part of the frame which comes into direct contact with the ropes from inevitable wear, pieces of angle iron are riveted to the frame. It would not be useless to use a gasket between the frame and this angle made of such materials as rubber or leather. Impacts when the cage is seated would thereby be damped to a considerable extent. Drawing 18 shows the part of the cam projecting on the right and the cage angle at the part where it seats on the cam. The places that are ... are a very responsible part of the cage, ...

...by which the cage is suspended from the suspension ropes. Holes are bored in the ends of the posts, through which there must be passed a pin or bolt serving to connect the flexible part of the suspension with the cage. To protect this part of the posts from rapid wear, massive cast or forged plates are used in these places, with a widened head, of the same thickness as the internal height of the web of the channel iron. From the round head, with cut vertical sides, the tail part of the plate runs downward, tapering almost to nothing. To fasten the plate shown in section on the right near Fig. 6, four rivets of smaller diameter, 18 mm, and one rivet 20 mm in diameter are used. The pin of the plate is calculated for shear with a large margin of strength.

The cage guides have more or less established dimensions, determined partly by the dimensions of the conductor they embrace and by the condition that they be securely fastened by rivets to the frame of the cage. An indispensable condition for a rational design of such guides is the absence of projections by which these guides might cut into the body of the conductor, whether it be wooden or metal. Fig. 19 shows such guides, used in Belgian shafts.

What a rational design of a cage can yield as a result, when proper attention is paid by the designer to each of its parts—that is, when the calculation takes into account all unfavorable working conditions—can be seen from the following.

If it is assumed that the normal weight of coal placed in a mine wagonette is \(N = 550\) kg, and if \(Q\), in kg, is the weight of the cage, then, according to the data available in the extensive monograph on the Westphalian hard-coal industry, the following table can be compiled.

Designations: Cage	Designations: Weight	Designations: Shaft	\(Q\)	\(\dfrac{Q}{N}\%\)
One wagonette.	Average weight.	—	650	118
Two wagonettes.	Average weight.	—	2000	181
Two wagonettes.	Lighter.	Möller-Scherbeck	1400	127
4 wagonettes.	General average.	—	3000	136
2 wagonettes side by side.	Average.	—	2600	117
2 wagonettes side by side.	Lighter.	Dalbusch III, Nornen I.	2000	91
2 wagonettes one behind another.	—	Konstantin II.	2250	102
4 decks.	Heaviest.	Prinzregent.	3250	140
4 decks.	Lighter.	Polfornem III.	2390	105
4 decks.	Heavier.	Kaiserstuhl.	3658	166
6 wagonettes.	Lighter.	Wilhelmin Viktoria I.	3500	106
6 wagonettes.	Heavier.	Deutscher Kaiser.	5610	170
8 wagonettes.	Lighter.	Consolidation.	3980	90
8 wagonettes.	Average.	—	4500	102
8 wagonettes.	Heavier.	Schlagel und Eisen.	6650	152
8 wagonettes.	Heavier.	—	—	—
1 wagonette.	According to Dessauer.	Maria.	195	36
4 wagonettes.	—	—	1440	66
4 wagonettes.	—	—	1640	75

Approximately twenty years have passed since the appearance of the collective work of German engineers on the Westphalian hard-coal industry. It is therefore interesting to take more recent data on the weight of cages built by German machine-building plants and compare them with the former data.

I take later information from the “Maschenen-bau, Aktien-Gesellschaft Tigler” works (Duisburg—Meiderich, Rhine). This works has produced a large number of cages for the needs of mining.

Number of cages	Number of decks	Weight \(Q\), kg	Weight \(N\), kg	\(\dfrac{Q}{N}\), %
32	1	25500	550	125
30	2	85600	2800	130
47	3	235000	3300	150
46	4	239000	4400	120

As can be seen, no tendency toward lightening the cage in its weight is being produced. The ratio \(Q:N\) still far exceeds the results obtained by engineer Dessar. The very same conclusion will be obtained if one turns to the data of K. Teiwes, placed in the third volume of the work: “Die Bergwerksmaschinen” (Die Schachtfördermaschinen, p. 11).

Here an interesting table is given, in which the individual weights of the cage, mine cars, payload, total dead load, the ratio of the weight of the cage to the payload, the ratio of the weight of the dead load to the payload, and the number of people raised by the cage at one time are considered in greater detail:

Number of mine cars	Number of decks and arrangement of mine cars	\(G\), weight of cage	\(W\), weight of mine cars	\(N\), payload	\(L = G + W\)	\(G:N\), %	\(L:N\), %	Number of people lowered at one time
1	—	650	300	550	950	120	170	—
2	—	2000	600	1100	2600	180	240	—
4	2; mine cars side by side.	2500	1200	2200	3700	110	170	—
4	2; mine cars one behind another.	3000	1200	2200	4200	150	190	—
4	4	3000	1200	2200	4200	150	190	—
6	3	4500	1800	3300	6300	140	190	30
8	4	6000	2400	4400	8400	140	190	40–50

Here the ratio \(G:N\) is still more unfavorable than that given by the Tigler works. Thus, the construction of the cage calculated by Dessar’s method deserves the most serious attention.

If, further, attention is paid to how the weight is distributed in Dessar’s cage among the individual items, then we see the following: with a cage weight of 1440 kg, 827 goes to the additional appurtenances, while the frame, uprights, etc., weigh only 613 kg, which amounts to 47% of the total weight of the cage. The task of further reducing the weight of the cage, both by reducing the weight of its appurtenances and by reducing the weight of the frames, uprights, etc., is next in line, and its successful execution could only be welcomed.

Here it is necessary to pay attention to the possibility of applying to the cage materials which, although more expensive, nevertheless possess greater resistance to mechanical actions.

Dessar approaches this question cautiously, pointing to nickel steel, which has a tensile strength of 85 kg/mm² and an elongation of 14%. Calculation suggests that, using such steel, it would be possible to make a cage for a shaft 1500 m deep with a total weight of only 1000 kg. The weight of the rope would then be only 6750, and the moment of resistance only 23000 kg·m.

We boldly join in this conclusion. For a whole series of years we have already pointed out to the cadre of future workers in mining the extreme necessity of taking up the question of replacing the former construction materials for the cage with new ones. Among such materials we include aluminum alloys, which were mentioned briefly at the beginning of the article.

The practice of aeronautics shows that the use of these new materials in every kind of construction work can yield results favorable to the highest degree.

One hundred years ago the chemist Wöhler first obtained aluminum; seventy-five years have passed since Sainte-Claire Deville obtained aluminum in a quantity sufficient for laboratory investigations.

If in 1860 the price of one kilogram of aluminum reached 300 francs, then already in 1908 it had decreased 150-fold.

Quite recently a cheerful thought was expressed in the technical literature, that we must enter the aluminum age. The mining engineer Alfred Wilm, who after many years of experiments gave humanity duralumin, practically realized the aspirations of technicians. Possessing considerable tensile strength, sufficient hardness and ductility, duralumin well resists the influence of the atmosphere, dampness, the action of nitric and sulfuric acid and their vapors, mercury vapor, and only yields noticeably to the action of hydrochloric acid and alkalis.

There is no reason to fear that duralumin is still expensive. Demand will increase, and its price will be reduced.

In conclusion to the question of the construction of the cage and its accessories, we must point out that, by some strange misunderstanding, cages are still lowered and raised unilluminated. And yet it would be possible to prepare a hoisting rope with a conductor inside and use this conductor to supply light energy.

This problem is on the agenda and must be solved by mankind. Let us illuminate the path of our working life in every respect; by doing so we shall only be fulfilling our technical duty.

2. Dynamic Calculation of a Single-Story Cage

Dynamic approaches to the solution of problems of a practical character are gaining more and more adherents, despite the fact that the use of dynamics considerably complicates the solution of questions. The reason for this turn lies in the fact that we have ceased to be satisfied with small velocities and small masses; and in such a case statics is not able to give satisfactory results, and it becomes necessary to introduce correction coefficients into the calculations, which in the end diminishes the value of scientific conclusions to the limits of “rule-of-thumb calculation” and quite naturally brings “practice,” rather than science, to the fore. Practitioners proudly like to repeat: “I built this and that”; but if one examines the matter in substance, then the whole construction amounts to a repetition of what has already existed and what undoubtedly possesses shortcomings. Shortcomings must be taken into account, not repeated. In this lies progress.

Professor S. A. Zaborovsky of the Ekaterinoslav Mining Institute was the first to take up the question of the dynamic calculation of the cage; he wrote a dis—

... dissertation entitled “Dynamic Calculation of a Two-Deck Mine Cage.” He based the calculation on the so-called potential method in structural mechanics. According to this theory, the potential energy possessed by an elastic system at the greatest deformation is equal to the kinetic energy of the masses contained in it, combined with the work of all the acting forces over the course of the deformation.

