S. Yu. Doborzhinsky.

Materials and Research

on the Theory and Calculation

of Headframe Structures.

Part II.

Tomsk.

Typolithography of the Siberian Association of Printing Works,
corner of Dvoryanskaya Street and Yamskoy Lane, own building.

1912.

Preface to Part II

To clarify the theoretical arguments presented in the first part of the present work, I consider it necessary to publish several numerical calculations of headframes—calculations that differ to a rather considerable degree from those available in the literature, not excluding the most recent work by Mörle.

The present calculations form parts of student projects in mine mechanics, for which I served as supervisor in the mining department of the Tomsk Technological Institute. I selected the calculations to be varied both in the method of their execution and in the type of headframes designed.

In view of the objections that I have encountered concerning the calculation of headframes for the breaking force of the hoisting rope, I consider it necessary to state the following: it cannot be denied that, in reality, breaks of new, entirely sound ropes possessing a normal degree of safety are a rather rare occurrence. Nevertheless, in view of the presence of so-called extreme stresses that appear in the ropes at the beginning of each hoist, it is necessary to take into account stresses exceeding the static ones. If \(s\) is the static load of the rope, and the dynamic load is \(S\), then, according to the well-known expression:

\[ S = s\left(\frac{v}{gt} + 1\right), \]

where \(v\) is the hoisting velocity after the lapse of the time interval \(t\). With a sufficiently small time interval in which, owing to the accelerated motion of the machine, the cage acquires the velocity \(v\), the stress \(S\) may attain the magnitude of the breaking force. Of course, in normal operation of the hoist this does not occur; but in any case:

\[ S > s, \]

i.e., the tension in the rope during one hoist fluctuates from \(S\) to \(s = G + P\), the useful and dead load. The value \(S\) should be taken into account

is very difficult, for it depends on a mass of factors, including the personality of the hoisting engineer and, of course, is subject to all sorts of fluctuations. Since the indicated dynamic forces are transmitted to the headframe, its calculation should be carried out for \(S+s\), for the dynamic force arising in one rope and the simultaneous full load of the other. Such a combination of stresses is quite possible because extraordinary forces may arise not only during hoisting but also during lowering, that is, in the rope with the empty cage; on the other hand, their simultaneous occurrence in both ropes is hardly admissible.

In view of the difficulty of an exact accounting of \(S\), it is necessary to take into consideration its maximum possible value:

\[ \operatorname{Max} S = \text{the breaking force.} \]

In view of what has been said, the strictly consistent performance of the calculation for the breaking forces will not give, and cannot give, results exaggerated in the sense of strength, despite the seemingly excessively substantial dimensions of the parts of the structure. To be convinced of this, it is enough to observe the swaying of the headframe while standing, during hoisting, on the platform near the sheaves. In particular, headframes of the so-called light construction are subject to swaying to a high degree.

Calculation of a Wooden Headframe

By student N. N. Pavlov.

For the stability of the headframe it is necessary that the resultant of the tensions of the upper and lower ropes pass between the legs of the headframe. This condition is satisfied by the present headframe (drawings 1 and 2).

The headframe consists of two essential parts: the sheave frame and three pairs of legs—trusses.

Sheave Frame

The sheave frame is a system of intersecting longitudinal and transverse beams, by means of which the pressure produced by the load being lifted is transmitted to the legs of the headframe. We decompose the resultant of the forces into vertical and horizontal components, which are equal in magnitude to the vertical and horizontal stresses. The greatest vertical stresses, transmitted directly to the sheave beams, will develop in the case where the lower rope breaks near the mouth of the shaft.

From the diagram it is seen that the vertical component in this case is equal to 52800 kg; to this is added the weight of the sheave with its bearing, equal to 1200 kg. Thus the sheave beams of the headframe take a load

$P_2 = 52800 + 1200 = 54000$ kg.

One sheave beam will bear one half of this load, i.e. 27000 kg. For this load I therefore make the calculation of the sheave beams. In the present case the sheave beam is a continuous beam resting on three supports. To determine the safe dimensions of such a beam, Clapeyron gives the following formula:

$$ M_0 l_1 + 2 M_1 (l_1 + l_2) + M_2 l_2 = \frac{P_1 a_1 (l_1^2 - a_1^2)}{l_1} + \frac{P_2 b_2 (l_2^2 - b_1^2)}{l_2} $$

$$ \begin{aligned} l_1 &= 155\ \text{cm},\\ l_2 &= 340\ \text{cm},\\ a_2 &= 185\ \text{cm},\\ b_2 &= 155\ \text{cm},\\ P_2 &= 27000\ \text{kg}. \end{aligned} $$

In this formula \(M_1\), \(M_1\), and \(M_2\) denote the moments over the corresponding supports; \(l_1\) and \(l_2\) are the distances between supports; \(a\) and \(b\) are the distances from the concentrated load to the nearest supports.

For \(M_0 = 0\), \(M_2 = 0\), and \(P_1 = 0\), Clapeyron’s formula takes the form:

\[ 2 M_1 (l_1 + l_2) = \frac{P_2 b_2 (l_2^2 - b_2^2)}{l_2}, \]

whence

\[ M_1 = \frac{P_2 b_2 (l_2^2 - b_2^2)}{2 l_2 (l_1 + l_2)}. \]

Substituting, in place of \(P_2, b_2, l_2, l_1\), their numerical values, I obtain:

\[ M_1 = \frac{27000 \cdot 155 \cdot 185}{340 \cdot 2} = 1138566. \]

Calling the support reactions from loads on the right-hand side \(S_1, S_2\), and \(S_3\), and, correspondingly, from loads on the left-hand side \(S'_1, S'_2\), and \(S'_3\), we have the equalities:

\[ S_1 = \frac{1}{l_1}(M_0 - M_1) = -\frac{M_1}{l_1} = -\frac{1138566}{155} \approx -7342 \]

\[ S'_2 = \frac{1}{l_1}(M_1 - M_0) = \frac{M_1}{l_1} = \frac{1138566}{155} = 7342 \]

\[ S_2 = \frac{1}{l_2}(M_1 - M_2 + P_2 b_2) = \frac{M_1 + P_2 b_2}{l_2} = \]

\[ = \frac{1138566 + 27000 \cdot 155}{340} = 15658. \]

\[ S'_3 = \frac{1}{l_2}(-M_1 + P_2 a_2) = \frac{-1138566 + 27000 \cdot 185}{340} = 11342. \]

The total support reactions are:

\[ R_0 = S'_1 + S_1 = 0 + (-7342) = -7342. \]

\[ R_1 = S'_2 + S_2 = 7342 + 15658 = 23000. \]

\[ R_2 = S'_3 + S_3 = \frac{11342 + 0}{\Sigma R} = \frac{11342}{27000}. \]

I calculate the beam on the basis of the formulas:

\[ M = Wkb; \qquad \text{for oak } kb = 80. \]

\[ W = \frac{a h^2}{6}; \qquad \text{I take } a = 40 \text{ cm.} \]

\[ h = \sqrt{\frac{113856 \cdot 6}{40 \cdot 80}} = \sim 51 \text{ cm.} \]

Since a beam with a cross-section of \(40 \times 51\) cm is difficult to find, instead of one beam I take two, each of dimensions \(40 \times 40\) cm, and connect them along their length with keys.

\[ \begin{aligned} R_1 &= 23000 \ \text{kg},\\ l_1 &= 185 \ \text{cm},\\ l_2 &= 155 \ \text{cm},\\ l &= 340 \ \text{cm}. \end{aligned} \]

The greatest shearing forces will be in the part \(l_2 = 155\) cm. The shearing force acting in the neutral layer is

\[ S = K_s h l_2 \quad (\text{where } h \text{ is the width of the beam } = 40 \text{ cm}.) \]

For a rectangular beam,

\[ K_s = \frac{\sum P \int_0^h y\, d\omega}{hY} = \frac{h^3 \cdot 12 \sum P}{2h \cdot h(2h)^3} = \frac{h^3 \cdot 12 \cdot 2 \cdot 23000}{2 \cdot 8 \cdot h^5} = \frac{3 \cdot 23000}{4 \cdot h^2}. \]

\(y\) is the distance from the center of gravity of the section to the axis of the beam.

Consequently,

\[ S = \frac{3 \cdot 23000 \cdot h \cdot l_2}{4h^2} = \frac{3 \cdot 23000 \cdot l_2}{4h}; \]

\(l_2 \simeq \frac{1}{2}l\). Substituting its value for \(l_2\), I find that

\[ S = \frac{3 \cdot 23000 \cdot l}{8h}. \]

This shearing force must be balanced by the resistance of the keys, i.e.,

\[ S = b \cdot a K_s' n; \]

where \(b\) is the length of a key; \(a\) is the width of a key.

\(K_s'\) is the shearing stress—\(15\) kg per cm\(^2\). The height of the key is

\[ c = 0.3h = 12 \text{ cm}. \]

The length of the key is \(b = 2c = 24\) cm.

The width of the key is \(a = h = 40\) cm; \(n\) is their number.

Substituting the numerical values, I find that

\[ n = \frac{S}{b \cdot a \cdot K_s'} = \frac{3 \cdot 23000 \cdot 340}{8 \cdot 40 \cdot 24 \cdot 40 \cdot 15} = 5 \text{ pieces}. \]

Distance between keys:

\[ l' = \frac{l}{5} = \frac{340}{5} = 68 \text{ cm}. \]

The force acting on the pulleys in the event of rupture of the upper rope near the mouth of the shaft will be somewhat smaller; however, I take the second

a pair of sheave beams of the same dimensions as those just calculated. To determine the dimensions of the transverse beams we shall apply the following method. The vertical pressure experienced by the sheave beams from the upper rope, as is seen from the diagram, is \(=3600\) kg. Adding to this the weight of the sheave \(=1200\) kg, I find that the total load on one of the sheave beams (of the second pair) will be:

\[ \frac{3600+1200}{2}=2400\ \text{kg}. \]

Taking the same notation and applying the same method of calculation, I find that the reaction produced in the upper transverse beams \((Z_0, Z_1, Z_2)\) by the second pair of sheave beams will be

\[ R'_0=-652,\quad R'_1=2043,\quad R'_2=1009. \]

Beam \(Z_2\) is simultaneously subjected to the action of the reaction from the sheave beams when the rope is in the lower position, \(R_2=11342\) kg, and to the action of the reaction from the sheave beams when the rope is in the upper position, \(R'_2=1009\) kg. Beam \(Z_2\) rests on three supports. In the first span the forces \(R_2\) and \(R'_2\) act. In view of the small distance between them, I replace them by their resultant \(P_1=R'_2+R'_2=2.1009=2018\) kg, the point of application of which lies between them.

\[ \begin{aligned} R'_2&=1009\ \text{kg}.\\ R_2&=11342\ \text{kg}.\\ a=f&=47\ \text{cm}.\\ b=e&=80\ \text{cm}.\\ c=d&=22\ \text{cm}. \end{aligned} \]

In the second span, the forces \(R_2\) and \(R_2\) are likewise replaced by their resultant \(P_2 \simeq 2268\) kg, with its point of application between these forces. Thus beam \(Z_2\) will be represented as resting on three supports, with one concentrated load in each span. By Clapeyron’s formula I have:

\[ 2M(l_1+l_2)=\frac{P_1 c(l_1^2-a_1^2)}{l_1}+\frac{P_2 b(l_2^2-a_2^2)}{l_2}; \]

\[ M=\frac{87\cdot(22201-7569)(2018+22684)}{88804}=314097; \]

\[ \begin{aligned} P_1&=2018\ \text{kg}.\\ P_2&=22684\ \text{kg}.\\ l_1=l_2&=149\ \text{cm}.\\ a_1=b_1&=87\ \text{cm}.\\ a_2=b_2&=62\ \text{cm}. \end{aligned} \]

If by \(m\), \(m_1\), and \(m_2\) we denote the moments due to the concentrated load, then the reactions of the supports are determined by the formulas:

\[ S_1=\frac{1}{l}(-M+m)=\frac{-314097+2018\cdot 62}{149}=-1268 \]

\[ S'_2=\frac{1}{l_1}(M+m.)=\frac{314097+2018\cdot 87}{149}=3286 \]

\[ S_2=\frac{1}{l_2}(M_1+m_1)=\frac{314097+22684\cdot 87}{149}=15353. \]

\[ S'_3=\frac{1}{l_2}(-M_1+m_2)=\frac{-314097+22684\cdot 62}{149}=-7331. \]

The full values of the reactions will be:

\[ \begin{aligned} R_1&=S_1=-1268\ \text{kg}.\\ R_2&=S'_2+S_2=18639\\ R_3&=S'_3=7331\\ \sum R&=24702 \end{aligned} \]

Since the forces acting on this beam are smaller than those acting on beam \(Z_1\), then, without calculating its dimensions (I take them to be the same as the dimensions of beam \(Z_1\)), I shall confine myself to calculating the reactions which it exerts on the longitudinal beams below.

Beam \(Z_1\) is under the action of the forces \(R_1, R_1, R'_1\), and \(R'_1\). Replacing, as before, the forces \(R_1\) and \(R_1\) by their resultant

\[ P'_1=R_1+R_1=23000+23000=4600\ \text{kg}. \]

and the forces \(R'_1+R'_1\) by their resultant

\[ P'_2=2043+2043=4086 \]

and taking the points of application of these resultants in the middle of the component forces constituting them, I shall again have the case of a beam resting on three supports with one concentrated load in each span.

\[ 2M_1(l_1+l_2)=\frac{P'_1 c\,(l_1^2-a_1^2)}{l_1}+\frac{P'_2 d\,(l_2^2-a_2^2)}{l_2}; \]

\[ M_1=\frac{87(2201-7569)(4086+46000)}{88804}=717974; \]

\[ S_1=\frac{1}{l_1}(-M_1+m)=\frac{-717974+4086\cdot 62}{149}=-3118\ \text{kg}. \]

\[ S'_2=\frac{1}{l_2}(M+m)=\frac{717974+4086\cdot 87}{149}=7204\ \text{kg}. \]

\[ S_2=\frac{1}{b}(M_1+m_1)=\frac{717974+46000\cdot 87}{149}=31677\ \text{kg}. \]

\[ S_3^1=\frac{1}{b}(-M+m_2)=\frac{-717974+46000\cdot 62}{149}=14323\ \mathrm{kg}. \]

\[ R'_1=-S_1=-3118;\quad R'_2=S'_2+S_2=38881;\quad R'_3=L_3=14323\ \mathrm{kg}. \]

\[ \Sigma R=40086\ \mathrm{kg}. \]

Similarly to the preceding, I find that:

\[ \begin{aligned} P'_1&=4086\ \mathrm{kg}.\\ P'_2&=46000\ \mathrm{kg}.\\ l_1&=l_2=149\ \mathrm{cm}.\\ a_1&=b_1=87\ \text{''}\\ a_2&=b_2=62\ \text{''} \end{aligned} \]

\[ h=\sqrt{\frac{M\cdot 6}{40\cdot 80}} =\sqrt{\frac{717974\cdot 6}{40\cdot 80}} =36\ \mathrm{cm}. \]

Since it will have to be cut in somewhat, I take beam \(Z_1\), and consequently also \(Z_2\), with dimensions \(40\times 40\ \mathrm{cm}\).

The lower longitudinal beam (\(p\)—the one nearest to the lower rope) is loaded at point \(D\) by the force \(R'_3\) (reaction from beam \(Z_1\)) \(=14323\ \mathrm{kg}\), and at point \(D_1\) is loaded by the force \(R_3\) (reaction from beam \(Z_2\)) \(=7331\ \mathrm{kg}\).

\[ 2M_1(l_1+l_2)= \frac{R'_3a(l_1^2-a^2)}{l_1} +\frac{R_3b(l_2^2-b^2)}{l_2}; \]

\[ M_1= \frac{ \frac{14323\cdot 110(295^2-110^2)}{295} + \frac{7331\cdot 35(190^2-35^2)}{190} }{ 295+190 } =922173. \]

\[ \begin{aligned} R'_3&=14323\ \mathrm{kg}.\\ R_3&=7331\ \text{''}\\ l_1&=295\ \mathrm{cm}.\\ l_2&=190\ \text{''}\\ a&=110\ \text{''}\\ l&=35\ \text{''} \end{aligned} \]

\[ S_1=\frac{-922173+14323\cdot 185}{295}=5856\ \mathrm{kg}. \]

\[ S'_2=\frac{922173+14323\cdot 110}{295}=8466\ \text{''} \]

\[ S_2=\frac{922173+7331\cdot 35}{190}=6203\ \text{''} \]

\[ S'_3=\frac{-922173+7331\cdot155}{190}=1129\ \text{kg}. \]

\[ \begin{aligned} R&=S_1=5856\ \text{kg}.\\ R_0&=S'_2+S_2=14669\ \text{kg}.\\ R_1&=S'_3=1129\ \text{kg}.\\ \Sigma R&=21654\ \text{kg}. \end{aligned} \]

Headframe legs.

