Pyramidal Headframes

The principal plane truss of pyramidal headframes is a trapezium which, if all joints are assumed to be hinged, constitutes a geometrically deformable figure. Under the action of active forces \(T\), this figure will skew until it assumes a form stable for active forces of the given direction (Fig. 54).

In construction practice such structures are used rarely (suspension bridges) and are not applicable to mine headframes, which are subjected to forces variable both in direction and in magnitude. If the ends of the legs \(AB\) and \(CD\) are fixed immovably into strong foundations, then skewing is eliminated, but the legs are then subjected to lateral bending, which is undesirable. In view of this, rigid joints are used and, in addition, at least one more bar is introduced, by which the truss becomes geometrically invariable with hinged joints. This purpose can be achieved by means of a single diagonal brace, but in practice this method is seldom resorted to, because in any fairly substantial structure such a brace would turn out to be too long. The simplest way of imparting rigidity to the truss is to connect the legs by means of a horizontal tie; in this case the legs remain uncut. The tie is connected to them quite rigidly, or in such a way that it can move through small angles. Let us determine the stresses acting in the members of such an ideal truss (Fig. 55). Let \(l_1\) and \(l_2\) be the distances of the point of application of the force \(T\), measured along the upper beam, to the tops of the legs; \(\sigma_0\) the stress in the horizontal brace; \(\sigma\) in the upper chord; \(\sigma_3\) in the lower tie—sill; \(\sigma'_1\) and \(\sigma'_2\) the stresses in the upper (up to the brace) parts of the legs, and \(\sigma_1, \sigma_2\) in the lower parts. The remaining notations are as before and are clear from the drawing.

Fig. 54.

The equations of the moments of the forces with respect to the poles \(A\) and \(B\), in the presence of sections \(ab\) and \(a'b'\), will be

\[ 1)\quad \sigma_1' l \cos \eta + T \cos \varepsilon\, l_2 = 0, \]

\[ 2)\quad \sigma_2' l \cos \xi + T \cos \varepsilon\, l_1 = 0; \]

whence we have

\[ \sigma_1' = -T\,\frac{l_2 \cos \varepsilon}{l \cos \eta}, \tag{47} \]

\[ \sigma_2' = -T\,\frac{l_1 \cos \varepsilon}{l \cos \xi}. \]

Fig. 55.

Section \(cd\) gives us the possibility of setting up the following equations of moments with respect to the poles \(C\) and \(D\):

\[ 3)\quad \sigma_1 L \cos \eta + T \cos \varepsilon\,(l_2 + H \operatorname{tg}\xi) - T \sin \varepsilon\, H = 0, \]

\[ 4)\quad \sigma_2 L \cos \xi + T \cos \varepsilon\,(l_1 + H \operatorname{tg}\eta) + T \sin \varepsilon\, H = 0, \]

whence

\[ \sigma_1 = -T\,\frac{(l_2 + H\operatorname{tg}\xi)\cos \varepsilon - H\sin \varepsilon}{L\cos \eta}, \tag{48} \]

\[ \sigma_2 = -T\,\frac{(l_1 + H\operatorname{tg}\eta)\cos \varepsilon - H\sin \varepsilon}{L\cos \xi}. \]

Section \(ef\) with respect to pole \(D\) gives the equation—

\[ 5)\quad \sigma_1 L \cos \eta + \sigma_0 H_1 + \sigma H + T \cos \varepsilon\,(l_1 + H\operatorname{tg}\xi) - T \sin \varepsilon\, H = 0, \]

but, in view of the presence of equation 3, equation 5 assumes the form:

\[ 6)\quad \sigma_0 H_1 + \sigma H = 0. \]

Section \(gh\) with respect to pole \(C\) gives the equation—

\[ 7)\quad \sigma_2' L \cos \xi + \sigma_0 H_1 + T \cos \varepsilon\,(l_1 + H\operatorname{tg}\eta) + T \sin \varepsilon\, H = 0, \]

which makes it possible to write:

\[ 8)\quad (\sigma_2 - \sigma_2')L\cos \xi - \sigma_0 H_1 = T\sin \varepsilon\, H. \]

Finally, equations 6 and 7 give, with the preliminary elimination of \(\sigma_2'\),

\[ \sigma_0=-\mathrm{T}\frac{H}{H_1}\left[\frac{l_2 \operatorname{tg}\eta-l_1 \operatorname{tg}\xi}{l}\cos\varepsilon+\sin\varepsilon\right]; \]

\[ \sigma_0=\mathrm{T}\left[\frac{l_2 \operatorname{tg}\eta-l_1 \operatorname{tg}\xi}{l}\cos\varepsilon+\sin\varepsilon\right], \tag{49} \]