Example: Let the hoisting device be a rod \(AB\), suspended at its upper end from a rope, and below carrying a mass \(P/g\). At the moment of descent the upper end of the rod strikes a stop suspended by means of the rod \(BC\), with point \(C\) completely fixed (Fig. 1).

Denote by:

\(F\)—the cross section of the rods \(AB\) and \(BC\),

\(l\)—the length of the rod \(AB\),

\(G\)—the weight of this rod,

\(n\)—the length of the rod \(BC\),

\(V_0\)—the velocity of the cage’s impact against the stop,

\(W\)—the acceleration during impact.

Concentrate the mass \((P+G):g\) at point \(A\).

At the moment of impact a force will act on point \(B\):

\[ \frac{(P+G)(g-w)}{g}=Q'. \]

This force \(Q'\) will produce a certain tension in the rope. But the same force will also act from the side of the rope on the cage, only in the opposite direction. In order to change the magnitude of the rope’s action on the cage, it is necessary to change the elongation of the rope, for example by unwinding the rope from the drum.

It is possible to calculate this change in elongation of the rope for the time interval from the moment of impact to the moment of greatest deformation.

Let the depth of the shaft be 400 meters, the breaking stress \(60\ \mathrm{kg/mm^2}\), and the stress in the rope \(8\ \mathrm{kg/mm^2}\). In that case the elongation will be

\[ \varepsilon = 8:20000 = 0.0004. \]

The elongation of the rope will be

\[ \Delta H = 400 \cdot 1000 \cdot 0.0004 = 160\ \mathrm{mm} = 16\ \mathrm{cm}. \]

If the cage itself weighs \(800\ \mathrm{kg}\), the weight of the two cars is \(2200\ \mathrm{kg}\), and the length

\[ AB = BC = l+n = 400\ \mathrm{cm}, \]

then, taking the greatest stress in the rod \(\sigma\) equal to three quarters of the elastic limit, one can find the elongation,

\[ \varepsilon=\frac{\sigma}{E} =\frac{{}^{3}\!/_{4}\cdot 2000}{2000.000} =0.00075, \]

The elongation of the rod \(AC\) will be

\[ \lambda = (l+n)\varepsilon = 400 \cdot 0.00075 = 0.30\ \mathrm{cm}. \]

We shall now write the equation for the rectilinear motion of a heavy point oscillating under the influence of a central elastic force.

In general form this equation is written as

\[ m\frac{d^2 x}{dt^2}=-K \]

Here \(K\) is the elastic force that causes the mass \(m\) to be attracted to the center of oscillation.

Using the well-known formula of strength of materials,

\[ \lambda=\frac{Pl}{E\omega}, \]

one may write that

\[ P=\frac{\lambda}{l}E\omega . \]

In our case this expression is rewritten in the following form:

\[ m\frac{d^{2}x}{dt^{2}}=-\varepsilon E\cdot F=-E\cdot F\cdot\frac{\lambda}{l+n}. \]

Hence we have

\[ \frac{d^{2}x}{dt^{2}}=-\frac{EF\cdot g\cdot\lambda}{(l+n)(P+G)}=-K\lambda . \]

This differential equation is quite analogous to the differential equation of motion of a mathematical pendulum executing small oscillations. For such a pendulum, the time of one oscillation will be

\[ t=\pi\sqrt{\frac{l}{g}} . \]

Applied to the case under consideration, the time of oscillation will be

\[ t=\pi\sqrt{\frac{1}{K}} . \]

Let us now consider the rod—the stand—as a system consisting of two parts: \(AB\) and \(BC\).

Under the action of the applied forces, the rod \(AB\) will elongate by \(\lambda_1\), and the second by \(\lambda_2\). These elongations will occur under the action of the forces \(R_1\) and \(R_2\).

It is not difficult to see that

\[ R_1=\sigma_{\max}\cdot F \]

and the elongation

\[ \lambda_1=\frac{R_1l}{EF}=\frac{\sigma_{\max}\cdot l\cdot F}{EF}=\frac{\sigma_{\max}l}{E}; \]

on the other hand,

\[ R_2=R_1-Q'=\sigma_{\max}F-\frac{(P+G)(g-w)}{g} \]

and the corresponding elongation will be

\[ \lambda_2=\frac{\sigma_{\max}\cdot n}{E}-\frac{(P+G)(g-w)n}{EFg}. \]

Now the expression for the potential energy of the stretched stand is found without difficulty, and it will be equal to

\[ \frac{1}{2}(R_1\lambda_1+R_2\lambda_2) = \frac{1}{2}\cdot \frac{\sigma_{\max}\cdot F\cdot\sigma_{\max}l}{E} + \]

\[ \frac{1}{2} \left\{ \sigma_{\max}\cdot F - \frac{(P+G)(g-w)}{g} \left[ \frac{\sigma_{\max}n}{E} - \frac{(P+G)(g-w)}{EFg} \right] \right\} = \]

\[ = \frac{\sigma_{\max}^{2} F l}{2E} + \frac{\sigma_{\max}^{2} F n}{2E} - \frac{\sigma_{\max}(P+G)(g-w)n}{2gE} \]

\[ - \frac{\sigma_{\max}(P+G)(g-w)n}{2gE} + \frac{(P+G)^2(g-w)n^2}{2EFg^2} = \frac{\sigma_{\max}^{2}F}{2E}(l+n) \]

\[ - \frac{\sigma_{\max}(P+G)(g-w)n}{gF} + \frac{(P+G)^2(g-w)^2 n}{2EFg^2} \]

This potential energy is numerically equal to the kinetic energy and to the work of all the acting forces during the deformation.

If the speed of seating of the cage on the cams is denoted by \(V_0\), then the kinetic energy and the work of the forces will take the following form,

\[ \frac{mV_0^2}{2} = (P+G)(\lambda_1+\lambda_2) - (P+G)\left(1-\frac{w}{g}\right)\lambda_2 + \frac{1}{2}(P+G)\left(1-\frac{w}{g}\right)\lambda_0 \]

In this expression \(\lambda_0\) denotes the elongation which the load \(AB\) has before the impact, being subjected to the force

\[ (P+G)\left(1-\frac{w}{g}\right). \]

If \(P+G\) is replaced by \(Q\), and the mass \(m\) by \(\dfrac{Q}{g}\), then we obtain the following equation.

\[ \frac{F(1+n)}{2E}\sigma_{\max}^{2} - \frac{Q(g-w)n}{gE} + \frac{Q^2(g-w)^2 n}{2EFg^2} = \frac{QV_0^2}{2g} + \]

\[ + Q\left[ \frac{\sigma_{\max}l}{E} + \frac{\sigma_{\max}n}{E} - \frac{Q(g-w)n}{EFg} \right] - \frac{Q(g-w)}{g}\left[ \frac{\sigma_{\max}n}{E} - \frac{Q(g-w)n}{EFg} \right] + \]

\[ + \frac{Q(g-w)}{2g}\cdot \frac{Q(g-w)}{g}\cdot \frac{1}{EF}. \]

If we take into account that the speed of seating of the cage on the cams is generally not significant, then its change per unit time, i.e. the acceleration, will also not be large in comparison with the acceleration due to gravity. This consideration makes it possible, for the sake of simplifying the conclusions, to discard, in the fraction \(\dfrac{(g-w)^2}{g^2}\), the term containing \(\dfrac{w^2}{g^2}\). Then we obtain the following quadratic equation with respect to \(\sigma_{\max}\).

\[ \frac{F(1+n)}{2E}\sigma_{\max}^{2} - \frac{Q(1+n)}{E}\sigma_{\max} - \frac{QV_0^2}{2g} + \frac{Q^2}{2EF}\left[n-l\left(1-\frac{2w}{g}\right)\right] = 0 \]

Solving this equation, we obtain the value of \(\sigma_{\max}\)

\[ \sigma_{\max} = \frac{Q}{F} + \sqrt{ \frac{Q^2}{F^2}\cdot \frac{2l}{1+n}\left(1-\frac{w}{g}\right) + \frac{QV_0^2E}{gF(1+n)} } \]

For practice, the question of ensuring that \(\sigma_{\max}\) does not reach the elastic limit of the material is of great importance. Therefore we assume that

\[ \sigma_{\max}=\frac{3}{4}\cdot 2000 = 1500\ \mathrm{kg/cm^2}. \]

Taking this standard as our basis, we shall use the same quadratic equation and solve it with respect to \(F\).