The reaction \(R\) acts directly on the headframe leg nearest to the trestles; the reaction \(R_0\) acts directly on the middle leg of the headframe, and the reaction \(R_1\) acts on the one nearest to the machine. I take this beam, as well as the other outer longitudinal beam (the one nearest to the upper rope), to be of the same dimensions as the middle longitudinal beam, the calculation of which follows, since it bears the greater load. The middle longitudinal beam, similarly to the preceding one, will be loaded at point \(D_1\) by the force \(R'_2\) (reaction from beam \(Z_1\)) \(=38881\) kg, and at point \(D'_1\) by another force \(R_2\) (reaction from beam \(Z_2\)) \(=18639\) kg.

\[ 2M_1(l_1+l_2)=\frac{R'_2a(l_1^2-a^2)}{l_1}+\frac{R_2b(l_2^2-b^2)}{l_2}; \]

\[ M=\frac{\dfrac{38881\cdot110(295^2-110^2)}{295}+\dfrac{18639\cdot35(190^2-35^2)}{190}}{295+190}=2465332; \]

\[ S'_1=\frac{-2465332+38881\cdot185}{295}=16025. \]

\[ S'_2=\frac{2465332+38881\cdot110}{295}=22854. \]

\[ S_2=\frac{2465332+18639\cdot35}{190}=16406. \]

\[ S'_3=\frac{-2465332+18639\cdot155}{190}=2232. \]

\[ \begin{aligned} l_1&=295\ \text{cm}.\\ l_2&=190\ \text{cm}.\\ a&=110\ \text{cm}.\\ b&=35\ \text{cm}.\\ R'_2&=38881\ \text{kg}. \end{aligned} \]

\[ R_2 = 18639\ \text{kg}. \]

\[ \begin{aligned} R' &= S_1 = 16025,\\ R'_0 &= S'_2 + S_2 = 39260,\\ R'_1 &= S'_3 = -2235,\\ \sum R &= 57520, \end{aligned} \]

\[ h=\sqrt{\frac{M_1\,6}{40\cdot 80}} =\sqrt{\frac{2465332\cdot 6}{40\cdot 80}} =64\ \text{cm}. \]

Since such a beam is difficult to find and a built-up one will have to be used, instead of increasing \(h\) (the height of the beam), I shall take two beams \(40\times40\) and place them side by side. The lower transverse beams \(gg\) and \(zz\) serve as the upper bracing of the legs of the headframe.

The beams \(ZZ\) take at midspan the action of the force \(R'\) (the reaction from the middle longitudinal beam) \(=16025\ \text{kg}\); the beams \(qq\) take at midspan the action of the force \(R'_0\) (the reaction from the middle longitudinal beam) \(=39260\ \text{kg}\). The beams \(SS\) take at midspan the action of the reaction \(R'_1=2235\).

Of these beams I calculate only the beams \(qq\), as those taking the greatest load. The others I take of the same dimensions as these.

The force \(R'_0\), acting on the beam perpendicular to its supports, shears it. I take the beam as double, with a cross section of \(30\times30\) cm each.

Denoting the total shearing stress by \(Q\), and the area of the cross section of the double beam by \(F\), one may compute the magnitude of the shearing stress:

\[ \frac{Q}{F}=\frac{39260}{30\cdot 30\cdot 2}=18=R_2. \]

The permissible value of \(R_2\) for oak is 20. Consequently, the beams \(qq\) are safe.

The same may be said with respect to the beams \(rr\) and \(ss\), since a smaller force acts on them. Having checked these beams for bending, I find that they likewise will not be broken under the influence of a load concentrated at the middle.

From the diagram in Figs. 3 and 4 it is evident that the horizontal component is greater; at the rupture of the upper rope this force \((R=21800\ \text{kg})\) tends to stretch the headframe. To it is also added the horizontal component from the lower rope, \(R_1=1400\ \text{kg}\). Thus the forces

\[ R+R_1=21800+1400=23200\ \text{kg} \]

will stretch the headframe. This is resisted by braces 2, 6, 10.

It must be borne in mind that on the supporting truss (the one nearest to the machine) there also acts the force \(R_0\) (the reaction from the outer longitudinal beam) \(=1129\) kg. Resolving it into components, I find that one part of it, \(s=1300\) kg, will go along the leg, while the other, \(S_1=700\) kg, will compress the longitudinal beam, i.e., will act in the opposite direction to the forces \(R+R_1\) (after transferring them to the end of the headframe).

Thus the force tearing the truss apart will be

\[ \frac{R+R_1-S_1}{2} = \]

\[ =\frac{21800+1400-700}{2}=11250\ \mathrm{kg.}=Rs. \]

In the further calculation I shall consider only the part of the headframe \(abcd\); I neglect the part \(abfl\), Fig. 1. The forces \(Rs\) tend to separate the leg \(cd\) from \(ba\) and \(ef\). The legs \(cd\), \(ba\), \(ef\), however, are connected by longitudinal beams 4, 7, 12 and by braces 2, 6, 10 into one whole. Consequently, for the stability of the headframe it is necessary that the ties 2, 6, 10 and 4, 7, 12 fully withstand the action of the force \(Rs\).

On the truss \(abcd\) there also acts the force \(R_0=1466\) kg (the reaction from the lower longitudinal beam p.). The force \(R_0\) has no influence on tension; it acts in another plane, in compression of another truss; therefore, in considering the truss \(abcd\), I shall take into account only the force \(Rs\), which tends to tear apart the ties 2, 6, 10 and 4, 7, 12 (see Fig. 6).

In calculating the leg by means of the truss diagram, they are divided into spans. The lateral supporting leg is divided into 3 spans, 3, 5, 11. Of these, spans 3 and 11 are calculated for longitudinal bending by Euler’s formula, as beams having one end fixed and the other free. Of these I shall confine myself to calculating only span 11, as the greatest in length and subjected to the greatest load. By Euler’s formula I have

\[ Pk=\frac{2\pi^2 I\varepsilon}{l^2}, \]

whence

\[ I=\frac{Pk\,l^2}{2\pi^2\varepsilon},\quad \varepsilon \text{ for oak }=108000\ \mathrm{kg}. \]

Such a moment of inertia is satisfied by a beam with cross-section \(\sim 21\times 21\) cm.

I calculate span 5 by Euler’s formula as a beam fixed at both ends.

\[ P_k=4\pi^2\frac{\varepsilon I}{l^2}, \]

whence

\[ Y=\frac{Pk\,l^2}{4\pi^2\varepsilon} =\frac{14700\cdot 360^2}{4\cdot 9{,}8596\cdot 188000}=424. \]

Thus, the beam is safe. Of spans 8–9, only span 9 need be considered, as it is the larger in length and in the forces acting on it.

Participating at the same time in another truss, they are there subjected to compression.

The force tearing it apart will be \(P = 12700\) kg. The compressive force is 31500 kg. (It enters the other truss as the 12th span.) Thus the tensile force is neutralized, and there remains only the force

\[ P_1 = 31500 - 12700 = 18800 \text{ kg}, \]

which will compress this span. Its calculation will be analyzed below.

Struts 2, 6, 10 are subjected to tension. Of these I shall consider only strut 2.

This strut is subjected to the greatest tensile force (2¼ times greater than strut 6, and 3¼ times greater than strut 10).

The tensile force is \(P = 1090\) kg.

\[ P = \omega k_z; \quad k_z \text{ for oak} = 100, \]

whence

\[ \omega = \frac{P}{k_z} = \frac{1090}{100} = 11 \text{ cm}^2. \]

In my case this same strut has a cross-section of 484 cm².

Braces 4, 7, 12 are subjected to compression. Both of their ends are fixed. Braces 4 will experience the greatest compressive force. In length, however, brace 12 will be the largest. We shall consider it and confine ourselves to it. By Euler’s 4th formula I have

\[ P = \frac{4\pi^2 \varepsilon Y}{l^2}, \]

whence

\[ Y = \frac{P l^2}{4\pi^2 \varepsilon} = \frac{2700 \cdot 750^2}{4 \cdot 9.8596 \cdot 108000} = 3570. \]

The middle legs constitute a truss in which the side spans 2, 6, 10, 14, 18 and 4, 8, 12, 16, 20 are subjected to compression, Fig. 5. Spans 2, 4, 18, 20 I regard as beams, of which one end is fixed and the other free. Of these spans I shall consider only the 4th, as having the greatest load \(P = 33200\) kg. By Euler’s formula I have

\[ P = \frac{2\pi^2 \varepsilon Y}{l^2}, \]

whence

\[ Y = \frac{P l^2}{2\pi^2 \varepsilon} = \frac{33200 \cdot 300^2}{2 \cdot 9.8596 \cdot 108000} = 1415. \]

Beams with a cross-section of approx. \(21 \times 21\) cm satisfy this moment of inertia.

Spans 6, 8, 10, 12, 14, 16 are also subjected to compression, but they are fixed at both ends. Spans 8 and 12, in the truss considered earlier, were subjected to tension; the compressive force is greater. Therefore span 8 (formerly 3) is subjected to compression by a force

\[ R = 32300 - 9400 = 22900 \text{ kg}. \]

Span 12 (formerly 5) is subjected to compression by a force \(P_1 = 31500 - 12700 = 18800\) kg, i.e., a force insignificant in comparison with the force of span 16, which has the greatest magnitude of load \(= 30900\) and a length, large in comparison with the others, \(l = 350\).

By Euler’s 4th formula, for this case I have

\[ P = \frac{4\pi^2 \varepsilon Y}{l^2}; \quad \text{whence} \]

\[ Y = \frac{P \cdot l^2}{4\pi^2 \varepsilon} = \frac{30900 \cdot 350^2}{4 \cdot 9{,}85 \cdot 96 \cdot 108000} = 895. \]

A beam of mine with dimensions \(40 \times 40\) cm. Its moment of inertia will be 213333.

The supporting beams 3, 7, 11, 15, 19 are also subjected to compression. Both their ends are fixed.

The greatest compressive force \(P = 1700\) kg will be at beam 3, length \(l = 440\) cm. Although a smaller force \(P_1 = 800\) kg will act on beam 19, its length \(l\) is considerably greater, \(l = 620\) cm. Therefore I shall consider only this beam, assigning to the remaining supporting beams dimensions identical with it. By Euler’s 4th formula I have:

\[ P = \frac{4\pi^2 \varepsilon Y}{l^2}; \quad Y = \frac{P \cdot l^2}{4\pi^2 \varepsilon} = \frac{800 \cdot 620^2}{4 \cdot 9{,}85 \cdot 96 \cdot 108000} = 72. \]

A beam with a cross-section of approx. \(5{,}5 \times 5{,}5\) cm satisfies this moment of inertia.

Beams 5, 9, 13, 17 are subjected to tension.

The greatest tensile force will be at beam 5. I shall confine myself only to its consideration, assigning to the remaining beams dimensions identical with it.

\(P = \omega k_z\), whence

\[ \omega = \frac{P}{k_z} = \frac{1100}{100} = 11 \text{ cm}. \]

Calculation of a Two-Legged Iron Headframe

By student K. M. Golubentsev.

We shall carry out its calculation for a rope 300 sazhen long. The cross-sectional area of the rope is \(w = 0.0006\ \mathrm{m.^2}\); the diameter of the rope is 41 mm; the diameter of the wires is 2.0 mm; their number is 180; the breaking force is 67860 kg; the weight of 1 m is 5.7 kg (Hütte).

The headframe is two-legged, metallic, drawing 7. Its trusses are directed in such a way that the resultant obtained from the summation of the forces breaking the rope passes within the legs, i.e., between the trusses, which is necessary, since otherwise an overturning moment may arise owing to the large horizontal component of the rope tension.

The load on the headframe reaches its maximum when, owing to extreme stresses or the jamming of one of the cages in the guides, the tension of the rope reaches the value of the breaking force; the headframe must be calculated for this latter condition, and two cases are possible: 1) the upper rope breaks, 2) the lower rope breaks. In the graphical calculation of the headframe we shall assume that the jammed cage is at the mouth of the shaft while the other is near the mine yard; the breaking load of the rope, equal to 67860 kg, I increase to 80000 kg, so that in what follows the forces from the self-weight of the parts of the headframe and the pressure of the wind need not be introduced. The weight of the cage, the rope, and the empty cars is about 5500 kg.

Case I: the tension at the moment of rupture of the upper rope is 80000 kg.

The resultant \(S_0 = 145000\) kg. Resolving this resultant along the trusses of the headframe, drawing 8, we obtain that the component along truss \(A\) is equal to 138000 kg, and along truss \(B\)—11000 kg.

The resultant of the tensions of the lower rope is equal to 10100 kg; the component along truss \(A\)—9600 kg, and along truss \(B\)—1650 kg. Thus the forces acting on the sheaves are 145000 kg and 10100 kg, and in the plane of truss \(A\)—138000 kg and 9600 kg, of truss \(B\)—11000 kg and 1650 kg. Determining graphically the forces \(P_1\) and \(P_2\), acting at the joints, and the reactions \(R_1\) and \(R_2\), drawings 10, 12, we construct the Cremona diagram, drawings 16, 17, and also drawings 14, 15, by which we determine the stresses for this case in the various parts of trusses \(A\) and \(B\).

Truss B.

Scale: 100 kg in 1 mm.

Member Nos.	Load in kg
1	9400
2	2650
3	800
4	2400
5	1800
6	9800
7	9700
8	7300
9	13020
10	7800
11	1310
12	610
13 (14)	6140
16	1650
17 (19)	3250
20	4900
21	12300
22	1350
23	2740
24	2100
26	1100
27 (32)	3550
28	2100
29	850
30	450
33	12300
34	1400
35	6600
36	6900
37	8550
39	1550
40	2150
41	11470
42	500
43	2300
44	8060
2′, 3′, 15, 18, 25, 31, 45, 46	0

Truss A.

Scale: 100 kg in 1 mm.

Member Nos.	Load in kg
1	98500
2	19000
3	5600
4	15000
5	2100
6	92200
7	23000
8	2000
9	4800
10	23000
11	86500
12	62500
13	41500
14	26500
15	40000
16	22200
17	35000
18	52000
19	31000
20	3000
21	70000
22	20000
23	75000
24	1500
25	3060
26	8000
27	4500
28	10000
29	78000

Case II. The tension of the lower rope is 80,000 kg; the resultant is 149,000 kg. The components of this resultant along the headframe trusses, Fig. 9, are as follows: along truss \(A\)—134,000 kg; along truss \(B\)—23,500 kg. The tension of the upper rope is 5,500 kg; the resultant is 10,000 kg. The components of this resultant along the trusses are: along truss \(A\)—9,000 kg; along truss \(B\)—1,050 kg. Thus, in this case, forces act on the sheaves in the plane of truss \(A\)—134,000 kg and 900 kg; in the plane of truss \(B\)—23,500 kg and 1,050 kg. Determining graphically the forces \(P_1\) and \(P_2\), acting at the joints of the trusses, and the reactions \(R_1\) and \(R_2\), Figs. 11, 13, we construct the Cremona diagram, from which we determine the stresses in the members of the trusses for this case, Figs. 18, 19.

TRUSS \(B\). Scale: 200 kg per 1 mm. Nos. of members	Load, kg
1	14800
2	3920
3	1200
4	3500
5	3000
6	15600
7	16000
8	12700
9	21800
10	12900
11	1500
12	700
13	6500
14	6500
16	1840
17 (19)	3600
20	5100
21	23300
22	3000
23	9400
24	7000
26	3520
27	12000
29 (34)	3000
30	1400
32	12000

TRUSS \(A\). Scale: 1000 kg per 1 mm. Nos. of members	Load, kg
1	80000
2	25800
3	5000
4	12000
5	2000
6	75000
7	18600
8	4000
9	2200
10	22500
11	70000
12	35200
13	49000
14	29000
15	22000
16	38000
17	58500
18	41000
19	27000
20	4000
21	79000
22	22000
23	84500
24	1500
25	3000
26	9000

TRUSS B.

Scale: 200 kg in 1 mm.

Nos. of members	Load in kg
33	23300
35	14300
36	14650
37	16700
38	19280
39	3800
40	4230
41	24800
42	1000
43	4440
44	17840
2′, 3′, 15, 18, 25, 31, 45, 46	0

TRUSS A.

Scale: 1000 kg in 1 mm.

Nos. of members	Load in kg
27	2300
28	11200
29	87508

Sections of the truss members.

All the members composing trusses “A” and “B” I divide into five categories, namely \(S, H, D, D^1, D^2\). I determine the sections of the members according to the greatest load obtained from comparing the two stresses arising in one and the same member when the upper rope breaks and when the lower rope breaks. Thus I determine the sections of the stressed members of truss “A.”

The category of members \(S\) includes members—1, 6, 11, 12, 17, 21, 23, 29 with a maximum load of 98500 kg. The sectional area for this load, with \(K = 4\) kg per 1 sq. mm, is equal to

\[ \frac{98500}{4} = 24625\ \text{sq. mm.} \approx 247\ \text{sq. cm.} \]

It consists of four unequal-leg angles \(115 \times 170 \times 14\), with an area of 38.1 sq. cm each, and of two plates \(250 \times 19\), with an area of 95 sq. cm. The total area is 247.4 sq. cm.