The section \(km\) enables us to compose the equation of static equilibrium:

\[ 9)\quad \sigma_0+\sigma+\sigma_3=0, \]

and hence—

\[ \sigma_3=+\mathrm{T}\frac{H-H_1}{H_1}\left[\frac{l_2 \operatorname{tg}\eta-l_1 \operatorname{tg}\xi}{l}\cos\varepsilon+\sin\varepsilon\right]. \tag{50} \]

The legs transmit to the supports forces directed along them; the horizontal component of these forces will be \(\sigma_3'\)

\[ 10)\quad \sigma_3' + \sigma_2 \sin\xi-\sigma_1 \sin\eta=0. \]

The difference \((\sigma_3-\sigma_3')\) determines that horizontal pressure which is transmitted to the supports as a result of the lateral bending of the legs. If initially the truss has the form \(ACDF\) (Fig. 56), then from the action of the horizontal force \(\sigma\) it is deformed and assumes a form which we may represent by the figure \(AB_1C_1D_1E_1F\). Strictly speaking, for a complete explanation of the action of forces on a truss of the type under consideration, one should determine the stress due to the lateral bending of the legs, but we shall not analyze this question because of its complexity.

Fig. 56.

Finally, \(\sigma_3'\) is easily obtained with the aid of equation 10 and the already found values of \(\sigma_1\) and \(\sigma_2\)

\[ \sigma_3'=\mathrm{T}\frac{(l_1\operatorname{tg}\xi-l_2\operatorname{tg}\eta)\cos\varepsilon+H(\operatorname{tg}\xi+\operatorname{tg}\eta)\sin\varepsilon}{L}. \tag{51} \]

For the graphical determination of the same forces, the following simple method may be proposed:

Drawing 57.

Extend the sides \(AB\) and \(CD\) to their mutual intersection at \(G\) (Drawing 57). Resolve the force \(T\) along the direction \(GFE\) and the panel \(BC\). At the point \(G\), introduce a force \(GI\), equal to and opposite in direction to \(FS\), and a horizontal force \(MG\), whose moment with respect to one of the supports would be equal and opposite to the moment of the force \(FN\), i.e. the force \(GM\) is determined by the equation:

\[ GM\cdot (H+H') + FN\cdot H = 0. \]

For this purpose we set off \(PG = FE\) and join \(P\) with \(N\). The horizontal force \(GM_1\), in magnitude, satisfies this condition, since

\[ \frac{GM_1}{FN}=\frac{GP}{PF}=\frac{EF}{EG}=\frac{H}{H+H'}. \]

If \(MG = M_1G\), then it remains for us to resolve \(GI\) and \(MG\) along the directions \(AG\) and \(GD\), or, more precisely, along their extensions, and therefore we shall have:

\[ \sigma_2 = GK + Gu, \]

\[ \sigma_1 = GL - Gi. \]

The validity of the method is proved as follows: since \(NT = FS\), then from triangle \(FNT\) we obtain—

\[ \frac{FN}{\sin(\varepsilon-\delta)} = \frac{FS}{\sin(90-\varepsilon)} = \frac{T}{\sin(90+\delta)}, \]

whence—

\[ FN = T\frac{\sin(\varepsilon-\delta)}{\cos\delta}, \]

\[ FS = T\frac{\cos\varepsilon}{\cos\delta}. \]

To decompose the force \(-\mathrm{T}_1 = F S\), we can use formulas (37), changing the notation in them according to the given case; in exactly the same way, by formula (38), we determine the components \(GM\)

\[ GL = -\mathrm{T}_1 \frac{\sin(\xi-\delta)}{\sin(\eta+\xi)}, \]

\[ GK = -\mathrm{T}_1 \frac{\sin(\eta+\delta)}{\sin(\eta+\xi)}, \]

\[ Gi = -GM \frac{\cos \xi}{\sin(\eta+\xi)}, \]

\[ Gn = GM \frac{\cos \eta}{\sin(\eta+\xi)}; \]

but since

\[ GM = -FN \frac{H}{H+H'} = -FN \frac{L-l}{L}, \]

then, after substitution, we find that

\[ \sigma_2 = -\mathrm{T}\left[ \frac{\cos \varepsilon}{\cos \delta} \frac{\sin(\eta+\delta)}{\sin(\eta+\xi)} + \frac{L-l}{L} \frac{\cos \eta}{\cos \delta} \frac{\sin(\varepsilon-\delta)}{\sin(\eta+\xi)} \right], \]

\[ \sigma_1 = -\mathrm{T}\left[ \frac{\cos \varepsilon}{\cos \delta} \frac{\sin(\xi-\delta)}{\sin(\eta+\xi)} - \frac{L-l}{L} \frac{\cos \xi}{\cos \delta} \frac{\sin(\varepsilon-\delta)}{\sin(\eta+\xi)} \right]. \tag{52} \]