Thus, we have

\[ \frac{(1+n)}{2E}\sigma_{\max}^{2}F^{2} - Q\left[ \frac{(1+n)\sigma_{\max}}{E} + \frac{V_0^{2}}{2g} \right]F + Q^{2} \frac{\left[ n-1\left(1-\frac{2w}{g}\right) \right]}{2E} =0 \]

Solving this equation, we obtain

\[ F= \frac{Q}{\sigma_{\max}} \left\{ \frac{V_0^{2}E}{2g(1+n)\sigma_{\max}} +1 \pm \sqrt{ \left[ \frac{V_0^{2}E}{2g\sigma_{\max}(1+n)} +1 \right]^{2} - \frac{n-1\left(1-\frac{2w}{g}\right)}{1+n} } \right\}. \]

If the radical expression is greater than zero, then both roots are real and positive. Thus, it must be that

\[ \frac{V_0^{2}E}{2g(1+n)\sigma_{\max}} +1 > \sqrt{ \frac{n-1\left(1-\frac{2w}{g}\right)}{1+n} }. \]

In the radical expression the second term is small in its numerical value in comparison with the first; and therefore, without great error, this term may be neglected. Then we obtain a considerably simplified expression for the area

\[ F= \frac{Q}{\sigma_{\max}} \left[ \frac{V_0^{2}E}{g\sigma_{\max}(1+n)} +2 \right]. \]

This formula will be used to determine \(F\).

For our simplest example we find

\[ F= \frac{3000}{1500} \left[ \frac{2500\cdot 2{,}000{,}000}{981\cdot 1500\cdot 400} +2 \right] = 2(8.5+2) = 21\ \mathrm{cm}^{2}. \]

Knowing \(F\), one can determine the time of one complete oscillation \(t\).

\[ t=\pi\sqrt{\frac{1}{k}} = 3.14\sqrt{ \frac{400\times 3000}{2{,}000{,}000\times 21\times 981} } = 0.0165\ \mathrm{sec}. \]

Half of this time, i.e. \(0.00825\) seconds, corresponds to the time of swing from the center of oscillation to the greatest deformation. Since the speed at which the cage lands on the dogs is allowed to be small \((V_0=0.5\ \text{meter})\), in \(0.00825\) seconds only about \(4.1\) millimeters unwind from the drum. This is only \(0.0256\) of the elongation of the rope. In view of this, it may practically be assumed that the elongation \(\Delta \Pi\) and the force \(Q'\) during \(0.00825\) sec. remain unchanged.

In turn, this proposition implies the assumption that the kinetic energy of the cage at the moment of impact is balanced by the tension of the rope, and consequently the cage falls freely with velocity \(V_0\) onto the dogs.

Let us imagine a single-deck cage consisting of two horizontal rectangular frames connected by four uprights. The entire wagonette is placed in the cage. The cross-members of the lower frame are located directly under the axles of the wagonette. Four dogs (Fig. 2) \(k\) rest on two beams \(a\), laid with their ends on absolutely rigid supports.

Let us assume the mass of the falling cage to be concentrated equally at the points of contact of the wagonette wheels with the rails.

Now let us solve the following problem: what forces \(R\) must be applied at the same points in order to produce a deformation of the cage equal to that experienced by the cage falling onto the cams with velocity \(V_0\)?

The short crosspieces of the lower rails, although they are subjected to bending, may be neglected in this bending. Then there remain the deflection \(\delta_1\) of the long side of the lower frame and the deflection \(\delta_2\) of beam \(a\), which supports the cams.

Let us determine these deflections.

For beam \(AB\) of length \(l_1\) (Fig. 3), freely resting on two supports and loaded with a concentrated force \(R_1 = 1\) at a distance \(C\) from the left support, we have the following equation for the deflection arrow

\[ \lambda_1=\frac{1}{EJ_1}\cdot \frac{C^2(l_1-C)^2}{3l_1}. \]

If the same beam is subjected to the action of the force \(R_2 = 1\) at a distance \(C\) from the right support, then the deflection arrow in the dangerous section will be

\[ \lambda_2=\frac{1}{EJ_1}\cdot \frac{C^2(l_1-C)^2}{6l_1} \left[\frac{2C}{l_1-C}+1+\frac{C^2}{(l_1-C)^2}\right]. \]

If both forces act simultaneously, the deflection will be

\[ \delta_1=R(\lambda_1+\lambda_2), \]

where

\[ R=R_1=R_2. \]

In exactly the same way we find the deflection of beam \(a\),

\[ \delta_2=R(\mu_1+\mu_2), \]

where

\[ \mu_1=\frac{d^2(l_2-d)^2}{3EJ_2l_2} \quad \text{and} \quad \mu_2=\frac{d^2(l_2-d)^2}{6EJ_2l_2} \left[\frac{2d}{l_2-d}+1+\frac{d^2}{(l_2-d)^2}\right]. \]

The point of application of each of the four forces \(R\) will travel the path

\[ \delta=R(\lambda_1+\lambda_2+\mu_1+\mu_2)=R\varepsilon. \]

And the work of the force \(4R\) will be equal to

\[ 4\cdot \frac{1}{2}\cdot R^2\rho=2\rho R^2. \]

On the other hand, this work will be measured by the sum

\[ \frac{QV_0^2}{2g}+Q\delta=\frac{QV_0^2}{2g}+Q\rho R. \]

Thus, we obtain the quadratic equation

\[ 2R^2\rho-Q\rho R-\frac{QV_0^2}{2g}=0 \]

\[ R^2-\frac{Q}{2}R-\frac{QV_0^2}{2g\rho}=0. \]

Solving this equation with respect to \(R\), we obtain

\[ R=\frac{Q}{4}\pm\sqrt{\frac{Q^2}{16}+\frac{QV_0^2}{2g\rho}}. \]

We shall of course take the larger root.

It is then possible, from the same quadratic equation, to determine the value \(V_0\), i.e. that limiting value of the velocity at which the stresses in the parts of the cage will not exceed the elastic limit:

\[ V_0 = \pm \sqrt{\frac{2\rho g R(2R-Q)}{Q}} \]

Since both values of \(V_0\) are identical in absolute magnitude, the question of the sign before the root is immaterial.

If we denote by \(\sigma\) the greatest stress in the frame allowable in the static calculation of the cage on the cams, if the cage is still under the load \(Q\), then one may write the proportion

\[ 4R : Q = \sigma_{\max} : \sigma = \alpha \]

Using the calculation equation of strength of materials, we write

\[ \sigma_{\max} = \frac{RC}{W_1} \quad \text{and} \quad \sigma = \frac{QC}{W_1} \]

Determining \(R\) and \(Q\) from these relations and substituting in the expression for \(V_0\), we obtain

\[ V_0 = \sqrt{\frac{2g\rho \cdot W_1 (2\alpha - 1)\sigma_{\max}}{C}} \]

If \(V_0\) is known, then \(\alpha\) can be determined, and consequently also \(\sigma\), i.e. the stress which may be allowed in the static calculation, with the condition that impact action at velocity \(V_0\) should not be dangerous for the strength of the cage.

Example. It is necessary to design a single-deck cage intended for raising two mine cars. Since a large cage is intended for servicing a shaft of a definite cross-section, the supporting area of the cage may also be considered known. The theoretical minimum of the internal floor area of the cage is fully determined by the dimensions of the mine cars accommodated on one deck, but usually a small clearance remains between the sides of the cars and the sides of the cage, about 10 centimeters wide. As for the height of the cage, it is chosen with the calculation that even a tall person can stand in the cage without stooping (about two meters).

These data are quite sufficient for determining the frame of the cage, which will consist of upper and lower frames, six vertical posts, and four diagonals (Fig. 4). The long walls of the cage are subjected to calculation, representing continuous through trusses. These trusses are completely identical, and identical forces act on them, obtained from the sum of the weight of the cage itself and the load. The load of the cage may be considered a known quantity, because a mine car of definite dimensions has a definite weight and accommodates a definite useful load whose weight is known. Let us suppose that a mine car with its load weighs 1600 kg. The weight of two mine cars will be 3200 kg.

The own weight of the cage is an indeterminate quantity. Practice indicates that iron cages with all accessories weigh from 0.8 to 1.5 of the hoisted load, while steel ones weigh 0.6–0.7 of the weight of iron cages. Single-deck cages for one car weigh from 400 to 800 kg, and for two from 1000 to 2400 kg.

The machine-building plant of the joint-stock company Tigler (Duisburg—Meiderich on the Rhine) gives the following data on the weight of the cages it has built:

		Weight
26 cages in 1 story weigh	. . . . . . .	15600 kg
3 » » »	. . . . . . .	5700 »
8 » » »	. . . . . . .	12800 »
3 » » »	. . . . . . .	4800 »
Average weight of one cage	. . . . . . .	1425 kg

The first 26 cages are, evidently, cages for one car, and the last 14 for two cars. It is therefore more correct to take the average weight as 1700 kg.