See the table of sections.

\[ I_x = \frac{250.600^3}{12} - \frac{250.544^3}{12} + \frac{28.544^3}{12} - \frac{28.222^3}{12} = 464718 \]

\[ I_y = 2 \cdot \frac{33.250^3}{12} + \left( \frac{534.28^3}{12} - \frac{220.28^3}{12} \right) \approx 8651 \]

The factor of safety against longitudinal bending of the longest member in category \(S\), i.e.

\[ 12\ \text{or}\ 17 \qquad n = \frac{8651.10.2000000}{98500.780^2} = 3.5. \]

Rods 10, 14, 19, with a maximum load of 31000 kg, belong to category of rods \(H\). The cross-sectional area for this load, with \(k = 4\) kg per 1 sq. mm, must be not less than

\[ \frac{31000}{4}=77.5\ \text{sq. cm.} \]

and consists of two channel irons No. 22 according to the Russian assortment, with a useful area of 38.94 sq. cm each.

\[ I_x = 5662 \]

\[ I_y = 280783 \]

The factor of safety against longitudinal bending at the greatest length of the 10th rod is

\[ n=\frac{5662\cdot 10\cdot 2000000}{31000\cdot 700^2}=8. \]

Rods 13, 15, 16, 18, with the greatest load of 52000 kg, belong to category of rods \(D\). The useful area at a stress of 4 kg per 1 sq. mm must be equal to

\[ \frac{52000}{4}=130\ \text{sq. cm.} \]

and consists of four angle irons \(160 \times 100 \times 13\), with an area of 32.9 sq. cm each. The total area of the four angle irons is 131.6 sq. cm, i.e., greater than required.

\[ I_y=\frac{200\cdot 600^3}{12}-2\frac{574^3\cdot 87}{12}-\frac{26\cdot 280^3}{12}=81387 \]

\[ I_x=2\frac{200^3\cdot 13}{12}+2\frac{26^3\cdot 147}{12}=1780. \]

The factor of safety against longitudinal bending at the greatest length of rods 16 or 18, equal to 560 cm, is

\[ n=\frac{1780\cdot 10\cdot 2000000}{52000\cdot 560^2}=2.5. \]

Rods 8, 9, 5, 3, 27, 20, 24, 25, with the greatest load of 5600 kg, belong to category of rods \(D^1\). With the allowable stress of 4 kg per 1 sq. mm, the section must have a useful area of

\[ \frac{5600}{4}=14\ \text{sq. cm.} \]

I make it of two angle irons \(100 \times 50 \times 5\), with an area of 7.28 sq. cm each; their total area is 14.56 sq. cm. After checking for longitudinal bending, these angle irons proved insufficient, and it was necessary to take angle irons \(75 \times 140 \times 11\), with an area of 22.5 sq. cm each.

\[ I_y = 45317.7 \]

\[ I_x = 902.2 \]

The factor of safety against longitudinal bending, for the greatest length of member 9, equal to 890 cm, is

\[ n=\frac{902{,}2\cdot 10\cdot 2000000}{5600\cdot 890^2}=4. \]

The category of members \(D^2\) includes members 2, 4, 7, 22, 26, 28, with the greatest load of 25800 kg. With an allowable stress of 4 kg per 1 sq. mm, the required cross-sectional area of the members for this load is

\[ \frac{25800}{4}=64{,}5 \text{ sq. cm.} \]

I take two angles \(100\times160\times13\), with an area of 32.25 kg each.

\[ I_y=48854{,}6, \]

\[ I_x=1672{,}8 \]

The factor of safety against longitudinal bending, for the greatest length of 530 cm, is

\[ n=\frac{1672{,}8\cdot 10\cdot 2000000}{25800\cdot 530^2}\simeq 4{,}5 \]

The category of members \(S\) of truss \(B\) includes—1, 6, 11, 17, 19, 24, 28, 29, 38, 44, with the greatest load of 19280 kg. With an allowable stress of 4 kg per 1 sq. mm, the required cross-sectional area of the members for this load is

\[ \frac{19280}{4}=48{,}2 \text{ sq. cm.} \]

I put in two angles \(115\times170\times14\), with a net area of \(38{,}1\ \text{cm}^2\) each.

\[ I_y=22894 \]

\[ I_x=2213{,}6,\quad \eta=55{,}3\ \text{mm.} \]

The factor of safety against longitudinal bending, for the greatest length of the member of 530 cm, is

\[ n=\frac{10\cdot 2000000\cdot 2213{,}6}{530^2\cdot 19280}=4. \]

The category of members “\(H\)” includes members—9, 10, 15, 18, 21, 25, 31, 36, 41, with the greatest load of 24860 kg. With a safe stress of 4 kg per 1 sq. mm, the required cross-sectional area of the members is equal to

\[ \frac{24860}{4}=62{,}15\ \text{cm}^2. \]

I put in two channel irons No. 16, with a net area of \(24{,}92\ \text{cm}^2\) each. With this area of the members, the allowable stress will be about 5 kg per 1 sq. mm, which is permissible.

\[ I_y = 89.0 + 2 \cdot 24.92 \cdot 28.14^2 = 39557.8 \]

\[ I_x = 2 I_\xi = 2 \cdot 954 = 1908. \]

The factor of safety against longitudinal bending, for the greatest length of the 9th member, is

\[ n = \frac{10 \cdot 2000000 \cdot 1908}{24860 \cdot 360^2} = 10.4. \]

The category of members “\(D\)” includes members 13, 14, 27, 32 with the greatest load of 12000 kg. With a safe stress of 4 kg per 1 sq. mm, the required cross-sectional area of the members is

\[ \frac{12000}{4} = 30 \ \mathrm{cm}^2. \]

I take two angles \(90 \times 100 \times 12\), with a useful cross-sectional area of \(21.5 \ \mathrm{cm}^2\) each.

\[ I_y = I_\xi + 2 \omega a^2 = 199.8 + 2 \cdot 21.5 \cdot 169.7^2 = 12527 \]

\[ I_x = 2 I_\eta = 2 \cdot 152.3 = 304.6. \]

The factor of safety against longitudinal bending, for the greatest length of the member, at 350 cm, is

\[ \text{we have } n = \frac{10 \cdot 2000000 \cdot 304.6}{350^2 \cdot 1200} = 4. \]

The category of members “\(D^1\)” includes members 7, 8, 35, 37 with the greatest load of 16700. With a safe stress of 4 kg per 1 sq. mm, the required cross-sectional area of the members is

\[ \frac{16700}{4} = 42 \ \mathrm{cm}^2. \]

I place two angles \(90 \times 120 \times 12\), with a useful area of \(23.9 \ \mathrm{cm}^2\) each.

\[ I_y = I_\xi + 2 \omega a^2 = 334.6 + 2 \cdot 23.9 \cdot 161.7^2 = 26591.4 \]

\[ I_x = 2 I_\eta = 2 \cdot 161.6 = 323.2. \]

The factor of safety against longitudinal bending, for the greatest length of the 35th member, equal to 305 cm, is

\[ \text{we have } n = \frac{10 \cdot 2000000 \cdot 323.2}{305^2 \cdot 16700} = 4, \]

The diagonal members \(D_2\) must have a cross-sectional area, at the greatest force of 9400 kg and at an allowable stress of 4 kg per 1 sq. mm, equal to

\[ \frac{9400}{4} = 24 \ \mathrm{cm}^2. \]

This category includes members 2, 2′, 3, 3′, 4, 5, 12, 16, 20, 22, 23, 26, 30, 34, 39,

40, 42, 43, 45, 46, I install two angles \(75 \times 100 \times 10\) cm, with a net area of \(16.6\ \mathrm{cm}^2\) each.

\[ I_y = I_e + 2\omega a^2 = 78.9 + 2 \cdot 16.6 \cdot 180.5^2 = 10835.7 \]

\[ I_x = 2I_\eta = 162.2 \cdot 2 = 324.4. \]

The factor of safety against longitudinal bending, for the greatest length of the member, \(420\) cm,

\[ \text{we have } n=\frac{10 \cdot 2000000 \cdot 324.4}{420^2 \cdot 9400}=4. \]

Examining the legs of the angles and the channel irons that make up the members of trusses “A” and “B,” we see that their thickness is equal to 10, 11, 12, 13, 14 mm; therefore, the packing plates required at the joints of the trusses, drawings 34, 35, 36, I take as \(1.5\) times the average thickness of the legs, i.e. \(13.5\) mm. The smallest diameter of the rivets used is

\[ d=\sqrt{5s}-0.2=1.34-0.2=12\ \mathrm{mm}. \]

Taking into consideration the size of the cover plates, which is determined by the number of rivets for each member entering the joint, the diameter of the rivets must be made 16, 22, and 26 mm, i.e. of such sizes as are always commercially available. In some joints it is necessary to increase the diameter of the rivets while also observing the condition that their length not exceed \(2.5\) diameters. With an allowable shear stress of \(500\) kg per \(1\ \mathrm{sq.\ cm}\), each rivet takes a force

\[ S=\frac{\pi d^2}{4}\,500\ \mathrm{kg}. \]

i.e. for a diameter of 12 mm — 650 kg,

    ”        ”        ”    16  ” — 1000  ”

    ”        ”        ”    22  ” — 2300  ”

    ”        ”        ”    26  ” — 3200  ”

Having these data, the number of rivets for each member at the joints is determined by the quotient obtained by dividing the corresponding load acting on the member by the allowable load for a rivet, which was also done in designing the details of the trusses on the drawing.

The force acting on the sheaves is transmitted to trusses “A” and “B” through supports and brackets by means of two beams riveted to the upper parts of the trusses (see detail drawing) 20, 21, 22, 23, 24.

Let us determine the section of these beams:

Beam No. 1 for truss “A.”

From the diagram of moments for truss “A,” the maximum moment is equal to \(69000 \cdot 200 = 13800000\) kg·cm. \(M_{mx} = Wkb = 13800000\); I take \(kb = 1000\) kg per \(1\ \mathrm{cm}^2\).

\[ I=\frac{M_{mx}H}{2\cdot 1000} \]

I determine the height of the beam from preliminary calculations to be 850 mm. Consequently, the moment of inertia of the beam is

\[ I=\frac{13800000\cdot 42{.}5}{1000}=586500. \]

To obtain such a moment of inertia, I make the section of the beam as follows—4 angles \(170 \times 115 \times 14\) and 4 cover plates \(850 \times 18\).

\[ I_x=\frac{85^3\cdot 90{.}2}{12} -\left( \frac{85^3\cdot 60}{12} +2\frac{82{.}2^3\cdot 10{.}1}{12} +2\frac{51^3\cdot 1{.}4}{12} \right)=587990. \]

Which is quite sufficient.

The rivets by which this beam is riveted to the truss I set at a diameter of 26 mm. Their number is equal, according to the greatest of the forces \(P_1\) and \(P_2\), i.e. 94000 kg,

\[ n=\frac{94000}{3200}=30 \]

or, on each side of the truss, 15 rivets.

Beam No. 2 for truss “B.”

From the diagram of bending moments for truss \(B\), the maximum moment is equal to

\[ M_x=11750\cdot 200=2350000\ \text{kg·cm}. \]

\[ Wkb=M_x,\qquad I=\frac{M_xH}{2kb} \]

I take \(Kb=1000\) kg per \(1\ \text{sq. cm}\) and \(H=500\) mm; then, for the beam to be strong, the moment of inertia of its section must be equal to

\[ I=\frac{2350000\cdot 25}{1000}=58750. \]

I make the section of the beam as follows:

4 angles \(170 \times 115 \times 14\) and 2 cover plates \(500 \times 20\).

The moment of inertia \(I_x\) of such a section is equal to

\[ I_x=\frac{50^3\cdot 67}{12} -\left( \frac{50^3\cdot 40}{12} +2\frac{47{.}2^3\cdot 10{.}1}{12} +2\frac{16^3\cdot 1{.}4}{12} \right)=103285 \]

i.e. greater than required.

The rivets by which this beam is riveted to truss “B” I set at a diameter of 22 mm. Their number is determined by the greatest of \(P_1\) and \(P_2\), i.e. 16000 kg, and is equal to

\[ \frac{160}{23}=7, \]

on each side of the truss I set 5 pcs., i.e. 10 in all, and not 7.

The brackets, as seen from the detailed drawing, consist of two shaped plates 18 mm thick, riveted to the angles and cover plates.

There is no need to doubt the strength of their dimensions, since their cross-section far exceeds the section of the members of category “S,” which are subjected to a greater load than they are. The rivets on the shaped plates are placed at intervals of 8 diameters.

Sheave beams.

Each of the bearings of the sheave shaft rests on a beam, through which the load is transmitted to the sheave brackets.

The resultant force of 14900 kg is transmitted to two supports, i.e., to the bearings of the sheave shaft; consequently each receives 74500 kg. Resolving this force into horizontal and vertical components, we obtain the first equal to 31000 kg, and the second—64000 kg. The horizontal force of 31000 kg tends to shift the beam from the brackets in the direction of the drum; therefore I fasten it to the brackets by means of angles. I use rivets of 22 mm diameter; consequently their number must be equal to

\[ \frac{31000}{2300}=14, \]

i.e., 7 on each side of the bracket.

Owing to the presence of the horizontal force of 31000 kg and an arm of 350 mm (center of the bearing), the beam is subjected to a moment of \(31000 . 35 = 1085000\), and to a vertical force of 64000 kg. Thus, in the beam there arise normal stresses, directed parallel to its axis, from bending, and shearing stresses, perpendicular to the axis of the beam, from the vertical force of 64000 kg. The height of the beam, for structural reasons, must be 320 mm; by a preliminary calculation I come to the conclusion that it must have a cross-section with moment of inertia

\[ I_x=\frac{32.31,5}{12}+419+4.31,82.12,52^2=24684. \]

In the following calculations I take only \(I_x\), since we have plane bending, where the plane of action of the force coincides with the plane of bending.

Thus, let us determine what the stress of the material is under the existing loads on the beam with the above-described cross-section. From the presence of the bending moment there arise normal stresses \(k_z'\) and \(k_z''\), equal to 12.

\[ k_z'=\frac{P}{\omega}=\frac{31000}{175.3}=172.4\ \text{kg per 1 sq. cm, and } k_z''=\frac{M_{mx}e}{I} \]

\[ =\frac{1085000\cdot 16}{24684}=703\ \text{kg}. \]

From the presence of a vertical force of 64000 kg, a tangential stress \(k_s\) arises, equal to

\[ k_s=\frac{64000}{175.3}=365\ \text{kg}. \]

The resultant stress I determine by the Sch—Yenan formula:

\[ k=\frac{3}{8}(k_z'+k_z'')+\frac{5}{8}\sqrt{(k_z'+k_z'')^2+4k_s^2} =\frac{3}{8}875^2+\frac{5}{8}\sqrt{875^2+4365^2} \]

\[ =328+712=1040\ \text{kg per 1 sq. cm.} \]

In view of the enormous margin in the overall calculation of the headframe, I consider such a stress permissible. From the considerable vertical force of 64000 kg and the height of the beam of 320 mm, buckling of the vertical web of the beam about the vertical axis is possible; therefore I consider it necessary to check this web for longitudinal bending. The load is transmitted to the beam through the bearing tension plate; therefore the part of the beam subjected to this load I take as equal to the length of the tension plate, i.e., consequently, the moment of inertia of the section of this web relative to the vertical axis is equal to

\[ I_y=\frac{110\cdot 1.5^3}{12}=31 \]

the safety factor is

\[ n=\frac{2\cdot \pi^2\cdot \varepsilon\cdot I}{P\cdot l^2} =\frac{20\cdot 2000000\cdot 31}{64000\cdot 32}\approx 20. \]

Thus, in this respect also, the beam is strong.

Pulleys and their axle.

I calculate the axle as a beam freely resting on supports and subjected to a force of 149000 kg.

\[ \frac{149000\cdot 30}{2}=0.1d^3k_b. \]

\(k_b\)—I take 1500 kg for steel. I take this limiting allowable stress in view of the fact that the force of 149000 kg is possible in an exceptional case:

\[ d=\sqrt[3]{\frac{149000\cdot 30}{1500\cdot 2}}=246\ \text{mm}. \]

I adopt \(d=250\) mm. The journal:

\[ M_m=\frac{149000\cdot 15}{15}=0.1d^3k_b. \]

\[ d=\sqrt[3]{7450}=195\ \text{mm}, \quad \text{I adopt } d_1=200\ \text{mm}. \text{ The length under normal-} \]

is equal to \(\dfrac{106000}{11}=9650\ \mathrm{cm}^2\); taking structural considerations into account, I make it \(120 \times 120 = 14400\ \mathrm{cm}^2\).

I determine the average thickness of the bearing plate in such a way that I consider half the length of the plate as a freely supported beam loaded by one half of the bearing pressure, uniformly distributed (Hütte, p. 412, vol. II).

\[ \frac{53000 \cdot 60}{4} = M_{mx} = 744000 \quad W k_b = 79500\, k_b - 1000\ \mathrm{kg}. \]

For steel,

\[ \frac{a h^2}{6} = 795 \qquad h^2 = 79.5 \qquad h = 90\ \mathrm{mm}; \]

I make it \(100\ \mathrm{mm}\).

The legs of the headframe rest on a foundation, which I make of concrete. Since the load is transmitted to the foundation vertically, whereas our force is directed along the headframe legs, i.e. obliquely, there consequently arises in the foundation a horizontal component equal to \(51000\ \mathrm{kg}\), which tends to shift the upper part of the foundation relative to the lower part along the plane of least resistance. Taking the horizontal plane passing through the axis of the plate as the most dangerous, owing to its smaller dimensions in comparison with the others, let us see what stress in the concrete will occur if the horizontal component appears in this plane. The area is \(134 \times 160 = 21440\ \mathrm{cm}^2\); the stress will be

\[ \frac{51000}{21440} = 2.5\ \mathrm{kg}, \]

which is entirely permissible. The dimensions of the foundation base are governed by the necessity that the direction of the force acting along the legs should not pass outside the core of the section of the base; the height of the foundation in the ground is \(1.5\ \mathrm{m}\).