In order to bring these expressions to the form (48), we shall use the geometric properties of Figure 57, namely:

\[ l_1 = H'(\operatorname{tg}\eta+\operatorname{tg}\delta), \quad L = l + H(\operatorname{tg}\eta+\operatorname{tg}\xi) \quad \text{and} \quad H' = H\frac{l}{L-l}, \]

which make it possible to determine the angle \(\delta\):

\[ \operatorname{tg}\delta = \frac{l_1\operatorname{tg}\xi - l_2\operatorname{tg}\eta}{l}. \]

We shall now write the expressions for the stresses as follows:

\[ \sigma_2 = -\mathrm{T}\left[ \frac{\cos \varepsilon}{\operatorname{tg}\eta+\operatorname{tg}\xi} \left( \frac{\sin\eta+\cos\eta\,\operatorname{tg}\delta}{\cos\xi\cos\eta} \right) + \frac{H(\operatorname{tg}\eta+\operatorname{tg}\xi)}{L} \frac{\cos\eta}{\operatorname{tg}\eta+\operatorname{tg}\xi} \cdot \left( \frac{\sin\varepsilon-\cos\varepsilon\,\operatorname{tg}\delta}{\cos\xi\cos\eta} \right) \right]; \]

\[ \sigma_1=-T\left[ \frac{\cos \varepsilon}{\operatorname{tg}\eta+\operatorname{tg}\xi} \left(\frac{\sin \xi-\cos \xi\,\operatorname{tg}\delta}{\cos \xi \cos \eta}\right) - \frac{H(\operatorname{tg}\eta+\operatorname{tg}\xi)}{L} \frac{\cos \xi}{\operatorname{tg}\eta+\operatorname{tg}\xi} \cdot \left(\frac{\sin \varepsilon-\cos \varepsilon\,\operatorname{tg}\delta}{\cos \xi \cos \eta}\right) \right]; \]

after reduction these expressions assume the following form:

\[ \sigma_2= \frac{-T}{L\cos \xi} \left[ \frac{(\operatorname{tg}\eta+\operatorname{tg}\delta)L}{\operatorname{tg}\eta+\operatorname{tg}\xi} \cos \varepsilon - H\operatorname{tg}\delta \cos \varepsilon + H\sin \varepsilon \right], \]

\[ \sigma_1= \frac{-T}{L\cos \eta} \left[ \frac{(\operatorname{tg}\xi-\operatorname{tg}\delta)L}{\operatorname{tg}\eta+\operatorname{tg}\xi} \cos \varepsilon + H\operatorname{tg}\delta \cos \varepsilon - H\sin \varepsilon \right]. \]

Substituting into these expressions the value found for \(\operatorname{tg}\delta\), let us first write them in the following way:

\[ \sigma_2= \frac{-T}{L\cos \xi} \left[ \frac{l(\operatorname{tg}\eta+\operatorname{tg}\delta)}{\operatorname{tg}\eta+\operatorname{tg}\xi} \cos \varepsilon + H\operatorname{tg}\eta \cos \varepsilon + H\sin \varepsilon \right]; \]

\[ \sigma_1= \frac{-T}{L\cos \eta} \left[ \frac{l(\operatorname{tg}\xi-\operatorname{tg}\delta)}{\operatorname{tg}\eta+\operatorname{tg}\xi} \cos \varepsilon + H\operatorname{tg}\xi \cos \varepsilon - H\sin \varepsilon \right], \]

and then they take exactly the form of equation (48). The remaining forces \(\sigma_0\), \(\sigma\), \(\sigma_3\) are readily obtained by means of the same general methods of geometrical construction, so that we shall consider this question exhausted.

Formulas (48) enable us to draw the following conclusion: in pyramidal headframes of the type considered, in the rear legs (facing the machines) compression will always take place; in the front legs, however, only so long as

\[ (l_2+H\operatorname{tg}\xi)\cos \varepsilon-H\sin \varepsilon>0 \]

or, so long as

\[ l+H\operatorname{tg}\xi>H\operatorname{tg}\varepsilon, \tag{53} \]

or, geometrically, so long as the resultant of all the forces does not intersect the rear legs (see Fig. 57).

\[ UD>UW. \]

Formulas (49) show that in the part \(\sigma B\) of the upper panel there is a compressive force, and in the tie—a tensile one. Stress—

in the lower chord, the bearing, and the sum of the horizontal reactions of the supports are not equal, as in prismatic headframes, to the horizontal component of the active force \(T\), but differ from it by a certain amount

\[ \Delta \sigma = \sigma_3 - T \sin \varepsilon . \]