Thus, one truss will carry a load of 2450 kg. Let this load act as a concentrated load at the points k—k of the beam AB, the points k—k falling exactly on the lines passing vertically through the centers of gravity.

The next simplification in the calculation is that the cams are placed exactly under the corner posts. In reality the frame is somewhat longer than the distance between the outer posts.

Let us determine analytically the force \(X'\), acting in the middle post \(C1\), and the force \(X''\), acting in each diagonal.

Let the cross-section of each post be \(F_1\), and the cross-section of the diagonals \(F_2\). Let \(J\) denote the moment of inertia of the cross-section of the frames. Let the force compressing each corner post be \(N\).

Finally, let the following be denoted:

\(M_1\)—the bending moment for beam AB for the interval Ak;
\(M_2\)—the bending moment for beam AB for the interval KC;
\(M_3\)—the bending moment at any point of the upper frame.

The force \(N\), compressing any one of the outer posts, is determined without difficulty.

\[ N=-\frac{1}{2}\left(X'+2X''\sin\alpha\right), \]

where \(\alpha\) is the angle of inclination of the diagonal to the beam AB.

The moments are also found simply:

\[ M_1=\left(K-\frac{X'}{2}-X''\sin\alpha\right)x; \]

\[ M_2=M_1-K(x-a)=Ka-\left(\frac{X'}{2}+X''\sin\alpha\right)x;\qquad M_3=\frac{X'x}{2} \]

Let us apply to our truss the principle of Structural Mechanics concerning the “principle of least work.”

By virtue of this principle, “the partial derivatives of the work of deformation of the entire system with respect to each unknown force sought must be equal to zero.” Since the unknown forces are \(X'\) and \(X''\), it is necessary to write two equations of the following type

\[ 2\left( \frac{1}{EJ}\int_{0}^{a} M_1\frac{dM_1}{dX_n}\,dx + \frac{1}{EJ}\int_{x}^{l} M_2\frac{dM_2}{dX_n}\,dx + \frac{1}{EJ}\int_{0}^{l} M_3\frac{dM_3}{dX_n}\,dx \right) + \frac{1}{EF_1}X'\frac{dX'}{dX_n}h + \frac{2}{EF_2}X''\frac{dX''}{dX_n}d + \frac{2}{EF_1}N\frac{dN}{dX_n}h =0 \]

In this equation \(h\) denotes the height of the truss, \(2l\) the length of the beams \(AB\), \(d\) the length of the diagonal, \(a\) the distance from points \(A\) and \(B\) to the points of application of the force \(K\), and \(X_n\) each of the forces sought, i.e. \(X' \ldots X''\).

Taking the partial derivatives of the moments with respect to \(X'\) and \(X''\), we obtain

\[ \frac{dM_1}{dX'}=-\frac{x}{2};\qquad \frac{dM_2}{dX'}=-\frac{x}{2};\qquad \frac{dM_3}{dX'}=\frac{x}{2}; \]

\[ \frac{dM_1}{dX''}=-x\sin\alpha;\qquad \frac{dM_2}{dX''}=-x\sin\alpha;\qquad \frac{dM_3}{dX''}=0; \]

The partial derivatives of \(N\) with respect to \(X'\) and \(X''\) will be

\[ \frac{dN}{dX'}=-\frac{1}{2};\qquad \frac{dN}{dX''}=-\sin\alpha; \]

Finally, the partial derivatives of \(X'\) and \(X''\) with respect to \(X'\) and \(X''\) give

\[ \frac{dX'}{dX'}=1;\quad \frac{dX'}{dX''}=0;\quad \frac{dX''}{dX''}=1;\quad \frac{dX''}{dX'}=0 \]

Introducing these values of the partial derivatives and the magnitudes of the moments, we give the integral equations the following form

\[ \int_0^a\left( -\frac{kx^2}{2} +\frac{X'x^2}{4} +\frac{X''\sin\alpha\,x^2}{2} \right)\,dx + \int_a^l\left( -\frac{kax}{2} +\frac{X'x^2}{4} +\frac{X''\sin\alpha\,x^2}{2} \right)\,dx + \int_0^l \frac{X'x^2}{4}\,dx + \frac{J}{2F_1}X'h + \frac{J}{F_1}\left(\frac{X'}{4}+X''\sin\alpha\right)h =0 \]

After integration and reduction of like terms, this equation can be put in the following form

\[ \left(\frac{l^3}{3J}+\frac{3h}{2F_1}\right)X' + \left(\frac{l^3}{3J}+\frac{h}{F_1}\right)\sin\alpha\,X'' = \frac{a(3l^2-a^2)k}{6J} \]

In exactly the same way we obtain the second equation

\[ \int_0^a\left( -kx^2\sin\alpha +\frac{X'x^2\sin\alpha}{2} +X''\sin^2\alpha\,x^2 \right)\,dx + \int_a^l\left( -ka\sin\alpha\,x +\frac{X'\sin\alpha\,x^2}{2} +X''\sin^2\alpha\,x^2 \right)\,dx + \frac{J}{F^2}X'' + \frac{J\sin\alpha\,h}{2F_1}X' + \frac{J\sin^2\alpha\,h}{F_1}X'' \]

After integration and reduction we obtain

\[ \left(\frac{l^3}{6J}+\frac{h}{2F_1}\right)X' + \left(\frac{l^3}{3J}+\frac{h}{F_1}+\frac{d}{F_2\sin\alpha}\right)\sin\alpha\,X'' = \frac{a(3l^2-a^2)k}{6J} \]

These two equations are the fundamental ones for the calculation of the cage. In them the known quantities are \(l\), \(h\), \(d\), the angle \(\alpha\), and the magnitude of the force \(K\). Thus the following quantities remain unknown: \(J\), \(F_1\), \(F_2\), \(X'\), and \(X''\).

If the design of a cage working in some shaft is given, then it is easy to check to what extent this design is satisfactory with respect to strength under the dynamic action of the force. It is only necessary to specify the velocity with which the cage lands on the cams.

If, however, only the cage layout is available, and the dimensions of the cross-section of the frame, posts, and diagonals are unknown, then one must find ways to determine these quantities. The problem cannot be solved at once; rather, one must gradually approach a satisfactory result.

Let us first solve the first problem.

Let a cage be given for one story, intended for lifting two mine cars. We shall take the load on the cage as equal to \(=1600 \times 2 = 3200\) kilograms. The cage’s own weight may be determined by calculation. As the basic data we shall take the following quantities:

\[ l = 124\ \text{cm.}; \qquad a = 58\ \text{cm.}; \qquad h = 176\ \text{cm.}; \qquad d = 215.3\ \text{cm.}; \]

\[ J = 925\ \text{cm.}^2; \qquad F_1 = 9\ \text{cm.}^2; \qquad F_2 = 5.6\ \text{cm.}^2. \]

We shall take the speed at which the cage seats on the catches as equal to \(= V_0 = 40\ \text{cm./sec.}\)

The cage consists of the following parts.

PARTS OF THE CAGE	Quantity in meters	Weight of one meter, kg.	Total weight in kg.
1. Upper and lower frame of channel iron \(160 \times 65\) mm.; total linear meters \(2(2.800 \times 2 + 0.900 \times 2)=\) ............	15	18.7	285
2. Cage floor made of 5 cross-members of channel iron \(120 \times 55\) mm.; linear meters \(0.900 \times 5=\) ....	4.5	13.3	60
3. Instead of rails, angle iron \(75 \times 75 \times 12\) mm.; total length \(2.800 \times 2=\) ............	5.6	13.1	74
4. Six posts of channel iron \(65 \times 42\) mm.; total length \(1.920 \times 6=\) ............	11.5	7.05	81
5. Four diagonals of strip iron \(70 \times 8\) mm.; total length \(=1.950 \times 4=\) ............	7.8	4.37	34
6. Floor, roof, and sheathing of the cage on two sides, of perforated sheet iron 4 mm thick; area \(=(2.800 \times 0.900 + 2.800 \times 1.200)2=\) ........	11.76	25	294
7. Ten gusset plates, 8 mm thick. .....	0.6	63	38
8. Springs, canopy, doors, guides, fastenings of cross-members to frames, etc. ........	»	»	300
TOTAL ........			1166

We shall round this number to 1200 kilograms.

Thus, the load on the truss will be 2200 kg. (whereas, by the approximate calculation, this load had been determined as 2450 kg., i.e. 11.4% greater).