The bearing surface of the shoe of the truss legs \(B\) is \(9 \times 40 = 360\ \mathrm{sq.\ cm}\). The acting force is \(17800 + 7200 = 25000\ \mathrm{kg}\); the stress per \(1\ \mathrm{sq.\ cm}\) is

\[ \frac{25000}{360} \approx 70\ \mathrm{kg}. \]

The area of the plate, with an allowable stress on cement of \(11\ \mathrm{kg}\) per \(1\ \mathrm{cm}^2\), will be

\[ \frac{25000}{11} = 2280\ \mathrm{cm}^2; \]

I make it \(80 \times 80\), with a thickness of \(80\ \mathrm{mm}\). The dimensions of the foundation are determined on the basis of the same considerations as for truss \(A\).

…causing the headframe to overturn, cannot be expected in the present headframe, since the resultant of all the forces passes between the headframe trusses, and the headframe, being above the mine building, is not subjected to the action of the wind. The perpendicularity of the bearing plates to the planes of the headframe trusses eliminates the occurrence of a horizontal component at the bearings, by virtue of which the spreading apart of the trusses, and consequently the separation of the headframe legs from the bearing plates, would be possible. A horizontal component at the supports, situated in the plane of the trusses—although of a higher order and insignificant—does, however, exist in my truss design and may therefore cause slipping off the supports. Such an assumption is almost impossible to make with respect to truss “A,” first because it is the heaviest in construction and, chiefly, because the resultant of all the forces is directed almost in the plane of this truss; and second, because in this truss there is member 10, which is a tie, taking upon itself the horizontal thrust of the arch-truss. Taking all this into consideration, I make the hinges at truss A free. Drawings 25, 26, 27, 28, 29, 30.

For truss “B,” however, I make the hinges anchored, i.e., connected with the plate, drawings 31, 32, 33, bearing in mind its special construction and the insignificant load, the component of which, lying in the plane of the truss, is not capable of producing, in the shoe-plate, a friction force sufficient to prevent the truss from sliding under the horizontal component situated in the plane of the truss and equal to about 900 kg. All the dimensions of the parts of the anchor connection—such as the protruding anchor pin at the shoe, the bolts, and the other parts—in view of the insignificance of the force acting on them, 900 kg, and of the steel of which they are made, which permits a large load, I determine not by calculation, but exclusively according to structural considerations. The bearing area in the shoe for truss A in my case is equal to \(60 \times 12 = 720\ \mathrm{cm.^2}\); the greatest stress in members 1 and 29 is equal to 98,000 kg. Adding another 8,000 kg from the truss’s own weight, we obtain the load acting on the shoe, equal to 106,000 kg. The stress over the bearing area will be

\[ \frac{106000}{72000} \sim 1.5\ \text{kg per 1 sq. mm.} \]

I make the foundation of the headframe legs of cement; with an allowable stress of 11 kg per 1 cm², therefore, in order to transmit to the support the acting load of 106,000 kg, the area of the plates must be

the bearing is taken as \(1.5d = 300\) mm. The pressure on the liners during operation is

\[ \frac{149000}{8.2}\ \mathrm{kg} = 30.20\,q, \]

where \(q\) is the allowable stress for bronze,

\[ q = \frac{149000}{16 \cdot 30.20} \simeq 15\ \mathrm{kg}, \]

which is possible.

All dimensions of a normal bearing are taken from Berlov and are indicated in the detailed drawing. Pulley hub: the thickness is equal to

\[ \frac{1}{4}\left(\frac{3}{2}d\right) + 1\ \mathrm{cm}. \]

(Hütte, p. 661)

\[ \frac{3}{8}d + 1\ \mathrm{cm} = 10.5\ \mathrm{cm}. \]

I take \(120\) mm; the outside diameter is \(d_0 = 240 + 250 = 490\) mm; the length of the hub is \(580\) mm.

I calculate the spokes of the pulleys according to “Tyme,” i.e., proceeding from the condition that two spokes must offer strong resistance to the resultant of the tensile forces:

\[ 149000 = 2Fk,\qquad F = \frac{149000}{2 \cdot 950} = 80\ \mathrm{cm}^2. \]

I take \(k = 950\) kg per sq. cm. I make the thickness of the spokes \(40\) mm and the width \(200\) mm; the number of spokes is \(8\). The bolts fastening the spokes to the rim and to the hub I calculate for shear under the force

\[ \frac{149000}{2}\ \mathrm{kg}. \]

For structural reasons it is possible to place 2 bolts each; consequently, the area of each, with \(k = 1200\) kg, is equal to

\[ \frac{149000}{48000} = 31,\qquad d \simeq 63\ \mathrm{mm}, \]

or \(2\frac{3}{4}\).

Supports.

In the graphical calculation of the headframe trusses, the connection at the supports was assumed to be hinged, as a result of which the reactions of the supports were also placed in the plane of the trusses, i.e., had a direction perpendicular to the bearing plates. This was, first; secondly, in the case of the trusses of my headframe, which closely approximate arches with a span of up to 15 meters, the elongation of the members from temperature changes will be considerable; therefore, in order not to cause additional stresses in the members owing to their elongation and shortening, I must make the supports of the headframe hinged. Every headframe is designed in such a way that the appearance of an overturning moment is impossible, i.e., in other words, separation of the headframe legs from the bearing plates must not occur under any circumstances. Indeed, the appearance of a horizontal component situated in the plane of the pulleys, and

Iron Six-Legged Headframe

by student V. P. Selivanov.

It is required to determine the dimensions of an iron headframe for the extraction of coal in the amount of 15,000,000 poods per year. Given:

1) Depth of the mine . . . . . \(H = 210\) m.

2) Diameter of the drum . . . . \(D_1 = 3.2\) m. (Determined from the initial dimensions of the rope).

3) Diameter of the rope sheave . . \(D_2 = 3.2\) m.

4) Height of the axis of the rope sheave above ground level . . . . . \(\simeq 20\) m.

5) Distance from the center of the drum to the vertical axis passing through the center of the rope sheave (determined from the condition of convenience in winding the rope onto the drum); it is equal to . . . \(\simeq 20\) m.

6) Dead load:

1) Weight of a two-deck cage . . . 3000 kg.
2) “ of 4 empty cars . . . 1200 “
3) “ of the rope . . . . . 520 “

————————————————
We take, in round numbers . . . 4800 kg.

7) Useful load:

4) In 4 cars, 500 kg each \(= 2000\) kg.

8) Total load:

\[ 4800 \text{ kg.} + 2000 \text{ kg.} = 6800 \text{ kg.} \]

9) The breaking load for the rope selected from the tables with a sixfold factor of safety is \(40800\) kg.; I take 42000 kg.

10) For the calculation I choose a headframe structure consisting of 4 vertical and 2 inclined legs, drawings 57, 58, 59. The inclined supporting leg, representing in its entirety a braced truss directed along the resultant obtained from the composition of the forces caused by the tension of the upper rope. When the resultant is directed along the braced truss, it obtains more substantial dimensions, while the vertical truss becomes lighter.

11) Hoisting is carried out by cylindrical drums, which is quite permissible at a mine depth of 210 met.

The headframe must be constructed so strongly that, in the most unfavorable position of the cage being jammed at the moment of rope breakage, the stress in the material does not exceed the elastic limit—for cast iron.

In this calculation, the stresses caused by wind and weight are taken into account.

Here, as is known, there may be two most dangerous cases.

1) The upper rope breaks when its loaded cage is near the level of the ground surface, while the cage of the lower rope has descended to the depth of the mine shaft yard.

2) The lower rope breaks when its cage is near the level of the ground surface, while the cage of the upper rope reaches the depth of the mine shaft yard.

The upper rope breaks.

We take the scale of forces as \(1\ \mathrm{mm} = 500\ \mathrm{kg}\); then the stress from the breakage of the upper rope will be expressed by the vector:

\[ 42000 : 500 = 84\ \mathrm{mm}. \]

Setting off from the vertex of the angle formed by the rope over the sheave, in the two directions of the rope toward the shaft and toward the drum, by 84 mm, we obtain the magnitude of the resultant, directed in the plane of the strut truss, Fig. 37, 38; it is equal to

\[ R_1 \simeq 152\ \mathrm{mm} \simeq 76000\ \mathrm{kg}. \]

The resultant from the action of the lower rope is obtained equal to

\[ S_1' \simeq 18\ \mathrm{mm} \simeq 9000\ \mathrm{kg}. \]

Let us resolve it in two directions:

1) Vertically, \(V_1' = 2.5\ \mathrm{mm} = 1250\ \mathrm{kg}\).

2) In the direction of the strut truss, \(R_1' = 15.7\ \mathrm{mm} \simeq 8000\ \mathrm{kg}\).

Thus, in the plane of the strut truss, two forces act on the sheave supports: 76000 kg and 8000 kg; i.e., for each bearing of the sheave with the lower rope there is:

\[ A\ 76000 : 2 = 38000\ \mathrm{kg}. \]

and for each bearing of the sheave with the lower rope:

\[ B = 8000 : 2 = 4000\ \mathrm{kg}. \]

In order to determine the stresses in the members of the (supporting) strut truss, we must first find the magnitudes of the reactions acting at the upper joints of the truss. The cap of the strut truss, consisting of two shaped plates, forms a beam lying on three supports (the three upper joints of the truss).

When selecting the section of the shaped plates representing the upper beam, we must take the local deflection into account.

There are three methods for determining the reactions of a three-support beam:

1) The support reactions are determined by Castigliano’s theorem, which states:

“The derivative of the work with respect to a force is equal to the displacement of the point of application of the force.”

Since at the point of the middle support there is no displacement of the member, the derivative of the work with respect to the reaction is therefore \(=0\);

\[ \frac{dG}{R}=0 \]

From this condition the magnitudes of the reactions can be determined.

2) The second method is based on Clapeyron’s theorem, according to which we first find the support moments; from the moments we determine the shearing forces, and from the latter we determine the support reactions, as equal to \(R+S\), the sum of the shearing forces about the support.

3) The third method consists in first assuming the middle support to be absent; finding the deflection produced by the given load at the point of application of the middle support; from it one can easily find the concentrated force applied at the same point and producing the same deflection. This force will give us the magnitude of the reaction of the middle support, since there must be no deflection at the point of application of the middle support.

The last method is especially convenient graphically; therefore it is the one adopted for the present calculation. Thus, first assuming the middle support to be absent and constructing the curve of deflection on any convenient scale (the construction is shown in drawing 39), we obtain the magnitude of the deflection under the absent middle support,

\[ f = 22 \ \mathrm{mm}, \]

corresponding to the unknown middle reaction \(X\).

Next, we assign an arbitrary force, usually one producing the greatest moment under the given support, in our case:

\[ Q = 42000 \ \mathrm{kg}. \]

This concentrated force gives the deflection:

\[ f = 19 \ \mathrm{mm}. \]

to check whether the total stress from the action of: 1) the cable, 2) the wind, and 3) the weight does not exceed the permissible limit.

If the permissible stress is exceeded, we again carry out the calculation, but with a greater margin of strength, until the desired result is obtained.

The first condition of static determinacy of the strut truss is that the joints of the truss be hinged; this means that we must not obtain bending moments in the joints that would alter our calculation. In practice, as in our case, the joints are more often made non-hinged, and in order in such a case to avoid, as far as possible, bending moment in the joints and to determine the forces graphically, we must compose our truss of rods as long as possible. That is why I divide our strut truss, consisting of two legs, by horizontal and diagonal rods into only five panels.

The right leg of the strut truss consists of rods 22, 14, 10, 6, 2.

And the left leg of Nos. 18, 15, 11, 7, 3.

The length of the leg within one panel is equal to: $l = 360\ \mathrm{cm}$.

As is seen from the table, the greatest force in the leg of the strut truss occurs in panel No. 15; it is equal to: $55400\ \mathrm{kg}$.

Within each panel the leg works in longitudinal bending, caused by the tension of the cable, by the wind, and by the weight.

First of all we choose the form of the cross-section of the leg; for longitudinal bending it is important to obtain as large a minimum moment of inertia as possible; moreover, the following conditions must be observed:

1) ease of fabrication,
2) minimum riveting,
3) a form convenient for scrapings.

For our case, therefore, the most suitable section will be an I-section, with wide flanges in order to increase the minimum moment of inertia. The height of the web of the I-beam must be sufficient so that it does not buckle, and so that the beam can work in bending from the component force of the weight.

For small bending forces the height of the beam is taken as not less than 0.03 of the span.

The necessary minimum moment of inertia is determined by Euler’s formula for the second case of end fixing of a rod:

\[ I = \frac{P \cdot l^{2} \cdot \eta}{\pi \cdot E}; \]

The resultant of the tension of the upper rope is directed in the plane of the diagonal truss; it is equal to

\[ R_2 = 17.5 \ \text{mm.} \cong 9000 \ \text{kg.} \]

Thus, in the plane of the diagonal truss, two forces act on the pulley supports: 68000 kg and 9000 kg.

The magnitudes of the reactions of the three supports of the imaginary head beam are found by the graphical method described above; in drawings 40 and 42 they are obtained as equal to:

\[ 1_1' = 900 \ \text{kg.} \]

\[ 2_1' = \frac{39 \cdot f'}{f} = \frac{39 \cdot 20.5}{17.5} = 45700 \ \text{kg.} \]

\[ 3_1' = 30400 \ \text{kg.} \]

Adding, for verification, the three reactions obtained, we get the numerical value of the load:

\[ 900 \ \text{kg.} + 45700 \ \text{kg.} + 30400 \ \text{kg.} = 77000 \ \text{kg.} \]

The check shows us the correctness of the data obtained by the graphical method.

The stresses acting in the members of the diagonal truss are found from Maxwell’s diagram, drawings 45 and 46, for the second case of rope breakage. Tension and compression are indicated by the same conventional signs as before.

The numerical magnitudes of the forces are entered in the same table in which the forces in the members of the diagonal truss for the first case of the breaking rope are placed. Table II.

The same members are placed in one row, for the convenience of comparing the forces acting in them in both cases of rope breakage. Comparing these forces, it is seen that for the case of breakage of the upper rope the forces obtained are greater; therefore we adopt them for the calculation.

Diagonal truss.

Let us proceed to determine the dimensions of the parts of the diagonal truss. The stress in the members must not exceed the limit adopted by us, 875 kg per 1 sq. cm.

We shall calculate all cross-sections of the members of the diagonal truss with a somewhat larger factor of safety; having obtained the dimensions, we obtain also the weight of the diagonal truss, after which we shall have the possibility of checking—

But since the forces are proportional to the deflection ordinates, we obtain the following equality:

\[ X : 42000 = 22 : 19. \]

Whence:

\[ X = 42000 \cdot \frac{22}{19} = 48000\ \text{kg}. \]

This, then, will be the magnitude of the reaction of the middle support. It is now easy to find, by the usual graphical method, the reactions of the end supports, taking the reaction of the middle support as a concentrated force acting on the beam from below. Fig. 41. The magnitudes of the reactions of the end supports were obtained as follows:

\[ 1' = 34700\ \text{kg}. \]

\[ 3' = 700\ \text{kg}. \]

Adding, as a check, all three reactions obtained,

\[ 34700 + 48600 + 700 = 84000\ \text{kg}, \]

we obtain the magnitude of the given load.

Having obtained the reactions of the supports, we find the stresses in the members of the inclined truss by the graphical method according to Maxwell’s method (Cremona). Figs. 43, 44.

On the members of the schematic truss, compression and tension are shown by arrows directed toward the joints and away from the joints.

The numerical results of the stresses in the members of the inclined truss are entered in a special Table I placed on the drawing. In selecting the sections of the members of the inclined truss, we shall make use of the data of this table.

The lower rope breaks.

The tension of the lower rope is

\[ 42000 : 500 = 84\ \text{mm}. \]

Again we find the resultant:

\[ S'_2 = 158{,}5\ \text{mm} = 79250\ \text{kg}. \]

We resolve it:

1) In the vertical direction

\[ V'_2 = 23{,}5\ \text{mm} = 11750\ \text{kg}. \]

2) In the direction of the inclined truss

\[ R_2 = 136{,}2\ \text{mm} = 68000\ \text{kg}. \]

The tension of the upper rope is equal to:

\[ Q = 4800\ \text{kg} \sim 9{,}6\ \text{mm}. \]

Here:

1) \(P = 55400\ \mathrm{kg} \simeq 56000\ \mathrm{kg}\)—the force producing longitudinal bending.

2) \(l^2 = 360^2 = 129600\)—the square of the panel length.

3) \(E = 2150000\)—Young’s modulus for wrought iron.

4) \(\eta = 6\); the factor of safety was taken equal to 5; in our case, in view of the forces from wind and from weight, we increased it.

Thus:

\[ I=\frac{56000\cdot129600\cdot6}{10\cdot2150000}=2025. \]

We turn to the handbook, in the section “Construction of bridges” (Hütte, p. 400); the dimensions of riveted beams are given. After a preliminary selection in sketches, a beam was chosen for which: see Table III.

1) Size of angles \(= \simeq 90.90.11\ \mathrm{mm}\).

2) Size of flange plates \(= \simeq 200.11\ \mathrm{mm}\).

3) Diameter of rivets \(= 2\ \mathrm{cm}\).

4) Height of web \(= \simeq 60\ \mathrm{cm}\).

5) Weight of 1 running meter of the beam \(= 104.8\ \mathrm{kg} + 68.64\ \mathrm{kg} = 173.5\ \mathrm{kg}\).

6) Weight of the leg within the panel \(= 173.5 \times 3.6 = 724.6\ \mathrm{kg}\).

7) Cross-sectional area of the beam \(= 179\ \mathrm{cm}^2\).