Again, if we turn to the method of calculation by means of construction, it will turn out that the force to be taken up directly by the supports is:

\[ \frac{H'}{H+H'}\,FN=\frac{l}{L}\,FN . \]

If the active force \(T_2\) were applied not to the upper chord, but to the bracing, then on the basis of an analogous calculation we would obtain:

\[ \sigma_2^{0}=-T_2\frac{(l_1' + H_1 \operatorname{tg}\eta)\cos\varepsilon_2 + H_1\sin\varepsilon_2}{L\cos\xi}, \]

\[ \sigma_1^{0}=-T_2\frac{(l_2' + H_1 \operatorname{tg}\xi)\cos\varepsilon_2 - H_1\sin\varepsilon_2}{L\cos\eta}, \tag{54} \]

and for the horizontal forces—

\[ \sigma^{0}=-T_2\frac{H_1}{H}\left[\frac{l_2'\operatorname{tg}\eta-l_1'\operatorname{tg}\xi}{l'}\cos\varepsilon_2+\sin\varepsilon_2\right], \]

\[ \sigma_0^{0}=-T_2\left[\frac{l_2'\operatorname{tg}\eta-l_1'\operatorname{tg}\xi}{l'}\cos\varepsilon_2+\sin\varepsilon_2\right]. \tag{55} \]

and so forth. In the case of a headframe with compartments arranged one behind another, when in the main plane of the truss there simultaneously act the forces \(T_1\) and \(T_2\), inclined to the verticals at the angles \(\varepsilon_1\) and \(\varepsilon_2\), in order to obtain the total stresses they should be summed:

\[ (\sigma_2+\sigma_2^{0}),\quad (\sigma_1+\sigma_1^{0}),\quad (\sigma+\sigma^{0}),\quad (\sigma_0+\sigma_0^{0}), \]

remembering that

\[ L=l+H(\operatorname{tg}\eta+\operatorname{tg}\xi) =l' + H_1(\operatorname{tg}\eta+\operatorname{tg}\xi). \]

Only the type of main truss we have examined is often characteristic of small wooden headframes. For larger headframes, especially iron ones, it is preferred to use trusses with diagonal crossing braces;

whereby the legs remain continuous. Often, in wooden headframes, in addition to the horizontal tie, diagonal rods are also used—in the part between the upper panel and the tie (Fig. 58).

Fig. 58.

This case is easily reduced to the one considered earlier by means of a simple resolution of forces, indicated in the drawing. The force \(T\), transferred to some arbitrary point \(G\), is resolved along the directions \(GB\) and \(GE\), and then the forces \(Gh'\) and \(Be'\)—along the directions of the legs and the diagonal rods.

Of course, the resistances of the legs will be equal and opposite to the forces: \((Bd - Ee)\) and \((Ca - Fb)\).

The stresses in the rods are: \(Ei = Cn\) and \(Bl = Fm\). There remains the force \((Ef - gF)\), which causes stresses in the legs, easily computed by formulas (48–51), or by means of a fictitious force, whose moment, taken with respect to any one of the supports, is equal to \((Ef - gF)H_1\); the force producing this moment is applied at the point of intersection of the legs \(BA\) and \(CD\). We resolve this force along the directions of the legs.

In the panel \(BC\) and the rod \(EF\), the stresses are conveniently computed with the aid of the simple formulas (49). However, by this method we determine the stresses in \(BC\) and \(EF\) only approximately, since we have not introduced the stresses in the rods, which play some role in the transmission of the horizontal force. In the relation under consideration, the part of the truss \(FBCF\) is statically indeterminate: either the horizontal or the diagonal rods are superfluous. If the horizontal force \(Ef - gF\) is denoted by \(\rho\), where \(\rho = -\sigma_0\) (formula 49), the stresses in the rods by \(s_1\) and \(s_2\), and the angles of their inclination to the horizon by \(\varphi\) and \(\psi\), then, for equilibrium, the following equations must hold (of course, in the absence of the horizontal tie):

\[ s_1 \cos \varphi - s_2 \cos \psi + \rho = 0, \]

\[ s_1 \sin \varphi + s_2 \sin \psi = 0, \]

whence—

\[ s_1 = -\rho \frac{\sin \psi}{\sin(\psi + \varphi)}, \]

\[ s_2 = -\rho \frac{\sin \varphi}{\sin(\psi - \varphi)}. \tag{56} \]

If for some reason it is desirable to have expressions for the stresses as functions of the heights of the points at which they are attached to the legs, then one may use the relation

\[ \tg \varphi = \frac{H-H_1}{l+(H-H_1)\tg \eta} \quad \text{and} \quad \tg \psi = \frac{H-H_1}{l+(H-H_1)\tg \xi}. \]

A more exact calculation will be obtained if we divide \(\rho\) in half, that is, assign \(\frac{1}{2}\rho\) to the horizontal ties and \(\frac{1}{2}\rho\) to the inclined ones.