Let us solve the integrated equations for \(X'\) and \(X''\) as functions of the quantity \(K\).

\[ \left(\frac{l^3}{3J}+\frac{3h}{2F_1}\right)X' + \left(\frac{l^3}{3J}+\frac{h}{F_1}\right)\sin \alpha\,X'' = \frac{(3l^2-a^2)a}{6J}\,K \]

\[ \left(\frac{l^3}{6J}+\frac{h}{2F_1}\right)X' + \left(\frac{l^3}{3J}+\frac{h}{F_1}+\frac{d}{F_2\sin^2\alpha}\right)\sin\alpha\,X'' = \frac{(3l^2-a^2)a}{6J}K \]

Since the right-hand sides of these equations are equal, it is simpler to subtract the first equation from the second. Then we shall be able to find the relation between \(X'\) and \(X''\).

The subtraction gives the following result

\[ \frac{d}{F_2\sin\alpha}X'' = \left(\frac{l^3}{6J}+\frac{h}{F_1}\right)X' \]

Or, substituting the numerical quantities,

\[ 46.9X''=363.09X' \]

Whence

\[ X''=7.74X' \]

Substituting this value of \(X''\) into the first equation, after calculating the coefficients we obtain

\[ 716.36X' + 5146.27X' = 446.90K \]

Whence

\[ X'=0.087K \]

and

\[ X''=0.673K \]

Let us now take up the question of the deformations of the cage and of the wooden beams under the shoes.

At the point of application of the force \(K\), the cage will receive a displacement \(\delta_1\), while the displacement of the wooden support under the shoes will be \(\delta_2\). Thus, the total displacement of point \(k\) will be equal to \(\delta=\delta_1+\delta_2\).

The supporting wooden beams have transverse sections of \(26\times26\ \mathrm{cm}\) and \(20\times20\ \mathrm{cm}\), and are laid one above the other (Fig. 5). The connections between these beams are cast-iron consoles. The schematic arrangement of the shoes and wooden supports is shown in Fig. 6.

The moments of inertia of the wooden beams will be

\[ J'=38081\ \mathrm{cm}^4 \quad\text{and}\quad J''=13333\ \mathrm{cm}^4 \]

If there were only one beam under the shoes with moment of inertia \(J\), then point \(k\) would receive a displacement

\[ y= \frac{Kc_1^2c^2}{6EJl_1} \left(2+\frac{c_1}{c}-\frac{c_1^3}{c_1^2c}\right) + \frac{Kc^2c_1^2}{6EJl_1} \left(2\frac{c_1}{c}+1-\frac{c_1^3}{c_1c^2}\right) = \]

\[ = \frac{Kc^2c_1^2}{6EJl_1} \left[ 3+3\frac{c_1}{c} -\frac{c_1^3}{cc_1} \left(\frac{1}{c_1}+\frac{1}{c}\right) \right] = \frac{Kc^2c_1^2}{6EJl_1} \left(3+2\frac{c_1}{c}-\frac{c_1^2}{c^2}\right) = \frac{mK}{J} \]

(see Hütte, Vol. I, Strength of Materials, beams of constant section)

With two beams, under the action of the same forces, the deflection in the upper beam will be

\[ y_1=\theta\frac{mK}{J'} \]

And the deflection of the lower beam

\[ y_2=\frac{m}{J''}(1-\theta)K \]

since both beams are connected to each other, \(y_1=y_2\).

Consequently,

\[ \frac{\theta}{J'}=\frac{1-\theta}{J''} \quad \text{or} \quad \theta=\frac{J'}{J'+J''}=\frac{38081}{51414}=0.745 \]

If the beams are made of pine, for which \(E=100000 \dfrac{\mathrm{kg}}{\mathrm{cm}^{2}}\), then

\[ \delta_{2}= \frac{0.745K(43.5)^2(91.5)^2}{6\cdot100000\cdot38081\cdot135} \left[ 3+\frac{43.5}{91.5}-\left(\frac{43.5}{91.5}\right)^2 \right] =0.000012346K. \]

By the same method we determine the deflection \(\delta\) of an iron beam of box section with dimensions \(160\times65\ \mathrm{mm}\), freely resting on two supports and having a span \(2l=248\ \mathrm{cm}\).

This beam is subjected to the action of forces \(K\) at the points \(k\), and then to the action of the force \(X' + 2X''\sin\alpha\) at the middle, directed upward from below.

The deflection of the beam from the first system of forces will be

\[ \delta_k= \frac{Ka^2(2l-a)^2}{6EJ\cdot2l} \left[ 3+\frac{2a}{2l-a}-\left(\frac{a}{2l-a}\right)^2 \right] = \frac{K\cdot58^2 190^2}{6\cdot2.000.000\cdot925\cdot248} \]

\[ \left[ 3+2\cdot\frac{58}{190}-\left(\frac{58}{190}\right)^2 \right] =0.000156K \]

The deflection from the force \(X' + 2X''\sin\alpha\) will be

\[ \delta_x= \frac{1.18k\cdot(2l)^3}{16EJ} \left[ \frac{a}{2l}-\frac{4}{3}\left(\frac{a}{2l}\right)^3 \right] =0.000131k \]

As a result we obtain

\[ \delta=\delta_k-\delta_x+\delta_2=0.0000365K\ \mathrm{cm}. \]

To study the dynamic action of the load in comparison with the static one, we shall write its equation of work according to Clapeyron’s theorem.

Let the load \(4Q=4400\ \mathrm{kg}\) be assumed concentrated at 4 points \(k\), and let this load fall onto the cams with a velocity \(V_0=40\ \mathrm{cm/sec}\).

The path traversed by each point of application of the force \(K\) (or by the fictitious force \(R\)) will be, according to the calculation, \(0.0000365R\).

Then

\[ \frac{4Q}{2g}V_0^2+4Q(0.0000365R) = 4\cdot\frac{1}{2}R(0.0000365R) \]

Substituting the numerical values here and carrying out the reductions, we obtain

\[ R^2-2200R-50.000.000=0 \]

Whence

\[ R=\sim8256 \]

Consequently, the ratio \(R:Q=\sim7.5\), i.e. the dynamic action is 7.5 times greater than the static one.

To check the corner posts it is necessary to use Euler’s formula.

\[ \frac{P}{n}=\frac{\pi^2 EJ_{\min}}{nh^2}, \]

where \(P\) is the force crushing the post, and \(n\) is the safety factor. Substituting into this expression the numerical values:

\[ n=5,\quad E=2.000.000,\quad J_{\min}=14.1\ \mathrm{cm}^4,\quad h=176\ \mathrm{cm}, \]

we obtain

\[ \frac{P}{5}=\frac{9.86\cdot 2.000.000\cdot 14.1}{5\cdot 30976}\simeq 1800\text{ kg}. \]

And the entire force \(P=9000\) kg.

I do not know why Professor S. A. Zaborovskii computes this force \(P\) as equal to 37000 kg.

In order to arrive at this number, it is necessary to consider the upright as a beam with two fixed ends, i.e., to use the formula

\[ P=\frac{4\pi^2EJ_{\min}}{h}=36000 \]

Then the safe load will be 7200 kg.

The force acting on the upright will be calculated by the formula

\[ \frac{1}{2}(X' + 2X''\sin\alpha)=0.59R=4871\text{ kg}. \]

Since the section of the channel iron (see Hütte, 9th ed., 1916, p. 712, vol. I) \(=9.03\), then per sq. cm the load will be

\[ \frac{4871}{9.03}\simeq 540\text{ kg}. \]

The section of the diagonal, after deducting the rivet holes, will be

\[ (7-2)0.8=4\text{ sq. cm}. \]

The tensile force \(X''=\simeq 5556\), and per sq. cm, 1386 kg.

Next it is necessary to check the strength of the cage frame. In this case it is sufficient to confine oneself to checking only the long side of the frame.

The arrangement of the forces acting on the frame, and their magnitude, are shown schematically in Fig. 7.

The corresponding bending moments at the points of application of the forces: 1—2—3—4 will be as follows.

\[ M_1=8256\times 16=132096\text{ kg·cm}. \]

\[ M_2=8256\times 50-4871\times 34=247186 \]

\[ M_3=8256\cdot 98-4458\cdot 82-4128\cdot 48=245388 \]

\[ M_4=8256\times 140-4458\cdot 124-4128\times 90-4\cdot 128\times 42=58152 \]

Thus, the stress in section (2) will be

\[ \sigma_{\max}=\frac{M_2}{W}=\frac{247186}{116}=2131\text{ kg/cm}^2 \]

If the stress per 1 sq. cm in the uprights and diagonals did not exceed 2000, then here it exceeds 2000 by 6.55%, but such a stress should not be feared, for the calculation was made on the assumption of concentrated loads from the entire weight of the cage, whereas in reality this is not so.