The rivets are not deducted, since the beam does not work in tension.

The least moment of resistance with respect to the axis \(yy\) has to be determined by dividing our beam into rectangles, whose moments of inertia with respect to any axis are known to us; thus the moment of inertia of the I-beam is obtained as equal to:

\[ I_{min}=2\left[ \frac{2.2\cdot1.1\cdot8^3}{3} + \frac{2\cdot1.1\cdot7.5^3}{3} + \frac{2\cdot7.9\cdot1.6^3}{3} + \frac{42\cdot0.5^3}{3} \right] =2\cdot1084=2168\ \mathrm{cm}^4. \]

In this calculation the diameter of the rivet is subtracted from the height of the flanges, while we neglect the weakening by the rivets fastening the angles to the web. Thus we required a beam with \(I_{min}=2025\ \mathrm{cm}\), and the beam selected by us gives \(I_{min}=2168\ \mathrm{cm}\); we consider the beam suitable.

The height of the riveted beam, as it turns out, may be taken as 40 cm, without the risk of falling below the least required moment of inertia \(I_{min}=2025\ \mathrm{cm}^4\); in selecting the greater height of the beam, we had in mind that the legs of the sloping truss are subjected to bending from the component of the force of weight; moreover, in order that buckling of the web of the I-beam should not occur, the height of the latter must not be less than 0.03 of the span,

as we mentioned above. The span within which we consider the beam to be bending is equal to: \(360 \times 5 = 18\) m. (more precisely, 18.2 m.) Therefore, the depth of the web of the I-beam must not be less than:

\[ H > 18\,0.03 = 54\ \text{cm}, \]

whereas we take the depth of the web of the I-beam to be equal to 60 cm.

The head of the strut truss, representing members 17 and 23 of drawing 43 of the schematic truss, works in compression and local bending, since the members are short.

For the convenience of riveting the inclined leg to the vertical one, the head is made of two shaped iron plates, riveted to the upper and lower sides of the strut truss; slots for the sheaves are made in the plates.

In drawing 47, by the known graphical method, the moment curve for a three-support beam is plotted, where:

1) The magnitude of the arm of the greatest moment is \(\approx 0.5\) cm.
2) Scale of forces: \(1\ \text{cm} = 10000\ \text{kg}\).
3) Scale of lengths: \(\dfrac{1}{50}\).
4) Pole distance \(= 4\ \text{cm}\).

The numerical value of the moment is obtained as:

\[ M = 0.5 \cdot 4 \cdot 10000 \cdot 50 = 1000000. \]

After selection on the drafts, we take the shaped plates with dimensions from the slots for the sheave to the lower edge of the plates: \(= 75\) cm.

Then the section modulus of both plates, with the thickness of each being 1.1 cm, is obtained as equal to:

\[ W_x = \frac{2.2 \cdot 75^2}{6} \approx 2000\ \text{cm}^3 \]

The bending stress is obtained as equal to:

\[ K = \frac{M}{W} = \frac{1000000}{2000} = 500\ \text{kg}/\text{cm}^2. \]

Now let us determine the compressive stress and add it to the bending stress.

The compressive force in member No. 17 is greater than in member No. 23; it is equal to 5375 kg; from it we find the compressive stress. The cross-sectional area of the shaped plates at the dangerous place is equal to:

\[ F = 2.2 \times 75 = 165\ \text{cm}^2. \]

and the stress is obtained:

\[ K_1 = 5375 : 165 \approx 32\ \text{kg}/\text{cm}^2 \]

The total stress per \(1\ \text{cm}^2\) will be:

\[ K = K + K_1 = 500 + 32 = 532\ \mathrm{kg./cm.^2} \]

The supporting members Nos. 19 and 21 form an intermediate support and are both calculated for one and the same greatest force acting in member No. 21—33750 kg. ≈ 34000 kg., causing longitudinal bending.

Again we assume \(\eta = 5.5\) and obtain

\[ I_{min} = \frac{34000 \cdot 4.25^2 \cdot 5.5}{10.2150000} = 1537. \]

The nearest number in the German standard assortment for I-beams is No. 45. The dimensions and form of the cross-sections of all connecting members and beams are given in special Table III.

The length of the strut member, by direct measurement, is equal to 4.25 m. Weight of the strut member: \(115 \times 4.25 = 489\) kg. Weight of the 2 strut members: \(489 \times 2 = 978\) kg.

The diagonal members Nos. 4, 8, 12, 16 and Nos. 15, 11, 7, 3 of drawings 43, 45 cross each other and form a truss with redundant members, statically indeterminate; but we need only choose the cross-sections of the members so that they could not work in compression at all, i.e. would buckle easily under the smallest compressive forces; then, in both cases of a broken rope, one of the rows of the indicated diagonal members transmits zero forces. Thus, in the calculation, the row of diagonal members must be regarded as if nonexistent; for both cases of a broken rope, such a row is marked by a dotted line on the schematic trusses.

Thus, in order to satisfy the second condition of static determinacy of the strut truss, we must choose the form of the cross-section such that the minimum moment of inertia is as small as possible; a rectangular section satisfies this condition.

To obtain a constructive joint connecting the diagonal member with the leg of the strut truss, we must make our member consist of two parts; both these parts are riveted to a gusset plate, and the latter, so as not to cause eccentric tension, is riveted centrally to the leg of the strut truss, so that the axes of all members in the joint meet at one point.

The greatest length of the diagonal member is \(l = 9.5\) m. The greatest tensile force is equal to 4500 kg.; we take 4800 kg.

Consequently, each component part of the member carries \(4800\ \mathrm{kg.} : 2 = 2400\ \mathrm{kg.}\)

Taking the allowable stress per \(1\ \mathrm{mm}^2\) as \(8\ \mathrm{kg}\), we obtain the cross-sectional area of each part:

\[ 2400\ \mathrm{kg} : 8 = 300\ \mathrm{mm}^2. \]

We must now determine the thickness and height of the part (strip) forming the diagonal member.

The thickness is determined from the condition that the strip can be riveted to the gusset plate; we take the minimum thickness as \(10\ \mathrm{mm}\), and then the height of the strip must be

\[ 300 : 10 = 30\ \mathrm{mm}. \]

Since the strip must work in tension, we must add the rivet diameter to the height; thus the height is obtained as:

\[ 3 + 2 = 5\ \mathrm{cm}. \]

The form of the cross-section of the diagonal member is shown in Table III.

The weight of one running meter of such a strip is (Hütte) \(3.9\ \mathrm{kg}\). Therefore, the weight of the compound diagonal members will be:

\[ \begin{aligned} 1)\quad &\text{No. }16 = 2 \times 6.3 \times 3.9\ \mathrm{kg} = 49\ \mathrm{kg}.\\ 2)\quad &\text{No. }12 = 2 \times 7.3 \times 3.9\ \mathrm{kg} = 57\ \mathrm{kg}.\\ 3)\quad &\text{No. }8 = 2 \times 8.4 \times 3.9\ \mathrm{kg} = 66\ \mathrm{kg}.\\ 4)\quad &\text{No. }4 = 2 \times 9.5 \times 3.9\ \mathrm{kg} = 74\ \mathrm{kg}. \end{aligned} \]

As we mentioned, we had to obtain the diagonal member as working as unfavorably as possible in longitudinal bending; the criterion is the safety factor, which we shall now determine.

The minimum moment of inertia of the diagonal member is equal to

\[ I_{\min} = \frac{10 \cdot 2^3}{12} \simeq 6\ \mathrm{cm}^4. \]

and

\[ n = \frac{6 \cdot 10 \cdot 2150000}{4800 \cdot 630^2} \simeq 0.07, \]

i.e., precisely what is required.

The horizontal members No. 20 and No. 1, as the Maxwell diagrams show, work only in tension; therefore, similarly to the diagonal members, we make each member composed of two iron strips.

The greatest force in member No. 20 is \(9750\ \mathrm{kg}\); I take \(10000\ \mathrm{kg}\). Thus each strip carries

\[ 10000 : 2 = 5000\ \mathrm{kg}. \]

The cross-sectional area of the strip, with an allowable stress of \(8\ \mathrm{kg}\) per \(1\ \mathrm{mm}^2\), is obtained as equal to:

\[ 5000 : 8 = 6.25\ \mathrm{cm}^2. \]

Adding the diameter of the rivet, we obtain:

\[ 6.25 + 2 = 8.25\ \mathrm{cm}^2. \]

With a strip thickness of 10 mm, convenient for riveting, the height of the cross-section of the strip is obtained as:

\[ 825 : 10 = 82.5\ \mathrm{mm}. \]

The nearest size of strip iron according to the assortment has the following dimensions:

\[ 8.5\ \mathrm{cm} \times 1\ \mathrm{cm}. \]

The weight of one running meter of such a strip is equal to 6.63 kg.

Weight of member No. 20:

\[ 2 \times 4.6 \times 6.63 = 60\ \mathrm{kg}. \]

Weight of member No. 1:

\[ 2 \times 9.3 \times 6.63 = 124\ \mathrm{kg}. \]

The horizontal members Nos. 5, 9, 13, working in longitudinal bending, are calculated by Euler’s formula.

The greatest length of the member: \(l = 8.2\ \mathrm{m}\).

The greatest stress \(= 3200\ \mathrm{kg} \approx 3500\ \mathrm{kg}\).

\[ I_{min}=\frac{P.l^2 n}{10.\varepsilon} =\frac{3500.672400.5}{10.2150000} =550\ \mathrm{cm}^4. \]

For reasons of joint constructability, we compose the cross-section from two box beams.

Neglecting the packing plate in favor of strength, we, by simple division of the obtained minimum moment,

\[ I_{min}=550:2=275\ \mathrm{cm}^4, \]

obtain the moment of inertia of each box-beam strip relative to the base:

\[ I_h=\frac{I_{min}}{2}=275. \]

In the Russian assortment (Hütte) of box beams, the nearest number is No. 20. All dimensions of the members are given in the table.

The weight of one running meter of the composite beam is equal to:

\[ 26.64 \times 2 = 53.28\ \mathrm{kg}. \]

Weight of No. 5 will be:

\[ 8.2\ \mathrm{m} \times 53.3 = 436\ \mathrm{kg}. \]

Weight of No. 9:

\[ 7 \times 53.3 = 372\ \mathrm{kg}. \]

Weight of No. 13:

\[ 5.8 \times 53.3 = 309\ \mathrm{kg}. \]

In what follows, our work will consist in finding the stresses from the wind and from the weight of the bracing truss, after which we shall be able to check the members for the total stress.

Forces from the wind and the self-weight of the parts of the trusses.

The part of the headframe projecting above the pithead building is subjected to the action of the wind.

Of all possible cases of wind action on the headframe, we choose the two most dangerous: the first, when the wind, as shown in drawing 48, blows into the front part of the headframe, and the second, when the wind blows from the opposite side. The first case gives additional stresses in the braced truss, and the second in the vertical truss. In the drawing, the point of application of the resultant wind force is found by simple graphical methods. The magnitude of the resultant wind force is found from the following considerations.

The area of the headframe projecting above the surface and subjected to the action of the wind is, by direct measurement on drawing 41, \(m.^2\); taking the vertical truss as though it were solid and assuming the wind pressure on \(1 \ \mathrm{m}^2\) of vertical surface to be equal to \(130 \ \mathrm{kg}\), we obtain the magnitude of the resultant:

\[ 41 \times 130 = 5330 \ \mathrm{kg}. \sim 6000 \ \mathrm{kg}. \]

The inclined and vertical trusses together with the ground form a very simple truss with sides

1) vertical . . . . . . \(21.4 \ \mathrm{m}\).
2) inclined . . . . . . \(23.2 \ \mathrm{m}\).
3) horizontal . . . . . \(8.8 \ \mathrm{m}\).

In drawing 49 two Maxwell diagrams are placed for the two cases of wind action on the headframe (the simplest truss). By direct measurement on the diagram, the additional compressive force in the braced truss is equal to \(13900 \ \mathrm{kg}\).

Taking this force as distributed uniformly in the head of the braced truss, we obtain the magnitudes of the reactions of the three upper joints of the braced truss:

\[ \begin{aligned} 1 &= 13900 : 4 = 3475 \ \mathrm{kg}.\\ 2 &= 13900 : 2 \sim 7000 \ \mathrm{kg}.\\ 3 &= 13900 : 4 = 3475 \ \mathrm{kg}. \end{aligned} \]

Let us determine the weight of the braced truss.

Weight of the 1st segment

1) 2 leg panels Nos. 18, 22—\(724.6 \times 2\) . . . \(\sim 1449 \ \mathrm{kg}\).
2) 2 supporting members Nos. 19, 21 . . . . . . \(978\) »
3) \(1/2\) the weight of horizontal member No. 20 . . . . \(30\) »
4) Weight of part of head plates Nos. 17, 23 . . . . \(600\) »

\[ 3057 \ \mathrm{kg}. \]

5) \(\sim 3.0\%\) for rivets and packing plates . . . \(\sim 103 \ \mathrm{kg}\).

\[ \text{Total} \ . . . \ 3160 \ \mathrm{kg}. \]

Weight of the II-nd section

1) \(1/2\) the weight of horizontal member No. 20 . . . . 30 kg.
2) 2 leg panels Nos. 15, 14 . . . . 1449 ”
3) 2 diagonal members No. 16 \(= 49 \times 2\) . . . . 98 ”
4) \(1/2\) the weight of horizontal member No. 13 \(= 309 : 2\) . . . . 155 ”

\[ 1732\ \text{kg.} \]

5) \(\simeq 5.5\%\) for rivets and splice plates . . . . 93 kg.

\[ \text{Total} \ . . . .\ 1825\ \text{kg.} \]

Weight of the III-rd section

1) \(1/2\) the weight of horizontal member No. 13 . . . . 155 kg.
2) 2 leg panels Nos. 11, 10 . . . . 1449 ”
3) 2 diagonal members No. 12 \(= 57 \times 2\) . . . . 114 ”
4) \(1/2\) the horizontal member No. 9 \(= 372 : 2\) . . . . 186 ”

\[ 1904\ \text{kg.} \]

5) \(> 5.5\%\) of the weight for rivets and splice plates . . . . 126 kg.

\[ \text{Total} \ . . . .\ 2030\ \text{kg.} \]

Weight of the IV-th section

1) \(1/2\) the weight of horizontal member No. 9 . . . . 186 kg.
2) 2 leg panels Nos. 7 and 6 . . . . 1442 ”
3) 2 diagonal members No. 8 \(= 66 \times 2\) . . . . 132 ”
4) \(1/2\) the weight of horizontal member No. 5 \(= 436 : 2\) . . . . 218 ”

\[ 1985\ \text{kg.} \]

5) \(\simeq 6.5\%\) of the weight for rivets and splice plates . . . . 141 kg.

\[ \text{Total} \ . . . .\ 2126\ \text{kg.} \]

Weight of the V-th section

1) \(1/2\) the weight of horizontal member No. 5 . . . . 218 kg.
2) 2 leg panels Nos. 3, 2 . . . . 1449 ”
3) 2 diagonal members No. 4 \(= 74 \times 2\) . . . . 148 ”
4) Weight of horizontal member No. 1 . . . . 124 ”

\[ 1939\ \text{kg.} \]

5) \(\simeq 3.5\%\) of the weight for rivets and splice plates . . . . 68 kg.

\[ \text{Total} \ . . . .\ 2007\ \text{kg.} \]

Weight of the strut truss within the limits of the curved portion (18 m):

\[ 3160\ \text{kg.} + 1825\ \text{kg.} + 2030\ \text{kg.} + 2126 + 2007\ \text{kg.} \simeq 11100\ \text{kg.} \]

Weight of 1 running meter on average:

\[ 11100 : 18 \simeq 600\ \text{kg.} \]

The full length of the strut truss, counting the upper part of the head and the shoes, will be 2.5 m longer; the weight of 2.5 m is equal to:

\[ 2.5 \times 600 = 1500\ \text{kg.} \]

Thus, the whole weight of the strut truss is equal to:

\[ 11100\ \text{kg.} + 1500\ \text{kg.} = 12600\ \text{kg.} \]

\[ \begin{aligned} K_{18}&=42225:179 \approx 240\ \text{kg}.\\ K_{15}&=66000:179 \approx 375\ \text{kg}.\\ K_{11}&=63600:179 \approx 355\ \text{kg}.\\ K_{7}&=61600:179 \approx 345\ \text{kg}.\\ K_{3}&=51350:179 \approx 285\ \text{kg}. \end{aligned} \]

And the stresses in the middles of the panels are obtained as:

\[ K_{18}=\frac{P_{18}}{W}+\frac{P.a.l^{2}}{I} =240+\frac{42225.0{,}00008.360^{2}}{2168} =240+195 \approx 435\ \text{kg}. \]

\[ K_{15}=\frac{P_{18}}{W}+\frac{P.a.l^{2}}{I} =375+\frac{66000.10}{2168} =375+300 \approx 675\ \text{kg}. \]

\[ K_{11}=\frac{P_{18}}{W}+\frac{P.a.l^{2}}{I} =355+\frac{63600.10}{2168} =355+293=648\ \text{kg}. \]

\[ K_{7}=\frac{P_{18}}{W}+\frac{P.a.l^{2}}{I} =345+\frac{61600.10}{2168} =345+280=625\ \text{kg}. \]

\[ K_{3}=\frac{P_{18}}{W}+\frac{P.a.l^{2}}{I} =285+\frac{51360.10}{2168} =285+237=522\ \text{kg}. \]

Using the data obtained, we construct the stress diagram in the leg of the inclined truss from longitudinal bending, and then also the diagram of the total stresses from longitudinal bending and from the bending component of the weight force.