The trusses often used in practice, shown in Fig. 59, must, for an approximate calculation, be divided into two cases: (Figs. \(a\) and \(b\)).

Fig. 59.

If there are no horizontal ties, the calculation is very simple: it is only necessary to carry out the resolution of forces, by the method indicated above, as many times as the number of panels into which the truss is divided.

Combined Prismatic-Pyramidal Normal Headframes

It remains for us to consider the basic plane truss of headframes which, strictly speaking, cannot be classed as either prismatic or pyramidal, but constitute a combination of them. These are headframes which, perhaps, may be called six-legged; their drawings and description were given above.

The schematic drawing 60 represents the basic plane truss of this type. The resultant force \(T\) is applied at the point \(O\). The lower strut panel is usually absent; the legs bear against the foundations. We assume that all the joints are hinged. Let us compose the equations of moments of the active forces and stresses for all joints. For the upper joints, assuming that the force \(T\) is resolved into horizontal and vertical components, the equation of moments will be:

Drawing 60.

\[ 1)\quad \sigma_2 l\cos \xi+\sigma_3 l\cos \zeta+T l_1\cos \varepsilon=0, \]

\[ 2)\quad \sigma_1 l\cos \eta+T l_2\cos \varepsilon=0; \]

for the lower joints—

\[ 3)\quad \sigma_2 L\cos \xi+\sigma_3(L+L_1)\cos \zeta+T\cos \varepsilon(l_1+H\operatorname{tg}\eta)+T\sin \varepsilon\,H=0, \]

\[ 4)\quad \sigma_1(L+L_1)\cos \eta-\sigma_2 L_1\cos \xi-T\cos \varepsilon(l_2+H\operatorname{tg}\zeta)+T\sin \varepsilon\,H=0, \]

\[ 5)\quad \sigma_1 L\cos \eta+\sigma_3 L_1\cos \zeta-T\cos \varepsilon(l_2+H\operatorname{tg}\xi)+T\sin \varepsilon\,H=0. \]

In addition, we have the equations of static equilibrium at the supports:

\[ 6)\quad \sigma_1\sin \eta-\sigma_2\sin \xi-\sigma_3\sin \zeta-T\sin \varepsilon=0, \]

\[ 7)\quad \sigma_1\cos \eta+\sigma_2\cos \xi+\sigma_3\cos \zeta+T\cos \varepsilon=0. \]

Equations 1, 2, 6, and 7 are quite sufficient for determining all the unknowns; the roots obtained verify equations 3, 4, and 5; moreover, equation 1, added to equation 2, gives us equation 7, so that in reality we have only three equations with three unknowns. From equation 2 we have:

\[ \sigma_1=-\mathrm{T}\frac{l_2\cos\varepsilon}{l\cos\eta}. \tag{57} \]

If, by means of this expression, we eliminate \(\sigma_1\) from equation 6, then it assumes the form:

\[ 8)\quad +\sigma_2 l\sin\xi+\sigma_3 l\sin\zeta+\mathrm{T}(l\sin\varepsilon+l_2\cos\varepsilon\,\operatorname{tg}\eta)=0, \]

which, together with equation 1, gives us the possibility of determining \(\sigma_2\) and \(\sigma_3\):

\[ \sigma_2=-\mathrm{T}\frac{l_1\cos\eta\sin(\zeta-\varepsilon)-l_2\cos\zeta\sin(\eta+\varepsilon)} {l\cos\eta\sin(\zeta-\xi)} \]

\[ \sigma_3=-\mathrm{T}\frac{-\cos\eta\sin(\xi-\varepsilon)+l_2\cos\xi\sin(\eta+\varepsilon)} {l\cos\eta\sin(\zeta-\xi)}. \tag{58} \]

For the graphical determination of the stresses one may propose the following simple method. Since the resultant of the stresses directed along \(DB\) and \(EB\) must intersect at one point with the stress \(\sigma_1\), directed along \(AC\), and with the resultant \(\mathrm{T}\), therefore, in order to find the direction of the resultant of the forces \(\sigma_2\) and \(\sigma_3\), we connect the point of intersection of the direction of the leg \(AC\) with the direction of the force \(\mathrm{T}\), i.e. point \(F\), with \(B\); then \(FB\) is the required direction. At point \(F\) we assume an applied force equal to \(\mathrm{T}\) and resolve it along the directions \(AF\) and \(BF\); the force \(-\sigma_1=Fc\); \(Fb=\mathrm{T}_0\) is the resultant of \(-\sigma_2\) and \(-\sigma_3\). We transfer its point of application to \(B\) and resolve it along the directions \(DE\) and \(EB\); then

\[ -\sigma_2=Bd\quad\text{and}\quad -\sigma_3=Be. \]