For calculating the strength of the riveted joints, it is assumed that the safe shearing stress does not exceed 1800 kg/cm², and the stress crushing the edge of the hole is \(=4800\) kg/cm². The diameter of the rivet shank is taken as 1.8 cm, so that the area of the shank is 2.5 sq. cm. The smallest thickness of iron in the riveted joints of the cage is ...

0.55 cm. In shear the rivet will withstand \(1800 \times 2.5 = 4500\) kg; the edge of the hole will withstand in bearing about \(0.90 \times 4800 = \sim 4750\) kg. Thus, the number of rivets has been taken with a margin.

It is also necessary to check the strength of the wooden beams under the cams.

Referring to Fig. 6, it may be seen that the bending moment at point \(k\) will be

\[ M_k = 0.745 K c_1 = 0.745 \cdot 8256 \cdot 43.5 = 267557 \text{ kg cm} \]

The section modulus of the beam is

\[ W = \frac{38061}{13} = \sim 2928 \text{ cm}^3 \]

Therefore

\[ \sigma_{\max} = \frac{267557}{2928} = \sim 91.4 \text{ kg/cm}^2 \]

And \(\sigma_{\max}\) may be allowed up to \(200 \text{ kg/cm}^2\).

The completed cage must be securely fastened to the hoisting rope, for, as experience has shown, insufficient attention to this fastening has more than once caused the rope to break, with all the consequences arising therefrom. The place where the rope is fastened to the cage represents the weakest side of the rope, owing to the local excessive stressing of the wires and their abrasion. The use of thimbles neutralizes to a certain extent the harmful action of the suspended load, but the weak fastening of the free end of the rope to the rope itself remains. In order to prevent this free end from moving, one has to resort to the use of bolted clamps, thereby ensuring the occurrence of considerable friction between the wires and, consequently, pinching of the free end. However, the continuous oscillation of the rope loosens this connection.

Among the negative aspects of a coupling device is the disturbance of the uniformity of the cross-section. This change in cross-section is affected by the following circumstances:

1) The rope is often made of crucible steel, for which the tensile strength often amounts to \(120\text{—}160 \text{ kg/mm}^2\), whereas for cast iron from a Siemens-Martin furnace \(k_z = 45\), and for steel from a Siemens-Martin furnace \(k_r = 55\); and these latter materials are also used for the harnessing device of the rope.

2) The rope works chiefly in tension, while in the coupling device a considerable bending along an arc of small radius appears at the wall.

3) Mining regulations prescribe that for the rope a \(7\text{—}8\)-fold factor of safety be taken, and for the coupling device at least a tenfold one (when lifting loads). During hoisting, the continuous oscillations of the rope do not have time to be transmitted to the coupling device, as a result of which, if this attachment is not designed rationally, additional stresses arise in the individual wires of the coupling; of these, the stresses due to acceleration of the coupling device in the horizontal direction are the most harmful.

It is therefore quite natural that designers should seek, as far as possible, to reduce these oscillations.

Good results in this respect are given by “Demag” clamps. The clamp consists (Fig. 7) of a box \(m\) of cast steel, into which enter two wedge-shaped bars \(C\) made of very strong material. The bars have a hole along the axis corresponding to the diameter of the rope. On the outside the bars have recesses \(n\), into which the short arms of the levers \(e\) enter; the axes of rotation of these levers have bearing points in the walls of the box. Chains are attached to the long ends of the levers (usually three on each side), running downward to the cage

(two toward the center of the cage and four toward the outer uprights), and at the top the inclined adjusting bolts \(O\), whose central nuts are provided with lugs to make rotation possible.

The action of the gripping device reduces to the following: when the rope begins to rise, it tightens the suspension chains, the upper links of which turn the long arms of the levers downward. At this time the short arms of the levers move the lower wedge upward, and this wedge in turn moves the upper wedge and thus clamps the rope. It is not difficult to understand that such clamping of the rope is possible only when the inner cones are composite, so that their parts will move closer together during upward travel and move apart during the reverse motion. The clamping force here is the friction between the rope and the inner surface of the cones. Let the taper of the wedge be expressed by the angle \(\alpha\). It is necessary to choose such a ratio of the lever arms

\[ \frac{l_1}{l_2}, \]

that the friction force \(K\) will be able to hold the loaded cage together with the lower counterbalancing rope and at the same time will not produce excessive pressure on the rope.

Let us denote the weight of the cage with the load and the lower counterbalancing rope by \(P\); let the angle of friction between the cones and the box of the device be denoted by \(\rho = \operatorname{tg}\mu \le 0.15\). Further, let the normal pressure of the wedge on the inner wall of the box be \(N\), and the normal pressure of the inner walls of the wedge on the rope be \(P_2\). Finally, suppose that at point \(A\) a force \(Q\) acts, and directly opposite it a reaction \(-Q\), and that at point \(C\) a force \(P_1 = \dfrac{P}{2}\) is applied.

Referring to the diagram (Fig. 8), it is not difficult to write the following equilibrium conditions for the forces applied to our body:

\[ P_1 + Q = N \operatorname{Sin}\alpha + N\mu \operatorname{Cos}\alpha \]

\[ P_2 = N \operatorname{Cos}\alpha - N\mu \operatorname{Sin}\alpha \]

Here \(\mu N\) denotes the friction force, directed downward along the generatrix of the cone, as shown by the arrow.

The angle \(\rho \le 8^\circ 30'\).

Dividing one equation by the other, we obtain

\[ \frac{P_1 + Q}{P_2} = \frac{\operatorname{Sin}\alpha + \operatorname{tg}\mu \cdot \operatorname{Cos}\alpha} {\operatorname{Cos}\alpha - \operatorname{tg}\mu \cdot \operatorname{Sin}\alpha} = \operatorname{tg}(\alpha + \rho) \]

Whence one may write

\[ 2P_2 = K = 2\frac{P_1 + Q}{\operatorname{tg}(\alpha + \rho)} \]

If \(CB = l_1\) and \(BA = l_2\), then

\[ P_1 l_1 = Q l_2 : Q = P_1\frac{l_1}{l_2} = \frac{P l_1}{2l_2} \]

After these additions one may write

\[ K = \frac{2P_1\left(1+\dfrac{l_1}{l_2}\right)} {\operatorname{tg}(\alpha+\rho)} = P\cdot \frac{\left(1+\dfrac{l_1}{l_2}\right)} {\operatorname{tg}(\alpha+\rho)} \]

This force \(K\), with the coefficient of friction \(\psi\) on the surface of the rope, produces a friction \(\psi K\), provided that

\[ \psi K > P \]

Thus, slipping of the cage is avoided.

It is further important to note the following. The force \(K\) is in a predetermined relation to the load \(P\), and therefore excessive slackening of the rope is avoided, which could occur with inept control of the hoist.

In ore hoisting it is often necessary to raise the cage above the cams. At the beginning of this operation the load \(P\) is equal to zero, and consequently so is the force \(K\). In this case the rope should only move in the cone, without producing the required lift. To avoid this, nuts \(g\) with locknuts are fitted on the bolts \(O\), ensuring constant pressing of the wedge against the rope. Further, the free end of the rope is clamped in the clamp \(k\) and is provided below with the block \(i\).

When cutting off the end of the rope for testing, the clamp has to be reset. This resetting is carried out simply. Having placed the cage on the cams, the nut \(g'\) is unscrewed and, by turning the nut \(h\), the lever is made to turn. After this it is easy to push the cone \(C\) out of the apparatus box. It is then only necessary to unscrew the nuts of the clamp \(k\), and the movement of the rope through the clamping device is performed without difficulty. Having cut off the end of the rope to the required length, the nuts \(h\) and \(g'\) are returned to their former position.

As is clear from this description, moving the cage along a straight rope is very convenient. Meanwhile, this operation is not so simple, especially if the rope is thick and has to be bent around a thimble. To avoid the superposition of different stresses in the rope at one and the same place, the inner cavity of the conical clamp is made slightly conical upward. This achieves a uniform distribution of pressure in proportion to the increase of the tensile force in the rope itself.

Finally, at the top of the entire clamping device there is a long neck, the purpose of which is to remove the place where the rope is clamped from the region of rope oscillations.

Many years of experiments with a large number of clamping devices of this system show that damage to the rope inside the clamp is very rare. Still more rarely does the rope get pulled out of the clamp.

The same firm, Demag (Duisburg), patented a rope suspension (Fig. 9) of the Eigen system. Here too there is a thimble around which the rope is wrapped, but, unlike the earlier suspension systems, here the free end of the rope is not tied at the top to the working rope, but lies directly against the upper part of the suspension, where it is gripped by a whole row of bolts. Thus the working rope freely emerges from the cast neck of the suspension, thereby avoiding considerable slackening of the rope wires. The slight stresses from the straightening of the rope wires are experienced in the lower part of the suspension neck, where there is a bend of the rope. However, owing to the great height of the neck, the oscillations of the rope are not reflected at this place in any noticeable way.