As is seen, in Fig. 52 the maximum stress is equal to 775 kg per 1 cm² and nowhere exceeds the maximum permissible limit of 800 kg/cm².

In addition, from the total stress diagram we conclude that the stresses in the middles of the three middle panels (15, 11, 17) are almost identical; in the end panels, however, No. 18 and No. 3, the stresses drop greatly; thus, in the middles of the panels:

1) No. 18 the stress will be: \(435+53=488\ \text{kg}\).

2) No. 3  "  "  \(522+43=563\ \text{kg}\).

Thus we see that from the middle of panels No. 15 and No. 7 toward the ends of the leg of the inclined truss there is much excess material; wishing, as far as possible, to equalize the stresses in all parts of the leg of the inclined truss, we shall have to narrow the leg of the inclined truss, beginning from the middle of panels No. 15 and No. 7 toward the ends of the leg.

We turn to the formula:

\[ K=\frac{P}{W}+\frac{P.a.l^{2}}{I_{min}}; \]

we see that we have the right to change only the first term, since we cannot reduce \(I_{min}\); it is necessary to us for longitudinal bending. Thus, in order to increase the stress, we can change

construct the stress diagram. The graphical constructions have been made on the drawing and require no explanation. In all the above-mentioned constructions the following numerical data are obtained:

Forces of the load acting in the panels of the leg of the strut truss. Scale: 1 mm = 100 kg	Components of the load forces producing bending of the leg of the strut truss. Scale: 1 mm = 50 kg
1st bay: \(3160 : 2 = 1580\) kg.	625 kg.
2nd bay: \(1825 : 2 = 912\) kg.	410 kg.
3rd bay: \(2030 : 2 = 1015\) kg.	380 kg.
4th bay: \(2126 : 2 = 1063\) kg.	445 kg.
5th bay: \(1973 : 2 = 986\) kg.	400 kg.
\(\Sigma 5560\) kg.	Total 2260 kg.

The moments are obtained as:

\[ \begin{aligned} M_{18} &= 5.5 \times 50 \times 4 \times 200 = 220000 \ \text{kg}\cdot\text{cm}.\\ M_{15} &= 10.5 \times 40000 = 420000 \quad " \\ M_{11} &= 12.0 \times 40000 = 480000 \quad " \\ M_{7} &= 10.5 \times 40000 = 420000 \quad " \\ M_{3} &= 4.5 \times 40000 = 180000 \quad " \end{aligned} \]

The stresses are obtained as:

\[ \begin{aligned} K_{18} &= M_{18} : W_x = 220000 : 4182 \approx 53 \ \text{kg}/\text{cm}^2.\\ K_{15} &= M_{15} : W_x = 420000 : 4182 = 100 \quad " \\ K_{11} &= M_{11} : W_x = 480000 : 4182 = 115 \quad " \\ K_{7} &= M_{7} : W_x = 420000 : 4182 = 100 \quad " \\ K_{3} &= M_{3} : W_x = 180000 : 4182 = 43 \quad " \end{aligned} \]

Now our task will be to construct a diagram of the stresses acting in each panel of the leg from longitudinal bending. The stresses at the ends of the panels (joints) are determined from the simple compression formula \(K = P : U\); the stress in the middle of each panel we shall determine by the Schwarz-Rankine formula:

\[ P = \frac{W \cdot K} {1 + \dfrac{\alpha \cdot w \cdot l^2}{I_{min}}}; \]

where: \(\alpha = 0.00008\) for cast iron.

Hence the stress will be:

\[ K = \frac{P}{W} + \frac{P \cdot \alpha \cdot l^2}{I_{min}}. \]

We determine the stresses at the ends of the panels of the leg.

In Drawing 50, on the simplest truss, 2 components of the weight force were found by ordinary graphical resolution.

One component is equal to 4800 kg; it produces bending of the strut truss and causes tension in it. We shall neglect the small magnitude of the tension in favor of strength and, in order not to complicate the calculation, we shall take the bending stress somewhat lower not as from a concentrated force, but as from a distributed one, since we have the weight of the separate portions of the strut truss.

The other component will produce compression of the strut truss; by directly measuring it and multiplying by the scale of forces it is obtained equal to 5800 kg, or in each leg of the strut truss there will be produced a compressive force equal to: \(5800 : 2 = 2900\) kg.

The resultant forces from wind and from weight at the three upper joints of the strut truss.

	At the 1st support	At the 2nd support	At the 3rd support
From wind	3475 kg	7000 kg	3475 kg
“ weight	2900 ”	0	2900 kg

In Drawing 51 we again construct the Maxwell diagram, showing the magnitude and sign of the forces acting in the members of the strut truss from the forces of weight and wind; here in the drawing the numerical magnitudes of the forces are set out in a separate Table IV.

Final check and determination of the dimensions of the members of the strut truss.

Reserve of strength against longitudinal bending in the legs of the strut truss:

1) Force from the rope  55400 kg.
2) “ from wind and weight 10600 ”
  Total . . . 66000 ”

\[ \eta=\frac{I\cdot 10\cdot E}{P\cdot l^2} =\frac{2168\cdot 10\cdot 2150000}{66000\cdot 360^2} =5{,}46. \]

The resultant stress in the legs of the strut truss from longitudinal bending and from bending due to the component of the weight force:

In Drawing 52 we set out the leg of the strut truss within the limits of its bending portion, and for it we construct the component forces of the weight for each panel by graphical means; from the components of the forces thus obtained we construct the diagram of moments, and, having obtained the latter, proceeding from the formula:

\[ M = W k, \]

only the first term of the given equality. The numerator \(P\) does not depend on us, and in order to increase the stress we must reduce \(W\)—the cross-sectional area of the leg; we cannot reduce the height of the I-section, since \(I_{min}\) will decrease; but the height of the web of the I-section can be reduced, since it is almost not reflected in the value of \(I_{min}\), as we mentioned above.

Thus, suppose that we wish to raise the stress in the middle of panel No. 3 to 775 kg; then we have:

\[ 775=\frac{51350}{W}+\frac{51350 \cdot 10}{2138}=\frac{51350}{W}+237; \]

taking the bending stress as 43 kg, \(\sim 50\) kg, we obtain that

\[ \frac{51350}{W}=775\ \text{kg.}-(237+5)=488\ \text{kg./cm}^2, \]

whence

\[ W=5130:488=105\ \text{cm}^2, \]

i.e., in order to satisfy a uniform stress along the whole leg, its cross-sectional area must be equal to \(105\ \text{cm}^2\).

We must reduce the cross-sectional area: \(179-105=74\ \text{cm}^2\); this subtraction must be made at the expense of the web of the beam, with the thickness of the latter being 1 cm; the height of the web would have to be reduced by 74 cm. Earlier we had that the height of the web is equal to 60 cm; therefore, for our case in the middle of panels 18 and 3 the web would disappear altogether. In fact it is not necessary to narrow the leg so much; the limit for narrowing in the 18th panel is the height of the supporting rods \(=45\) cm, and we narrow panel No. 3 symmetrically with panel No. 18.

In rods No. 17 and No. 23, an additional force of 1200 kg is caused by the wind and by the weight. The stress caused by this force will be:

\[ 1200:(75 \times 2)=8\ \text{kg./cm}^2 \]

Thus:

\[ \begin{aligned} &1)\ \text{Stress from bending and compression} &&—\ 532\ \text{kg./cm.}^3\\ &2)\ \text{“ ” additional compression} &&—\ \frac{8\ \text{ ”}}{\text{Total}\quad 540\ \text{kg./cm}^3} \end{aligned} \]

which is permissible.

In the supporting rods:

\[ \begin{aligned} &1)\ \text{Forces from the breaking rope} &&33750\ \text{kg.}\\ &2)\ \text{“ ” wind and weight} &&\frac{4200\ \text{kg.}}{\text{Total}\ 38000\ \text{kg.}} \end{aligned} \]

Therefore the strength condition will be written:

\[ M = W \cdot K = \frac{c(a-b)^2}{8}\cdot g = \frac{c\delta_1^2}{6}\cdot K, \]

whence:

\[ \delta_1=\sqrt{\frac{3}{4}\frac{g\cdot(a-b)^2}{K}} =\sqrt{\frac{3}{4}\frac{10\cdot(100-50)^2}{500}} =6\ \text{cm}. \]

It would be difficult to cast a plate of such thickness, and in the present case it is better to resort to strengthening the flanges of the plate with ribs; then the plate will be thinner. The number of ribs is taken as not less than three; with the double-T form of our leg and the corresponding dimensions of the plate, it is convenient for us to arrange 4 ribs, cast 2 together at a time.

Thus, we set a suitable plate thickness \(\delta_1=3\ \text{cm}\).

The thickness of the rib is taken as:

\[ \delta_2=0.7\delta_1=0.7\cdot 3 \approx 2\ \text{cm}. \]

It remains for us to determine the height of the rib, which we find from the condition of its safe resistance to bending.

The bending moment for the rib is equal to

\[ M=W\cdot\frac{l_1}{2}\cdot g = W\cdot K, \]

where:

\(W=35\ \text{cm}\times 50=1750\ \text{cm}^2\)—the area over which the action of one rib is distributed.

\[ \frac{l_1}{2}=\frac{35}{2}\ \text{cm}, \]

the distance from the center of gravity of the load to the axis of bending.

\(g=10\ \text{kg}/\text{cm}^2\)—foundation pressure;

\[ Wk=\frac{\delta_2 h^2}{6}\cdot K. \]

\(K \approx 300\ \text{kg}/\text{cm}^2\)—allowable stress for the ribs.

Thus, the strength condition will be written:

\[ M=1750\cdot\frac{35}{2}\cdot 10=\frac{2\cdot h^2\cdot 300}{6}; \]

whence

\[ h^2=\frac{1750\cdot 35\cdot 10\cdot 6}{2\cdot 2\cdot 300}=3062 \]

and

\[ h=\sqrt{3062}=58\approx 60\ \text{cm}. \]

Actually, the foundation plate is pressed against the foundation by a large normal pressure so strongly that, being provided with lower projections embedded in the foundation, it can hardly be displaced from its place by a comparatively small shearing force.

As can be seen, the beam has been calculated with a sufficient reserve of strength, and therefore its section need not be changed.

Supports of the inclined truss.

I. Normal pressure on the cap.

Reaction I from the rupture of the upper rope—\(96.5\) (mm) \(\times 500\) kg. — \(48250\) kg.
See the truss diagram.

Reaction I from wind and the component of the weight—\(26.5\) „ \(\times 400\) kg. — \(10600\) kg.

Second component of the weight along the leg:

\[ \frac{58 \times 200}{2} \]

\(5800\) kg.

\[ \text{Total } 64650\ \text{kg.} \]

II. Shearing force on the cap.

See the Maxwell diagram.

From the component of the weight—along the ground \(22\) (mm) \(\times 200\) kg. \(\quad 4400\) kg.

„ wind „ \(53\) „ \(\times 100\) kg. \(\quad 5300\) kg.

\[ \text{We take, in round numbers, } 10000\ \text{kg.} \]

III. Allowable stress on the foundation.

The foundation is made of tamped concrete (1 part cement + 3 parts sand + 6 parts crushed stone); then \(g = 10\ \text{kg}/\text{cm}^2\)—the pressure exerted by the foundation slab on the foundation.

The support of the leg of the inclined truss consists of a foundation slab. The area of the foundation slab is determined from the allowable pressure per \(1\ \text{cm}^2\) of foundation, \(g = 10\ \text{kg}/\text{cm}^2\); hence the area of the slab is obtained as:

\[ 64650 : 10 = 6465 \approx 7000\ \text{cm}^2. \]

The sides of the slab should not exceed a ratio of \(2:1\); best of all, having the trace of the leg on the slab, determine graphically the magnitudes of the sides, so that the slab would be obtained with the smallest possible flanges. The dimensions of the slab required by us are:

\[ 70 \times 100 = 7000\ \text{cm}^2. \]

Drawing 53.

If we were not to strengthen our slab with ribs, then its thickness \(\delta_1\) would be determined from the following considerations: the most dangerous place, where fracture of the slab may occur, is that section in which the flanges of the leg come into contact with the slab. The magnitude of the bending moment with respect to such a section is equal to

\[ M = W \cdot g \cdot \frac{l}{2} = \frac{c(a-b)}{2} \cdot g \frac{(a-b)}{4}, \]

and the section modulus:

\[ W = \frac{c\delta^2}{6}. \]

\[ \eta=\frac{I . 10 . E}{P . l^{2}}=\frac{1537 . 10 . 2150000}{38000 . 176625}\simeq 5, \]

i.e., the margin of strength is sufficient.

In the diagonal members no additional stresses arise, owing to the symmetry of the load; therefore we consider their calculation to have been made correctly, and the cross-sections of the members remain the same.

In the horizontal members Nos. 5, 9, 13, additional stresses likewise do not arise, and their sections therefore are not changed.

No. 20 receives an additional tensile force of 1600 kg. Thus:

1) Force from the rope . . . . 10000 kg.
2) “ ” wind and weight . . 1600 kg.
\[ \text{Total } .\ 11600\ \text{kg.} \]

For each strip composing the member there is:

\[ 11600:2=5800\ \text{kg.} \]

this force corresponds to the cross-section

\[ 8.5\ \text{cm}^{2}-2=6.6\ \text{cm}^{2}, \]

whence the stress is obtained as equal to:

\[ 5800:650=8.0\ \text{kg./cm}^{2}. \]

which is rather high; therefore we select, from the assortment, strip iron with a larger cross-section; if we take a cross-section of \(9.5\ \text{cm.}\times 2\), then the area working in tension will be:

\[ 9.5-2=7.5\ \text{cm}^{2}. \]

And the stress will be

\[ 5800:750=7.7\ \text{kg./mm}^{2}, \]

which is permissible. The dimensions of the resulting member are given in the table of cross-sectional forms. As for the additional weight thereby obtained, in view of its insignificance we neglect it, and we shall not redo the calculations in the corresponding places.

We calculated horizontal member No. 1 together with No. 20 and therefore exaggerated the tensile force acting in it, taking it equal to 10000 kg. In reality the following forces act in it:

1) From the action of the rope . . . . . . 5500 kg.
2) Additional forces from wind and weight . . 1800 kg.
\[ \text{Total } . .\ 7300\ \text{kg.} \]

… max. up to $\sim 10000\ \mathrm{kg}$. However, disregarding the frictional force of the plate against the foundation, we shall determine the diameter of the four foundation bolts by which the foundation plate is to be fastened to the foundation. The shearing force on each bolt is equal to

\[ 10000:4 = 2500\ \mathrm{kg}. \]

Taking the permissible shearing stress as $K_{\mathrm{sr}} = 800\ \mathrm{kg}/\mathrm{cm}^{2}$, we obtain the cross-sectional area of the bolt:

\[ 2500:800 = 3.14\ \mathrm{cm}^{2}, \]

which corresponds to a bolt diameter of

\[ d = 2\ \mathrm{cm}. \]

We take the smallest diameter adopted for a foundation bolt:

\[ d = 2.5\ \mathrm{cm}. \]

Since there is no overturning moment acting on the foundation plate, in our case the length of the foundation bolts is determined by the depth of the foundation.

Foundation.

As was mentioned above, the foundation is made of concrete. The upper platform of the foundation, on which the plate rests, is made with margins of $250\ \mathrm{mm}$.

The form and depth of the foundation are determined by a simple graphical selection on the drawing, while the following conditions must be observed:

1) The foundation must rest on a solid base and extend downward beyond the depth of freezing.

2) With a gradual widening of the foundation, the area of its lower base must be such that the forces transmitted to the soil do not exceed $2\ \mathrm{kg}/\mathrm{cm}^{2}$; in our case the area of the base of the foundation must be

\[ > \frac{64650}{2} = 32325\ \mathrm{cm}^{2}. \]

3) The resultant of all the forces on the foundation must pass through the core of the foundation section, so that the so-called opening of joints does not occur.

Details of the fastenings of the strut truss.

The greatest length of a riveted beam, forming the leg of the strut truss, available commercially, is $= 7.5\ \mathrm{m}$. The length of the leg, however, is on average $= 20.5\ \mathrm{m}$. Therefore our leg will consist of $20.5:7.5 \sim 3$ parts, and we shall have 2 joints not falling at the joints of the strut truss.

The thickness of the cover plates $\delta_{1}$ varies from $7 \sim 12\ \mathrm{mm}$ and is determined from the condition that their resistance should be no less than the resistan…

To satisfy this condition, it is necessary that the intersection of all the members composing the joint be at one point; in such a case the forces acting in the panels, the braces, and the external forces are balanced.

2) The thickness of the packing plate is usually taken as \(\delta = 2\delta_1\), where \(\delta_1\) in our case is the thickness of the web of the channel section, \(\delta = 2\).

3) \(K_s = 8 \sim 10\ \mathrm{kg./mm.^2}\)

4) \(d = 2\delta \sim 2\ \mathrm{cm.}\)—the diameter of the rivet.

In the braced truss there are four types of joints, according to which the remaining joints of the truss are also constructed.

1) Thus, for joint \(A\), drawing 54, the number of rivets fastening the supporting members Nos. 19 and 21 to the gusset plates is calculated according to the greatest force in member No. 21—37950 kg. Single shear will be denoted by the index \('\), and double shear by \(''\).