To prove the correctness of the construction, let us consider the ratio of the sides and the sines of the angles in the triangle \(aFb\); since

\[ \frac{\sigma_1}{\sin(\beta-\varepsilon)} = \frac{T_0}{\sin(\eta+\varepsilon)} = \frac{T}{\sin(\beta+\eta)}, \]

then

\[ \sigma_1=-T\frac{\sin(\beta-\varepsilon)}{\sin(\beta+\eta)} \]

and

\[ T_0=-T\frac{\sin(\eta+\varepsilon)}{\sin(\beta+\eta)}, \tag{59-a} \]

from the triangle \(gBe\) we obtain:

\[ \frac{T_0}{\sin(\zeta-\xi)} = \frac{\sigma_2}{\sin(\zeta-\beta)} = \frac{\sigma_3}{\sin(\beta-\xi)}, \]

and consequently, substituting the above expression for \(T_0\), we shall be able to write:

\[ \sigma_2 = -T \frac{\sin(\eta+\varepsilon)}{\sin(\eta+\beta)} \cdot \frac{\sin(\zeta-\beta)}{\sin(\zeta-\xi)}, \]

\[ \sigma_3 = -T \frac{\sin(\eta+\varepsilon)}{\sin(\eta+\beta)} \cdot \frac{\sin(\beta-\xi)}{\sin(\zeta-\xi)}. \tag{59-б} \]

The expressions for the stresses just derived are convenient for logarithmic calculations; in them the angle \(\beta\) is still unknown, and in order to determine it we shall consider the triangles \(AFO\) and \(AFB\); from these triangles we have:

\[ \frac{AF}{\sin(90-\varepsilon)} = \frac{l_1}{\sin(\eta+\varepsilon)}, \]

\[ \frac{AF}{\sin(90-\beta)} = \frac{l}{\sin(\eta+\beta)}. \]

Eliminating \(AF\), we obtain the equation:

\[ l\cos\beta\sin(\eta+\varepsilon) - l_1\cos\varepsilon\sin(\eta+\beta) = 0, \]

from which we find:

\[ \tg\beta = \frac{l_2}{l_1}\tg\eta + \frac{l}{l_1}\tg\varepsilon. \]

If, in expressions (59), we expand the sines of the sums and differences of the angles with \(\beta\) and write thus:

\[ \sigma_2=-T\frac{\sin(\eta+\varepsilon)}{\sin(\zeta-\xi)}\cdot \frac{\sin\zeta-\cos\zeta\,\operatorname{tg}\beta}{\sin\eta+\cos\eta\,\operatorname{tg}\beta}, \]

\[ \sigma_3=-T\frac{\sin(\eta-\varepsilon)}{\sin(\zeta-\xi)}\cdot \frac{\sin\xi+\cos\xi\,\operatorname{tg}\beta}{\sin\eta+\cos\eta\,\operatorname{tg}\beta}, \]

then, replacing \(\operatorname{tg}\beta\) by the expression given above, we arrive at formula (58); the same, of course, also applies to \(\sigma_1\).

Frames of this type are usually built so that \(\eta=\xi\); most often these angles are equal to zero, that is, the legs supporting the sheave beams are vertical. In this case the expressions for the stresses are simplified:

\[ \sigma_1=-T\frac{l_2}{l}\cos\varepsilon \]

\[ \sigma_2=-T\frac{l_1\sin(\zeta-\varepsilon)-l_2\cos\zeta\sin\varepsilon}{l\sin(\zeta-\xi)} =-T\frac{l_1\sin\zeta\cos\varepsilon-l\cos\zeta\sin\varepsilon}{l\sin(\zeta-\xi)} \tag{60} \]

\[ \sigma_3=-T\frac{l_1\sin(-\varepsilon)+l_2\sin\varepsilon}{l\sin(\zeta-\xi)} =-T\frac{\sin\varepsilon}{\sin(\zeta-\xi)}. \]

All forces are compressive; \(\sigma_1\) is always compressive, as is evident from formula (59a), and \(\sigma_2\) and \(\sigma_3\) are so as long as

\[ l_1\cos\eta\sin(\zeta-\varepsilon)-l_2\cos\zeta\sin(\eta+\varepsilon)>0, \]

\[ -l_1\cos\eta\sin(\xi-\varepsilon)+l_2\cos\varepsilon\sin(\eta+\varepsilon)>0. \]

We transform these inequalities into the following:

\[ l_1\operatorname{tg}\zeta-l_2\operatorname{tg}\eta-l\operatorname{tg}\varepsilon>0 \]

\[ -l_1\operatorname{tg}\xi+l_2\operatorname{tg}\eta+l\operatorname{tg}\varepsilon>0 \]

The second inequality is always satisfied in ordinary designs; the first, however, must be borne in mind when preparing preliminary drawings. Graphically, it expresses the condition under which the direction of the force \(T_0\) (in our drawing, \(FB\)) does not pass outside the angle \(DBE\).