Among the advantages of this suspension must be included the fact that the rope can be inspected in its most highly stressed parts. The thimble neck and the narrowed part of the suspension adjoining it have hinged shields on bolts \(n\).

Rope suspensions are built with adjustable straps or a spindle, which makes it possible to change the height at which the cage is set on the weighing device.

The rope suspensions are supported by means of plates \(O\), connected to one another by bolt \(p\), by the suspension lug \(n\), the lug \(r\), and the shackle. The presence of these parts makes it possible to avoid sharp impacts of the shackle on the cage. Small movements of the cage are produced by rotating the spindle \(C\), the coupling of which \(t\) has two trunnions \(f\), passing through the lugs of the plates \(O\). This device makes it possible to move the suspension and, consequently, the cage. If the spindle \(C\) breaks, the cage does not fall, since the nut \(f\) is caught in the lower part \(U\) of the rope clamp.

For fastening the cage of the lower balancing rope, simplified suspension designs are used, shown in Fig. 10. In these designs the free end of the rope is likewise pressed by bolts to an elongated neck, while the rope itself freely emerges from the corresponding rope in the wedge.

Since their appearance in mining practice, both the clamp and the “Demag” suspension have been continually improved; every year the firm provides new modifications, but the basic ideas in these designs remain untouched, which testifies to the rationality of the principles of construction. Several such types are shown in Figs. 7a and 7b.

An almost indispensable part of every cage serving for the hoisting and lowering of people is a parachute or device whose purpose is to arrest the cage in the event of breakage of the rope on which the cage is suspended.

In Fig. 11 is shown one such parachute, of the system of the machine-building factory A. Belen, Herne in Westphalia. The parachute belongs to the eccentric type.

The following idea underlies the construction of this parachute. Statistics show that most often the rope breaks when the cage is being hoisted. As for breakage of the rope during descent, this phenomenon is observed more rarely, and its cause is excessively rapid unwinding of the rope from the drum at the beginning of the cage’s descent, when the speed of its fall is still small. At this time the cage falls like a free body under the influence of gravity. Acquiring an ever greater speed, it outruns the rope and produces a jerky blow, sometimes quite sufficient to rupture the rope or tear it.

Calculation for such a stress in the rope can be carried out only if a large safety margin is provided, when the rope comes out very heavy. More rational is the possible reduction of the weight of the cage, the weight of the cars, and of the rope itself.

If the cage breaks away from the rope during hoisting, then for some time, under the influence of inertia, it will continue to move upward, will bring its speed to zero, and only after that will change the direction of its motion and will fall, like a free body or under conditions closely approaching such motion.

All parachutes designed for the instantaneous action of their organs are designed to catch the cage under the easiest conditions. How different the conditions are for catching the cage in ascending and descending motions can be seen from the following considerations.

Let us assume that, after the broken-off cage has acquired zero velocity during ascending motion, it has passed a distance \(S = 5\) cm, or 0.05 meter. In this case it will have the velocity

\[ U = \sqrt{2 g . s} = \sqrt{2 . 9{,}81 \times 0{,}05} = \sim 1 \text{ mt} \]

The falling cage, however, when the rope breaks during descending motion, has to be caught at a velocity of 15 mtr and more.

If the cage breaks away from the rope at a velocity of 10 meters directed downward (the cage is descending), then this is the same as if the cage were freely falling from a height of 5 meters:

\[ h = \frac{U^2}{2g} = \frac{100}{2 . 9{,}81} = \sim 5 \text{ mt} \]

If the weight of the cage is \(G = 12000\) kg, then at the moment of rupture it is capable of producing the work

\[ G . h = \frac{G}{g} \cdot \frac{U^2}{2} = \frac{12000}{9{,}81} \cdot \frac{100}{2} = \sim 60000 \text{ kg. mtr} \]

or \(6.000.000\) kg·cmt.

If this cage is acted upon by a force \(G\), equal to the weight of the cage, then the cage will fall with a uniform velocity of \(10\ \mathrm{mt/sec}\).

If the force acting on the cage is increased to \(2G\), then the cage will acquire uniformly retarded motion with a deceleration equal to \(w=-9.81\ \mathrm{mt/sec}\). This means that after \(5\ \mathrm{mtr}\) of travel it will stop. Thus the entire path traversed by the cage will be 10 meters, and the work possessed by the cage will be

\[ 2Gh = 2 \cdot 12000 \cdot 5 = 120000\ \mathrm{kg\cdot m}. \]

Increasing the acting force to \(3G\), we obtain a force working to retard motion equal to \(2G\), and the cage will stop after passing a distance of 2.5 meters,

\[ G(h+h_1)=3Gh_1;\quad h+h_1=3h_1;\quad h_1=2.5\ \mathrm{m}. \]

These considerations lead to the following principles for the construction of parachutes:

1. The force applied to the cage must have the character of a braking force, and in magnitude must be at least equal to twice the weight of the cage.

2. The braking force, or the force catching the falling cage, must act for a long enough time, until the cage has stopped.

3. All parts of the cage and of the guides participating in the work of catching must have at least a fourfold margin of strength relative to the acting forces.

4. The catching parts must be capable of withstanding, without bending or destruction, even such impacts as may be caused by a cage having a velocity of \(15\ \mathrm{mt/sec}\).

5. Simplicity and convenience in servicing.

These considerations have also been taken as the basis for the construction of parachutes of the Veis system, belonging to the eccentric type.

The surface of the eccentrics in contact with the guides is made either smooth or with small serrations.

Wooden beams or rails may be used as guides.

In Fig. 11 an eccentric parachute is shown as applied to wooden guides.

Here the positions of the eccentrics are visible when disengaged from the guide and in action, when the curved surfaces are pressed into the body of the guide from two sides.

The rotation of the eccentrics from the inactive position into the working position is effected by strong springs.

Let the cage weigh \(12000\ \mathrm{kg}\). Then the design weight must be taken as \(24000\ \mathrm{kg}\), so that each eccentric will have a load of \(6000\ \mathrm{kg}\).

The work of catching the cage will consist of the work of braking and of the work of crushing the guide.

If the width of the eccentric in the direction perpendicular to the plane of the drawing is \(7\ \mathrm{cm}\), and the length of the arc \(a\text{--}b\) is also \(7\ \mathrm{cm}\), then, allowing for oak a temporary compressive resistance equal to \(350\ \mathrm{kg/cm^2}\) (see Hütte, Part I, p. 547, 1915 edition), we obtain the crushing force

\[ 350 \times 7 \times 7 = 17150\ \mathrm{kg}. \]

Since the resultant friction force will act tangentially to the friction surface, approximately somewhere between points \(a\) and \(b\), and will have a direction differing little from the vertical, the coefficient of friction \(f\) will be equal to

\[ f=\frac{6000}{17150}=0.35 \]

If braking takes place over a path of 5 meters, then the work of deformation will be equal, when the eccentric is pressed in to a depth of \(2\ \mathrm{cm}\), to ...

\[ 350 \cdot 7 \cdot 2 \cdot 500 = 2.450.000 \text{ kg cm} = 24500 \text{ kg mtr} \]

For four eccentrics this amounts to 98,000 kg mtr. The braking action over a path of 5 meters gives the work

\[ 6000 \times 4 \cdot 5 = 120\,000 \text{ kg mtr} \]

Consequently, the total mechanical work will be equal to

\[ 98000 + 120000 = 218.000 \text{ kg mtr}. \]

But the work of catching the cage over a path of 5 meters should be only 120,000 kg mtr, which is about 60% of 218,000. This means that the cage should stop after traveling a distance of 3 meters.

If the cross-section of the conductor is \(12.5 \times 16\) sq. cm, if its permissible temporary resistance to the compressive force acting on it from two sides is \(660 \text{ kg/cm}^2\), and if it must withstand a stress from two sides equal to only 12,000 kg, then its factor of safety will be

\[ \frac{600 \times 200}{12000} = \sim 10 \]

At the beginning of the operation of catching the cage, the eccentrics rotate vigorously about their axes under the influence of the straightening of the springs and, with a force that is quite considerable, place their flat shelves onto the guides. In order that this jolt should not be too perceptible to the people in the cage and should not cause undesirable destruction in the striking parts, a soft gasket is arranged between the guides and the shelf of the eccentric: rubber or wooden, intended to damp the impact.

In Fig. 11a is shown the design of the same parachute for the case when rails serve as the guides.