	Rope.	From wind and weight.	Sum.
No. 19	24000	4200	28200 kg.
No. 21	33750	4200	37950 kg.

Thus, for our case:

\[ n'\frac{\pi d^2}{4}\,K_s = 37950\ \mathrm{kg.}; \]

taking

\[ \frac{\pi d}{4}\,K_s = 1256\ \mathrm{kg.}, \]

the shearing force on one rivet at a stress of \(8\ \mathrm{kg./mm.^2}\), we obtain:

\[ n'_{21} = 37950\ \mathrm{kg.} : 1256 \sim 27\ \text{rivets}. \]

I take 32 pieces; then for each flange of the I-beam there are \(32:2=16\) rivets, and they are arranged, as shown in the drawing, in two rows of 8 each.

Since the joint of the leg at joint 13 of drawing 55 is not present, the number of rivets along the panel is calculated from the difference of forces in panels 15 and 18, equal to

\[ 66000 - 42120 \sim 24000\ \mathrm{kg.}, \]

therefore,

\[ n'_{15-18} = 24000 : 1256 \sim 20\ \text{rivets}, \]

i.e., on each side of the packing plate there are \(20:2=10\) rivets.

	Rope.	From wind and weight.	Sum.
No. 15	55400	10600	66000 kg.
No. 18	35121	7000	42120 ”
No. 21	33750	4200	37950 ”
No. 20	10000	1600	11600 ”
No. 16	4500	0	4500 ”

It remains to mention here that the joints of the web, flanges, and angles must not lie in one plane.

The number of rivets connecting the flanges to each other, and the lower flange to the angles, is determined by the magnitude of the tangential forces.

If the distance between rivets is equal to \(p\), then the shearing force over this length is

\[ S = p \cdot k_s \cdot \delta . \]

The stress from the tangential forces is determined by the formula

\[ K_s=\frac{\sum P s \cdot e d w}{I \cdot U} \]

\(\sum P\), as was determined, is equal to \(1200\ \mathrm{kg}\).

Thus the stresses of all tangential forces are obtained as follows:

A) between the flanges:

\[ K_{s_1}=\frac{1200 \times 1{,}1 \cdot 20 \times 31{,}65}{20 \cdot 4182 \cdot 32{,}2}\sim 0{,}31\ \mathrm{kg}. \]

B) between the flange and the angle:

\[ K_{s_2}=\frac{1200 \cdot 2{,}2 \cdot 200 \cdot 31{,}1}{1 \cdot 4182 \cdot 32{,}2}\sim 127\ \mathrm{kg}. \]

Thus, the distance between the rivets connecting the flanges to each other will be:

\[ \frac{2\pi d^2}{4}\cdot K_z=p_1 K_{s_1}\delta_1, \]

whence, for \(K_z=400\), \(d=2\ \mathrm{cm}\), and \(\delta_1=20\ \mathrm{cm}\),

\[ p_1=\frac{2 \cdot 3{,}14 \cdot 2^2 \cdot 400}{4 \cdot 20 \cdot 0{,}31}=406\ \mathrm{cm}. \]

In exactly the same way, the distance between the rivets connecting the flanges to the angles will be:

\[ \frac{2\pi d^2}{4}\cdot K_z=p_2\cdot K_{s_2}\cdot \delta_2,\quad \text{where } \delta_2=1\ \mathrm{cm}. \]

we obtain:

\[ p_2=\frac{2 \cdot 3{,}14 \cdot 2^{-2}\cdot 400}{4 \cdot 1 \cdot 127}=21\ \mathrm{cm}. \]

C) By the same precise methods we find the distances between the rivets connecting the angles with the web of the beam; it is obtained equal to \(\sim 18\ \mathrm{cm}\).

In practice the distance between rivets is taken as

\[ p=(4\text{—}8)d=80\ \mathrm{mm}\text{—}160\ \mathrm{mm}. \]

We adopt the distance between rivets in our beam as

\[ p \sim 160\ \mathrm{mm}. \]

Joints of the inclined truss.

1) In arranging the joints in the inclined truss, the principal condition is the absence of eccentric tensions (compressions).

of the web, and since the cover plate in height is determined by the distance between the angles, we have the relation

\[ 2a\delta_1 = 2(H - 2c)\delta_1 = H\delta, \]

whence

\[ \delta_1 = \frac{H\delta}{2(H - 2c)} = \frac{60 \cdot 1}{2(60 - 2.9)} = 0.8\ \text{cm}. \]

The number of rivets is found from the condition of strength of the joint with respect to shearing and tensile (compressive) forces. The former, i.e. the shearing forces, under a uniformly (almost) bending load have the greatest values near the end supports and are equal to:

\[ \sum P = \frac{2250}{2} = 1125\ \text{kg} \sim 1200\ \text{kg}. \]

If, for a rivet of 2 cm diameter, a force of 1250 kg is allowed (stress 400 kg per 1 cm²), then the number of rivets compensating the shearing forces is obtained as:

\[ n = 1200 : 1250\ \text{kg} = 1\ \text{pc.};\quad \text{I adopt } 2\ \text{pcs.} \]

As for the compressive forces, the necessary number of rivets may be obtained from the following considerations:

Let the distance between rivets be \(p\).

The moment of the forces shearing the rivets with respect to the neutral layer will be:

\[ M_{\text{sr}} = \frac{\pi d^2}{4}\, k_s \{ n_1 p + (n_1 - 1)p \ldots p \} = \frac{\pi d^2}{4}\, k_s \frac{p n_1(n_1 + 1)}{2}, \]

where \(n_1\) is the number of rivets above or below the line of axes. The moment of the internal tensile (compressive) forces for the web will be

\[ \frac{1}{2}\cdot \delta \cdot \frac{H^2}{6}\cdot k_z. \]

We write the strength equation

\[ \frac{P n_1 \pi d^2 \cdot k_s}{4} = \frac{n + 1}{2} = \frac{1}{2}\frac{\delta H^2}{6}\cdot k_z; \]

but \(2pn_1 \sim H\), therefore:

\[ \frac{1}{2}\frac{\pi d^2}{4}(n_1 + 1)k_s = \frac{\delta H k_z}{6}, \]

whence

\[ n_1 = \frac{\delta H k_z \cdot 4 \cdot 2}{6\pi d^2 k_s} - 1 = \frac{1 \cdot 60 \cdot 800 \cdot 2}{6 \cdot 3.14 \cdot 400} - 1 \sim 13\ \text{rivets}. \]

And the total number of rivets in the joint is obtained as equal to:

\[ M = n + 2n_1 = 2 + 2 \cdot 13 = 28\ \text{rivets}. \]

The rivets are arranged with such a calculation that the pitch varies from 75 to 100 mm; if the pitch comes out smaller, then the rivets are placed in two rows in a staggered arrangement.

Number of rivets in member No. 20

\[ n_{20}''=\frac{11600}{2\cdot1256}\simeq 5 \text{ rivets.} \]

Number of rivets in member No. 16

\[ n_{16}''=\frac{4500}{2\cdot1256}\simeq 2 \text{ rivets.} \]

I take 3 rivets, since the distances between rivets in this joint are different in the various members, from \((3—5)d\), in order to avoid the incoming angles into the gusset plate.

In joint type \(C\), Fig. 56, as before, the number of rivets fastening the gusset plate to the legs is found from the difference of forces in panels 15 and 11:

\[ 66000-63600=2400\ \mathrm{kg}. \]

and

\[ n_{15-11}'=\frac{2400}{1256}\simeq 2 \text{ rivets.} \]

For the sake of the constructability of the joint I take 8 rivets, 4 on each side of the gusset location.

	Chord.	From wind and weight.	Total.
No. 13	3200	0	3200 kg.
No. 11	53000	10600	63600 kg.

The number of rivets in member No. 16 was determined above: \(n_{16}''\simeq 3\) rivets.

Number of rivets in member No. 13

\[ n_{13}''=\frac{3200}{2\cdot1256}\simeq 2 \text{ rivets;} \]

for the sake of the constructability of the gusset plate I take \(n_{13}''=4\) rivets.

For joint type \(D\), Fig. 53, as before,

\[ n_{k-3}'=\frac{59100-51350}{1256}\simeq 8 \text{ rivets,} \]

\[ n_{4}''=2400:\{1256\times 2\}=1 \text{ pc.; I take 2 rivets;} \]

\[ n_{1}''=\frac{7300}{1256\cdot2}\simeq 3 \text{ pcs.; I take 5 pcs.} \]

	Chord.	From wind and weight.	Total.
No. 4	2400 kg.	0	2400 kg.
No. 1	5500 ”	1800 kg.	7300 ”
No. 3	40750 ”	10600 ”	51350 ”
\(R\)	48500 ”	10600 ”	59100 ”

Vertical trusses.

After the calculation of the braced truss, the calculation of the vertical ones no longer presents any difficulties; in order not to complicate the work, the subsequent calculation will be carried out in a rougher form, with the assumption that the permissible errors will be on the side of the strength of the structure.

In our case, the vertical column consists of four angle-section legs, fastened by diagonal and horizontal members.

The distances between the legs are determined by the dimensions of the cages, by such a calculation that along the edges and between the cages there is a sufficient clearance for placing the guides.

Since the vertical column is acted upon chiefly by vertical forces, with the exception of small forces from the wind, in our diagonal members the vertical forces produce zero stresses, and therefore they have the character of wind bracing.

Head of the vertical trusses.

The head of the vertical part of the headframe is arranged, as shown in Drawing 57, of 6 box girders, riveted at their ends to 2 transverse girders of the same section form. From above, the girders are covered with a platform of sheet iron 2 cm thick, with slots for the sheaves. From below, the girders, in the corresponding places, are tied by strips of sheet iron. In practice it is customary to specify the form and dimensions of the head, and then to check the stresses in its parts.

Thus, in our case, we have the following data:

1) weight of the sheave . . . . . . . . . . 1500 kg.
2) “ roof . . . . . . . . . . . . . . . . 2650 “
3) “ platform and lower strips . . . . . . 1650 “
4) “ longitudinal girders No. 26 . . . . . 1320 “
5) “ transverse “ No. 26 . . . . . . . . . 280 “

We have to check the stresses in the girders, obtaining at the same time the reactions of the supports (shaped plates). We can calculate the front and rear shaped plates on which our girders rest; by obtaining the reactions of the shaped plates, we shall at the same time obtain the force acting on each vertical leg.

Assuming that, owing to the platform, the load is distributed equally between the two outer girders, and likewise between the middle girders, we have the stress in the outer girders from the following forces:

I. Uniformly distributed load:

1) 0.5 of the roof weight \(=\dfrac{2650}{2}\) . . . . 1325 kg.

2) 0.25 of the platform weight \(\dfrac{1500}{4}\) . . . \(\sim\) 400 ”

3) Weight of the 2 longitudinal beams . . . 440 ”

4) Heads of rivets and reserve . . . . 150 ”

Total . . . 2240 kg.

II. Concentrated load:

1) weight of the sheave . . . . . . . . 1500 kg.

2) \(0.5 V_2'\), vertical component when the rope breaks 5900 ”

Total . . . 7400 kg.

Thus, on each beam there falls

\[ 2240:2 \sim 1200\ \text{kg} \]

of uniformly distributed load, and

\[ 7400:2 \sim 3700\ \text{kg} \]

of concentrated load.

Determine the reactions of the supports of the outer beam; if one support is \(A\), and the other \(B\), then:

\[ A=\frac{1200}{2}+\frac{3700\cdot25}{365}=600+222\sim820\ \text{kg}. \]

\[ B=\frac{1200}{2}+\frac{3700\cdot340}{365}=600+3450\sim4000\ \text{kg}. \]

By the graphical method, on the appended drawing, we determine the magnitude of the maximum bending moment; it is equal to:

\[ M_{\max}=5.100.30.10=150000\ \text{in kg·cm}. \]

The section modulus for the box-shaped head beam adopted by us is equal to:

\[ W_x=388\ \text{cm}^3. \]

Determine the bending stress:

\[ K=M_{\max}:W_x=150000:388=390\ \text{kg}/\text{cm}^2. \]

Now determine the reactions of the supports of each of the two middle beams:

I. Uniformly distributed load:

1) 0.5 of the platform weight—\(1500:2=\) 750 kg.

2) Weight of the 2 beams . . . . 440 ”

We take as equal . . . 1200 ”

II. Concentrated load:

1) Weight of the sheave . . . . . 1500 kg.

2) \(0.5V_2=11750:2\) . . . . 5900 ”

Total . . . 7400 kg.

In this way, on each middle head beam there falls:

\[ 1200:2=600\ \text{kg} \]

of uniform load, and

\[ 7400:2=3700\ \text{kg} \]

of concentrated load.

The reactions of the supports are obtained:

\[ A_1=\frac{600}{2}+\frac{3700.25}{365}=555\ \text{kg}. \]

\[ B_2=\frac{600}{2}+\frac{3700\cdot 340}{365}=3745\ \text{kg}. \]

We see from the forces acting on the beams that the middle beams bend less than the outer ones, and therefore we shall not determine the bending stresses in them.

We must also determine the stresses in the beams due to compression, caused by the horizontal component of the force from the force \(R_2'\); this component tends to shift the beam with the head part of the headframe in the horizontal direction. The magnitude of the horizontal component is found from the equality:

\[ Q=R_2'\cdot \cos 67^\circ=68000\cdot 0.39=27200\ \text{kg}. \]

Taking this force to be distributed over one outer and one middle beam, we obtain the force acting on each beam:

\[ 27200:2=13600\ \text{kg}. \]

The stress caused by this force compressing the beam is determined by taking the cross-sectional area of the box beam as

\[ F=49.95\sim 50\ \text{cm}^2. \]

\[ k_z=13600\ \text{kg}:50=270\ \text{kg}/\text{cm}^2. \]

And the total stress acting in the head beam is obtained as equal to:

\[ K=k+k_z=390+270=660\ \text{kg}/\text{cm}^2. \]

The number of rivets holding each head beam against slipping is, for \((d=2\ \text{cm})\),

\[ 13600:1256\sim 10\ \text{pcs.} \]

All the head beams lie on two gusset plates; we already know the reactions of the supports of each beam, and now two problems stand before us—to determine the dimensions of the front and rear gusset plates.

In the appended drawings, the bending moments for these plates are determined graphically;

1) the maximum bending moment for the front plate is obtained as equal to:

\[ M_{max}=3.5\times 100\times 30\times 10=105000\ \text{kg}\cdot\text{cm}. \]

at

\[ K=800\ \text{kg}/\text{cm}^2 \]

\[ W_x=M_{max}:K=10500:800=131\ \text{cm}^2. \]

Taking the height of the gusset plate in the middle as \(40\ \text{cm}\), with its thickness \(1\ \text{cm}\), we obtain

\[ W_x=\frac{1\cdot 40^2}{6}=267\ \text{cm}^3, \]

i.e., the size of the plate is sufficient.

2) The maximum bending moment for the rear plate is equal to

\[ M_{\max} = 11.200.30.10 = 660000; \]

assuming \(K = 800\ \mathrm{kg}/\mathrm{cm}^2\), we obtain

\[ W_x = M_{\max} : K = 660000 : 800 = 825. \]

For structural reasons, and for the convenience of riveting the bracing truss, it is necessary to take the height of the plate equal to \(1.5\ \mathrm{m}\). Then, with its thickness equal to \(1\ \mathrm{cm}\),

\[ W_x = \frac{1.150^2}{6} = 3750\ \mathrm{cm}^3, \]

i.e., it is entirely sufficient.

The number of rivets fastening the rear fascia plate to the vertical legs is determined from the magnitude of the reactions; since in our case each reaction on the rear legs is equal to \(11300\ \mathrm{kg}\), then, with rivets of diameter \(2\ \mathrm{cm}\), we obtain the number of required rivets

\[ 11300\ \mathrm{kg} : 1256 = 9\ \text{pcs.} \]

Since the installation of a rivet usually costs no more than 5 kopecks, in practice it is customary to install a number of rivets considerably greater than the calculated number.

The forces acting in the wind bracing are insignificant when they are found provisionally on a rough sketch; therefore the sections of the members are chosen as the smallest suitable for riveting.

We have two types of wind bracing—diagonal and horizontal. The first, for stiffness, we make of angles No. 6, and the second, for the convenience of fastening the guides, of channel iron No. 12; in order to make the vertical part of the headframe as rigid as possible, we introduce additional diagonal members. The panel points on the legs of the vertical column are more frequent—every \(2\ \mathrm{m}\)—and the number of rivets is easily determined from the force by which the member can be torn apart.

To avoid a complicated calculation, we make all four vertical legs of the same size.

The greatest reactions from the fascia plates occur in the rear legs; the magnitude of each reaction is obtained as \(11300\ \mathrm{kg}\). We make the leg of uniform section over its entire height; the lowest part of the leg will be in the worst conditions, since stresses from the weight of the overlying parts of the vertical truss will arise in it.

Let us find the weight of the vertical truss from the head to the base shoes.

1) Weight of gusset plates (350 mm × 300 mm)—72 mm . . . 549 kg.
2) “ of all fascia sheets . . . 1083 “
3) “ of diagonal rods No. 6, total length 281 m . . . 1983 “
4) “ of horizontal rods No. 12, total length 141 m . . . 1910 “
5) “ of the 4 legs No. 16, total length \(21.2\,m \times 4 = 84.8\,m\) . . . 3425 “

Count as exactly . . . 9000 kg.