To determine the stresses acting in the upper panel, it is necessary to divide the vertical component of the force \(T\) among the joints; then it is clear that in the left part there will be the following force:

\[ \sigma' = T \frac{l_2}{l}\operatorname{tg}\eta \cos \varepsilon \tag{61a} \]

and for the right part (with respect to the point of application of the resultant),

\[ \sigma'' = T\left(\frac{l_2}{l}\operatorname{tg}\eta \cos \varepsilon + \sin \varepsilon\right). \tag{61b} \]

This, just like the left part of the vertical force, is received by the joint and transmitted by the legs to the supports.

Sometimes it is advised to move the supporting legs apart, and in this case the point of application of the active force \(T\) lies between \(B\) and \(C\) at \(O\) (Fig. 62). Let there be sections \(ab\) and \(cd\).

Fig. 62.

For the upper extreme half-belts the moment equations will be:

\[ 1)\quad \sigma_2 l \cos \xi + \sigma_3(l+l_0)\cos \xi' + T(l+l')\cos \varepsilon = 0, \]

\[ 2)\quad \sigma_1(l+l_0)\cos \eta + \sigma_2 l_0 \cos \xi + T l_2 \cos \varepsilon = 0; \]

for the lower extreme ones—

\[ 3)\quad \sigma_2 L \cos \xi + \sigma_3(L+L_1)\cos \xi' + T\cos \varepsilon\,(l+l_1+H\operatorname{tg}\eta) + T\sin \varepsilon\,H = 0, \]

\[ 4)\quad \sigma_1(L+L_1)\cos \eta + \sigma_2 L_1 \cos \xi + T\cos \varepsilon\,(l_2+H\operatorname{tg}\xi) - T\sin \varepsilon\,H = 0. \]

Moreover, it would also be possible to write the moment equations for the middle joints, but this is useless, since they will be identical with those obtained by eliminating \(\sigma_2\) from the 1st with the 2nd, or the 3rd with the 4th. The equations of static equilibrium of the supporting moments will be—

\[ 5)\quad -\sigma_1 \sin \eta + \sigma_2 \sin \xi + \sigma_3 \sin \xi' + T\sin \varepsilon = 0, \]

\[ 6)\quad \sigma_1 \cos \eta + \sigma_2 \cos \xi + \sigma_3 \cos \xi' + T\cos \varepsilon = 0; \]

these equations enable us, in determining the forces, to avoid the more complicated equations 3 and 4 and to use only

1, 2, and 5, and the 6th—as auxiliary, identical with the 1st and 2nd, or with the 3rd and 4th, considered jointly. Thus, if from equations 5 and 6 we eliminate the force \(\sigma_1\), we obtain the equation—

\[ 7)\quad \sigma_2 \sin(\xi+\eta)+\sigma_3 \sin(\zeta+\eta)+T\sin(\varepsilon+\eta)=0, \]

from which, with the aid of equation 1, we find:

\[ \sigma_3=-T\frac{(l+l_1)\sin(\varepsilon-\xi)\cos\eta+l_1\sin(\varepsilon+\eta)\cos\xi} {l\sin(\xi-\zeta)\cos\eta+l_0\sin(\eta+\xi)\cos\zeta}, \]

\[ \sigma_2=-T\frac{(l+l_1)\sin(\varepsilon-\zeta)\cos\eta+l_2\sin(\varepsilon+\eta)\cos\zeta} {l\sin(\xi-\zeta)\cos\eta+l_0\sin(\eta+\xi)\cos\zeta}; \tag{62} \]

we determine \(\sigma_1\) by substituting the expressions found into equation 6:

\[ \sigma_1=T\frac{l_1\sin(\varepsilon-\zeta)\cos\xi+l_2\sin(\varepsilon-\xi)\cos\zeta} {l\sin(\xi-\zeta)\cos\eta+l_0\sin(\eta+\xi)\cos\zeta}. \tag{62a} \]