In Fig. 12 is given the design of the parachute of the Weidich system, according to German patent No. 210869. (Machine factory of Emil Wolff, in Essen).

This parachute belongs to the type of eccentric brakes.

On both sides of the cage, where the conductors are located, there are several pairs of conical rollers (7), eccentrically mounted on the axle (5). The outer surface of these conical rollers is provided with small teeth to increase adhesion when pressed against the guides (10). The conical rollers are hollow and are provided inside with a screw thread. The connection between the conical roller and the axle (5) is achieved not directly, but by means of an eccentric (6) provided with a long bushing. On the lateral surface of the cylindrical eccentric there is a screw thread, the thread of which enters, like a nut, into the recess of the thread of the roller. When the rope breaks, a spring, of the leaf-spring type, causes the rollers eccentrically arranged on the axles to rotate, and the latter approach the axle, entering into closer contact with the guides.

Such is the idea of the design of the Weidich parachute. In tests with this parachute, a cage weighing 1800 kg with a load of 1000 kg was selected. When the cage was released from rest, the braking path was 1.4 meters, and the eccentric turned by 25 mm; in another test under the same conditions the braking path proved to be 3 meters, and the eccentric turned by 55 mm. In another case the same cage was released at a speed of 4 mtr/sec. The braking path was 4 meters, and the eccentric turned by 55 mm.

There are very many systems of parachutes; their theory is very interesting, but we shall not expound it, hoping to carry out this idea in our subsequent works on mine mechanics. Then we expect to touch in more detail upon the calculation of all parts of parachutes.

A. [[unclear: abbreviated title/rank]] Prof. S. V. Donskoi. I. Static analysis of the stand

Fig. 1.

Fig. 2.

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

For the article by Prof. S. N. [[unclear: author’s surname]]. 1. “Static Analysis of the Stand.”

Fig. 5.

Fig. 6.

Visible dimension markings: 1170; 1240; 800; 600.

[No visible translatable text on this page; the page consists only of a technical drawing.]

To the article by Prof. S. K. Komkov. I. “Static analysis of a stand.”

Fig. 11.

Visible member markings in the drawing: \(50 \times 25 / 6\), \(80 \times 47 / 8\), \(60 \times 25 / 6\).

Fig. 12.

Visible node labels and force markings in the diagram: \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), \(G\), \(H\), \(J\); \(0\), \(+1540\), \(-60\), \(+1180\), \(-400\), \(-790\), \(+793\), \(+1250\), \(+290\), \(-60\), \(-65\), \(-80\), \(-85\).

For the article by Prof. S. K. Kanyukov. I. “Static Analysis of the Stand.”

Fig. 14

Diagram of the stand with panels and member forces. Visible annotations include points \(A, B, C, D, E, F, G, H, I, J, K, L, M\), load/designation lines \(S_1, S_2, S_3, S_4\), and member labels with numerical values, including:

\[ a = 800,\quad b = 2000,\quad c = 300,\quad d = -150,\quad e = -150, \]

\[ f = 300,\quad g = 1820,\quad h = 300,\quad i = 150,\quad j = -150, \]

\[ k = 300,\quad l = 1040,\quad m = 300,\quad n = -150,\quad p = 150, \]

\[ q = 300,\quad r = 260,\quad s = 300,\quad t = -150,\quad u = -150. \]

Visible vertical member or panel-force annotations include:

\[ 1 = 258,\quad 2 = 258,\quad 3 = 258,\quad 4 = 258,\quad 5 = 258,\quad 6 = 258, \]

\[ 7 = 258,\quad 8 = 258,\quad 9 = 258,\quad 10 = 258,\quad 11 = 258,\quad 12 = 258,\quad 13 = 258. \]

A narrow auxiliary force diagram at the center repeats the sequence of member letters and numbers, including \(a, d, i, f, k, n, q, t, u, s, p, m, j\), with corresponding levels \(1\) through \(13\).

Fig. 15

Diagram of the stand with panels and member forces. Visible annotations include points \(A, B, C, D, E, F, G, H, I, J, K, L, M\), load/designation lines \(S_1, S_2, S_3, S_4\), and member labels with numerical values, including:

\[ a = 800,\quad b = 575,\quad c = 800, \]

\[ d = 258,\quad e = 258,\quad f = 300,\quad g = 258,\quad h = 300, \]

\[ i = -150,\quad j = -150,\quad k = 300,\quad l = -2060,\quad m = 300, \]

\[ n = -150,\quad p = -150,\quad q = 300,\quad r = 2340,\quad s = 300, \]

\[ t = -150,\quad u = 150. \]

Visible vertical member or panel-force annotations include:

\[ 4 = 258,\quad 5 = 258,\quad 6 = 258,\quad 7 = 258,\quad 8 = 258, \]

\[ 9 = 258,\quad 10 = 258,\quad 11 = 258,\quad 12 = 258. \]

A narrow auxiliary force diagram at the right repeats the sequence of member letters and numbers, including \(q, n, k, i, f, d, a, c, e, h, j, m, p, s, u\), with corresponding levels \(1\) through \(12\).

Fig. 17.

Fig. 18.

Fig. 19.

Fig. 7.

To the article by Prof. S. K. Kozhukhov. 2. “Dynamic Analysis of a Single-Rope Cage”

Fig. 1.

Visible notation in the figure:

\[ Q = P + G \]

Fig. 2.

Fig. 4.

Visible notation in the figure:

\[ a = 248\ \text{cm} \]

![Technical drawing of the stand, with dimensions and sectional views.]

Fig. 5.

Visible annotations and dimensions include: \(R_1 = 1\), \(r = 1\), \(1240\), \(880\), \(550\), \(650\), \(1700\), \(560\), \(1350\), \(430\), \(185\), and other partially unreadable dimension labels and symbols.

Plate 7.

Fig. 7-a.

Visible labels in the diagram include: \(P_1=\frac{P}{2}\), \(P\), \(P_2\), \(N\), \(N\sin\alpha\), \(S\sin\alpha\), \(N\cos\alpha\), \(\mu N\sin\alpha\), \(l_1\), \(l_2\), \(l\), \(A\), \(B\), \(C\), \(O\), \(Q\).

\[ P_1=\frac{P}{2} \]

\[ P_1=\frac{P}{2} \]

Fig. 7-b.

Drawing 7-c.

Drawing 8.

To the article by Prof. S. K. Kolmakov: “Dynamic analysis of single-post masts.”

Fig. 10.

Fig. 11.

Fig. 11-a.

For the article by Prof. S. K. Kolyada. 2. “Dynamic Analysis of a Rolling-Mill Stand.”

Fig. 12-a.

Fig. 13-a.

Fig. 13-b.

Fig. 12-b.

Fig. 12-c.

Fig. 12-e.

Fig. 14-a.

Fig. 14-b.

Fig. 14.

We shall now supplement the information on the manner in which the fastening of the free end of a rope is sometimes accomplished after forming a loop with the rope itself.

In Figs. 13 and 13b the design of an American clamp is given; in drawing 13a this clamp is shown separately, and in drawing 13b it is seen as used in practice. The design of the clamp is so simple that it does not need any supplementation. These clamps are cast from malleable iron, and usually from 4 to 6 of them are used.

In drawing 14 (a and b) thimbles are shown: on the left, an iron one, and on the right, a cast-iron one. To determine their dimensions and outlines, the following tables may be used.

Iron thimble for wire ropes

Rope diameter, mm	A	B	C	D	E	F	G	H	I	K	L
32	24	13	3	3	41	67	14	11	7	6	2
38	30	16	5	5	57	91	17	14	9	7	2
44.5	30	17	5	5	59	95	22	20	11	10	2
51	33	19	5	5	64	110	27	22	15	11	2
57–63.5	37	21	5	5	76	130	30	25	17	13	2
70–76	41	22	6	5	86	152	37	32	18	16	2
82.5–89	48	27	6	5	108	189	41	35	21	17	3
95–101.5	56	32	8	5	127	227	49	41	25	20	3
108–114.5	65	37	10	5	146	259	54	44	27	22	3
120.5–127	73	41	11	5	165	287	57	48	29	24	3

Cast-iron thimble

Rope diameter, mm	A	B	C	D	E	F	G	H	I	K
38	29	21	19	3	44	73	22	16	15	8
44	33	23	22	3	51	84	25	19	19	10
51	38	27	27	5	60	98	29	22	22	11
57–64	43	30	30	5	70	113	32	25	25	13
70–76	54	33	35	6	89	143	38	32	30	16
83–89	59	40	41	6	98	157	44	35	35	17
95–102	70	46	49	8	114	184	51	41	44	21
108–114	76	52	54	8	124	200	54	44	44	22
121–127	81	57	57	10	130	214	57	48	48	24