For one leg the force is

\[ 9000 : 4 = 2250\ \text{kg}, \]

allowing for rivets and additional angles, \(4\%\) ≈ 100 kg.

Total . . . 2350 kg.

The force in the rear leg from the wind is equal to:

\[ 12900\ \text{kg} : 2 = 6450\ \text{kg}. \]

See the diagrams of wind forces.

And the force from the weight of the strut truss is

\[ 6200\ \text{kg} : 2 = 3100\ \text{kg}. \]

In the lower panel of the rear leg the following forces arise:

1) From the weight of the head of the vertical column and the action of the rope . . . 11300 kg.
2) From the weight of the column itself . . . 2350 “
3) “ wind . . . 6450 “
4) “ weight of the strut truss . . . 3100 “

Count as exactly . . . 24000 kg.

With a cross-sectional area of the angle \(= 51.8\ \text{cm}^2\), the compressive stress is obtained as

\[ K = 24000 : 51.8 = 460\ \text{kg}/\text{cm}^2 . \]

The free length of the leg panel \(= 200\ \text{cm}\). The minimum moment of inertia of the adopted angle is

\[ I_{\min} = 506\ \text{cm}^4, \]

and the safety factor is obtained as

\[ \eta = \frac{506 \cdot 10 \cdot 2.150\,000}{24000 \cdot 200^2} = 21. \]

Allowing an error in favor of strength, we assume that the front and rear legs exert on the main beam the same force of 24000 kg; graphically we find the maximum bending moment, drawing II (6), which is equal to:

\[ M_{\max} = 8.6\ m \cdot 1000 \cdot 50.5 = 2050000, \]

taking \(K = 800\ \text{kg}/\text{cm}^2\), we obtain

\[ W_x = M_{\max} : K = 2050000 : 800 = 2563. \]

Allowing for weakening of the beam by rivets, I take the limiting number of a double-T beam in the German assortment No. 55 with \(W_x = 3602\).

Weight and cost of the headframe.

1) Weight of the strut truss . . . 12000 kg.
2) “ of the head of the vertical column (2 pulleys, roof, platform of the head beams)—9900 “
3) “ of the vertical column . . . 2000 “
4) “ of rivets and additional angles . . . 400 “
5) “ of the main beams, 4 pcs. . . . 3600 “
6) “ of foundation bolts . . . 2400 “

Count as exactly . . . 38000 kg.

Taking the erection cost at 30 kopecks per kilogram, we obtain the cost of an iron headframe:

\[ 38000 \times 30 \approx 11400 \text{ rubles.} \]

Data for a preliminary estimate of a designed iron headframe.

For a preliminary estimate of the cost of a headframe whose design is to be developed, one uses data from existing structures. Thomson’s table was given by me in the first part of the present work; I now give Gefer’s table.

Height of headframe.	Rope tension.	Actual weight of headframe.	Calculated weight.	Difference.
10 m.	1500 kg.	8500 kg.	9500 kg.	+ 1000 kg.
10 ”	2000 ”	11500 ”	10500 ”	− 1000 ”
12 ”	2500 ”	12170 ”	13800 ”	+ 1630 ”
13 ”	2000 ”	12400 ”	13650 ”	+ 1250 ”
14 ”	2000 ”	13000 ”	14700 ”	+ 1700 ”
14.5 ”	2000 ”	16000 ”	14925 ”	− 1075 ”
15 ”	1500 ”	12100 ”	14250 ”	+ 2150 ”
15 ”	2100 ”	16400 ”	16050 ”	− 350 ”
16 ”	1800 ”	16900 ”	16160 ”	− 740 ”
17 ”	1900 ”	18100 ”	17510 ”	− 590 ”
19 ”	1850 ”	18700 ”	19380 ”	+ 680 ”
20 ”	2000 ”	19600 ”	21001 ”	+ 1400 ”
21 ”	2150 ”	20700 ”	22680 ”	+ 1980 ”
21 ”	4450 ”	22200 ”	32340 ”	+ 10140 ”
22 ”	4200 ”	37000 ”	32780 ”	− 4220 ”
23 ”	2140 ”	26000 ”	24798 ”	− 1202 ”
23 ”	2260 ”	27200 ”	25346 ”	− 1854 ”
23 ”	4100 ”	31600 ”	33810 ”	+ 2210 ”
24 ”	3800 ”	29200 ”	33840 ”	+ 4640 ”
26 ”	4300 ”	39400 ”	39260 ”	− 140 ”
27 ”	4100 ”	42700 ”	39690 ”	− 3010 ”
28 ”	5000 ”	46000 ”	46200 ”	+ 200 ”

It is easy to see that the use of such tables is difficult, for if the height of the proposed headframe approaches one of those given in the table, the rope tension may differ greatly from the proposed one, or else the reverse.

If \(E\) is the weight of the headframe, then in general

\[ E = F(Hs) = A + BH + Cs + DH^{2} + KHs - Ls^{2} + \ldots \]

The empirical formula for \(E\) must satisfy the condition of simplicity. On the basis of a series of trial calculations I come to the conviction that one should stop at the expression:

\[ E = H(a + bs), \]

in which \(H\) is the height of the headframe in meters, \(s\) is the tension of the rope, and \(E\) is the weight of the headframe in kilograms; \(a\) and \(b\) are certain numerical coefficients that must be determined.

Let us write a series of numerical equalities

\[ \frac{E}{H} = a + bs; \]

substituting for \(s\) and \(\frac{E}{H}\) their values taken from the above table, compute the average from these equalities for one \(a\), and determine the differences between the actual equalities and the arithmetical mean. Using these differences, we compute the numerical values of \(b\); we take the mean of the determinations as the sought numerical coefficient, after which \(a\) is determined by substituting \(b\) into the arithmetical mean equality.

Here is the course of the calculation.

Equality	Difference equation	Value of \(b\)
\(850 = a + 1500b\)	\(301 = 1211b\)	\(b = 0.248\)
\(1150 = a + 2000b\)	\(1 = 711b\)	\(b = 0.001\)
\(1014 = a + 2500b\)	\(133 = 211b\)	\(b = 0.630\)
\(954 = a + 2000b\)	\(197 = 711b\)	\(b = 0.277\)
\(929 = a + 2000b\)	\(222 = 711b\)	\(b = 0.312\)
\(1103 = a + 2000b\)	\(48 = 711b\)	\(b = 0.067\)
\(807 = a + 1500b\)	\(344 = 1211b\)	\(b = 0.284\)
\(1093 = a + 2100b\)	\(58 = 611b\)	\(b = 0.095\)
\(1056 = a + 1800b\)	\(95 = 911b\)	\(b = 0.104\)
\(1065 = a + 1900b\)	\(86 = 811b\)	\(b = 0.106\)
\(984 = a + 1850b\)	\(167 = 961b\)	\(b = 0.173\)
\(980 = a + 2000b\)	\(171 = 711b\)	\(b = 0.240\)
\(986 = a + 2150b\)	\(165 = 561b\)	\(b = 0.294\)
\(1057 = a + 4450b\)	\(94 = -1739b\)	\(b = -0.054\)
\(1662 = a + 4200b\)	\(-511 = -1489b\)	\(b = 0.343\)
\(1130 = a + 2140b\)	\(21 = 571b\)	\(b = 0.037\)
\(1183 = a + 2260b\)	\(-32 = 451b\)	\(b = -0.071\)
\(1374 = a + 4100b\)	\(-223 = -1389b\)	\(b = 0.160\)
\(1217 = a + 3800b\)	\(-66 = 1089b\)	\(b = -0.060\)
\(1515 = a + 4300b\)	\(-364 = -1589b\)	\(b = 0.229\)
\(1581 = a + 4100b\)	\(-430 = -1389b\)	\(b = 0.309\)
\(1642 = a + 5000b\)	\(-491 = -2289b\)	\(b = 0.215\)
\(25332 = 22a + 59650b\)		\(22b = 4.184 - 0.125 = 4.059\)
\(1151 = a + 2711b\)		\(b = 0.185\)

\[ a = 1151 - 501 = 650. \]

And therefore our formula takes the form

\[ E = H(650 + 0.185s). \]

For simplicity we shall take:

\[ E = H(650 + 0.2s), \]

for such a simplification changes the result of the calculation only slightly, and in the direction of safety. The degree of suitability of the present formula may be judged from the differences between the actual weight of the headframes and that computed on the basis of the formula. The indicated data are given as additions in the last two columns of Gefer’s table.

The weight computed on the basis of the formula is multiplied, in order to obtain the cost of the headframe, by the price per kilogram of iron structures in the given locality.

If the height of the headframe \(H'\) is given in feet, and the rope tension \(s'\) in poods, in which it is likewise required to determine the weight of the structure \(E'\), then it is clear that

\[ 16.38E' = 0.304H'(650 + 0.185 . 16.38s'). \]

Whence we obtain

\[ E' = H'(12.063 + 0.056s') \]

or, in simplified form,

\[ E' = H'(12 + 0.06s'). \]

To the article by S. Yu. Doborzhinsky.—Materials and Research on the Theory and Calculation of Headframe Structures.

Scale
1:200.

Drawing 1.

Drawing 2.

Drawing 3.

Drawing 4.

Drawing 5.

Drawing 6.

1 mm. = 800 kg.

Drawing 13.

Drawing 12.

Force scale: 1 mm = 2000 kg.
Length scale: 1:200.

Drawing 14.

Drawing 8.

1 mm = 2000 kg.

Drawing 10.

[[unclear: handwritten label on the line]]

[[unclear: handwritten label on the inclined member]]

Drawing 16.

1 mm = 2000 kg.

Drawing 9.

Drawing 11.

Form B.

Form A.

Drawing 17.

Drawing 18.

Fig. 7.

Scale 1:200.

Plate 19.

Scale: 1 mm = 400 kg

Plate 20.

Plate 24.

Plate 15.

Scale 1:200

Scale: 1 mm = 2000 kg.

Table I of Sections

Truss A

Member Nos.	Actual load	Section	\(J_x,\ \mathrm{cm}^4\)	\(J_y,\ \mathrm{cm}^4\)
1, 6, 11, 12, 17, 21, 22, 23	28500	[[section sketch; visible dimensions include 600 and 140; composed section with axes \(x\), \(y\)]]	46478	1651
10, 14, 19	31000	[[section sketch; visible dimensions include 600 and 300; composed section with axes \(x\), \(y\)]]	60476	5662
13, 15, 16, 18	28000	[[section sketch; visible dimensions include 600 and 600; composed section with axes \(x\), \(y\)]]	81987	33570
3, 5, 8, 9, 20, 24, 25, 27	5600	[[section sketch; visible dimensions include 600 and 160; composed section with axes \(x\), \(y\)]]	19317	902
2, 4, 7, 22, 26, 28	28500	[[section sketch; visible dimensions include 600 and 140; composed section with axes \(x\), \(y\)]]	48854	1622

Truss B

Member Nos.	Actual load	Section	\(J_y,\ \mathrm{cm}^4\)	\(J_x,\ \mathrm{cm}^4\)
5, 9, 17, 21, 19, 24, 28, 29, 38, 44	19800	[[section sketch; visible dimensions include 400 and 120; composed section with axes \(x\), \(y\)]]	28904	2913
9, 10, 15, 18, 21, 25, 31, 36, 41	24860	[[section sketch; visible dimensions include 400 and 160; composed section with axes \(x\), \(y\)]]	39972	9213
13, 14, 27, 28	14000	[[section sketch; visible dimensions include 400 and 70; composed section with axes \(x\), \(y\)]]	10522	325
7, 8, 33, 37	16700	[[section sketch; visible dimensions include 400 and 90; composed section with axes \(x\), \(y\)]]	26391	323
22, 23, 35, 4, 5, 10, 15, 20, 25, 29, 30, 30, 34, 39, 40, 42, 43, 45, 46	9400	[[section sketch; visible dimensions include 400 and 100; composed section with axes \(x\), \(y\)]]	10835	324

Drawing 21.

Drawing 22.

Visible inscriptions on the drawing:

• Plan
• Machine room
• Unloading room
• Passenger building
• Shaft
• Building
• Shaft

Scale [[unclear: remainder of word/scale designation cut off]]

[[Engineering drawing; no readable caption or body text visible on this page.]]

[[unclear: handwritten label inside lower-right plan view]]

Scale 1:40.

To the article by S. Yu. Dobryninsky — “Materials and Research on the Theory and Calculation of Headframe Structures.”

Drawing 34.

Drawing 57.

Force scale: 1 mm = 250 kg.
Length scale: 1:50.

Drawing 38.

Drawing 35.

Scale, 1:40

Drawing 36.

Drawing 33.

Drawing 25.

Drawing 26.

Scale, 1:40

Drawing 59.

Drawing 31.

Drawing 32.

Scale 1:200

Visible drawing dimensions and labels:

• 7 m
• 2.50
• 1.50
• 1.25 m
• 0
• 5

Scale 1:200

Drawing 27.

Fig. 30.

[Technical drawing of a headframe structure with foundation details, bracing, platforms, and section/detail views. Visible callout numbers include 29, 30, 31, and 32; visible dimension annotations include 600 and 800.]

To the article by S. Yu. Dobrzynski, “Materials and Research on the Theory and Calculation of Headframe Structures.”

Drawing 50.

\(1\ \mathrm{mm} = 200\ \mathrm{kg}\).

Drawing 51.

\(1\ \mathrm{mm} = 400\ \mathrm{kg}\).

Drawing 54-a.

\(1\ \mathrm{mm} = 400\ \mathrm{kg}\).

Length scale \(1:100\).

Drawing 62.

Length scale \(1:200\).
Force scale: \(1\ \mathrm{mm} = 50\ \mathrm{kg}\).

Drawing 37.

Drawing 38.

Drawing 39.

Drawing 40.

Drawing 43-a.

Length scale \(1:100\).
\(1\ \mathrm{mm} = 1000\ \mathrm{kg}\).

Drawing 45-a.

Drawing 41-a.

Scale: 1 mm = 1000 kg.

Drawing 42-a.

Table IV.

No.	[[unclear: sign column]]	[[unclear: force/value column]]
1	−	1000
2		0
3		0
4		0
5		0
6	−	1130
7	−	1120
8	−	1650
9	−	1650
10	+	1000
11	+	1000
12	+	1600
13	+	1600
14	+	1060
15	+	1060
16		0
17		0
18		0
19	+	1120
20	+	1120

Scale of lengths: 1 : 500. Scale of forces: 1 mm = 250 kg.

Drawing 48.

Drawing 61.

Drawing 42.

Drawing 41.

Drawing 43.

Scale: 1 mm = 500 kg.

Drawing 44.

Drawing 46.

Drawing 45.

Table III of dimensions and forms of sections

Section group	Visible designation / heading	Diagram or section shown	Dimensions and notes
Economic sections	[[unclear: handwritten column heading]]	[[diagram: small built-up section]]	[[unclear: dimensions and notes illegible]]
Economic sections	I-beam sections	[[diagram: built-up I-section]]	[[unclear: dimensions and notes illegible]]
Economic sections	I-beam sections	[[diagram: built-up I-section]]	[[unclear: dimensions and notes illegible]]
Economic sections	Channel sections	[[diagram: built-up channel section]]	[[unclear: dimensions and notes illegible]]
Economic sections	[[unclear: heading]]	[[diagram: double-T or I-section]]	[[unclear: dimensions and notes illegible]]
Economic sections	Plane walls	[[diagram: plate-and-flange section]]	[[unclear: dimensions and notes illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
Economic sections	I-beam sections	[[diagram: I-section]]	[[unclear: dimensions and notes illegible]]
Economic sections	Channel sections	[[diagram: channel section]]	[[unclear: dimensions and notes illegible]]
Economic sections	Angle sections	[[diagram: angle section]]	[[unclear: dimensions and notes illegible]]
Economic sections	Angle sections	[[diagram: angle section]]	[[unclear: dimensions and notes illegible]]
Economic sections	Corner sections	[[diagram: large angle section]]	[[unclear: dimensions and notes illegible]]
Economic sections	Plane sections	[[diagram: plate section]]	[[unclear: dimensions and notes illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]
[[unclear: row heading]]	[[unclear]]	[[unclear]]	[[unclear: tabular numeric values illegible]]

Table II of forces in the members of the strut truss.

No. of member	Vertical loads	No. of member	Vertical loads	Additional forces from horizontal pressure
1	−3500	2	−3500	−1800
5	+1200	6	+2000
9	+2800	10	+2500
13	+4800	14	+2800
17	+12375	19	−9000	−1800
21	+75	23	+116	+1200
2.3	+175	18	+5000	+4200
4	+9600	4	+9800	+10600
6	+15600	8	+15800	+10600
10	+64000	12	+18400	+10600
14	+39200	16	+[[unclear]]	+10600
18	+[[unclear]]	17	+[[unclear]]	+7000
22	+37000	1	+30000	+10600
24	+31200	5	+28200	+10600
26	−1500	9	+25800	+10600
28	−3500	13	+[[unclear]]	+[[unclear]]
30	−4800	15	+30200	+10300
32	−8460	27	+930	+7000
34	−5515	3	−150
36	+4000	7	−4800
38	+3530	11	−3000	+4200
		15	−1000	+4200
		22	+[[unclear]]
		24	+[[unclear]]

Table II. Forces in the members of the strut truss under oblique load.

Drawing labels visible on the structural details:

• Drawing 54.
• Drawing [[unclear: number]].

Fig. 53.

Fig. 55.

Fig. 56.

Fig. 60.

1 mm = 40 kg.

Dynamic deflection and deformation

2000.

Line of action of forces