Since all these stresses must be produced by compressive forces, it is necessary that the following inequalities hold:

\[ -(l+l_1)(\varepsilon-\xi)\cos\eta+l_1\sin(\varepsilon+\eta)\cos\xi>0 \]

\[ (l+l_1)\sin(\varepsilon-\zeta)\cos\eta+l_2\sin(\varepsilon+\eta)\cos\zeta\gtrless 0, \]

\[ l_1\sin(\varepsilon-\zeta)\cos\xi+l_2(\varepsilon-\xi)\cos\zeta\gtrless 0. \tag{62b} \]

It is easy, however, to prove that the existence of the first inequality often excludes the possibility of the last. Let us write them as follows:

\[ (l+l_1)\cos\eta+l_2\frac{\sin(\varepsilon+\eta)}{\sin(\varepsilon-\zeta)}\cos\zeta\gtrless 0, \]

\[ -(l+l_1)\cos\eta+l_1\frac{\sin(\varepsilon+\eta)}{\sin(\varepsilon-\xi)}\cos\xi\gtrless 0; \]

after addition and reduction by \(\sin(\varepsilon+\eta)\), which, as a positive quantity, does not change the sign of the inequality, we obtain

\[ \frac{l_1\cos\xi}{\sin(\varepsilon-\xi)}+\frac{l_2\cos\zeta}{\sin(\varepsilon-\zeta)}\gtrless 0; \]

if

\[ \sin(\varepsilon-\xi)>0 \quad \text{and} \quad \sin(\varepsilon-\zeta)>0, \]

then, in the presence of the condition—

\[ l\sin(\xi-\zeta)\cos\eta + l_0\sin(\eta+l_0)\sin(\varepsilon-\xi)\cos\zeta>0, \]

we have, from the preceding,

\[ l_1\sin(\xi-\zeta)\cos\xi + l_2\sin(\varepsilon-\xi)\cos\zeta \gtreqless 0, \]

and consequently tension will occur in the rear legs.

If the front and middle legs are set vertically, then the expressions for the stresses in such cases are greatly simplified. In view of the frequent use of such a construction, let us consider this case separately. Putting—

\[ \eta=\xi=0, \]

we find

\[ \sigma_3=-\mathrm{T}\frac{\sin\varepsilon}{\sin\zeta}, \]

\[ \sigma_2= \mathrm{T} \frac{(l+l_1)\sin(\varepsilon-\zeta)+l_2\sin\varepsilon\cos\zeta} {l\sin\zeta}, \tag{63} \]

\[ \sigma_1= -\mathrm{T} \frac{l_1\sin(\varepsilon-\zeta)+l_2\cos\zeta\sin\varepsilon} {l\sin\zeta}. \]

In order that all the forces be compressive, it is necessary that

\[ (l+l_1)\sin(\varepsilon-\zeta)+l_2\sin\varepsilon\cos\zeta<0, \]

\[ l_1\sin(\varepsilon-\zeta)+l_2\cos\zeta\sin\varepsilon>0; \]

this is possible, however, only when—

\[ \varepsilon-\zeta<0. \]

Builders sometimes try to choose \(\zeta\) so that, at \(\operatorname{Max}\mathrm{T}\),

\[ \zeta=\varepsilon. \]

Then:

\[ \sigma_2=\mathrm{T}\frac{l_2}{l}\cos\zeta \quad \text{and} \quad \sigma_1=-\mathrm{T}\frac{l_2}{l}\cos\zeta, \]

and consequently tension will appear in the middle legs. Of course, there is no need to mention that if the beam \(ABC\) is continuous and all the connections in general are rigid, then the reasoning we have presented has only an approximate significance. However, this remark applies to all our reasoning in general.

The conditional inequalities \((62b)\), after transformation, may be written as follows:

\[ (l+l_0)\operatorname{tg}\xi+l_0\operatorname{tg}\eta-l\operatorname{tg}\zeta \gtrless 0, \]

and since, in general, most often \(\eta=\xi\), then—

\[ \left(1+2\frac{l_0}{l}\right)\operatorname{tg}\xi \gtrless \operatorname{tg}\zeta . \]

Since, however—

\[ \zeta>\xi, \]

which is observed in all existing constructions, the first inequality somewhat limits the possibility of reducing the angle \(\xi\).

In general, consideration of the stress expressions gives us an indication of the circumstance that, in selecting the principal dimensions of a headframe of this type being designed, one must proceed very cautiously. A graphical calculation can be performed by means of a method similar to those given above. We continue the directions of the force \(T\) and of the leg \(BC\) until they meet at \(F\), and then connect the point \(F\) with the point \(G\)—the intersection of the extensions of the legs \(AE\) and \(ID\). We transfer the point of application of the force \(T\) to \(F\) and resolve it along the directions \(FC\) and \(FG\). The segment \(Fa\) gives us the force directed along the leg \(BC\), so that

\[ \sigma_3=-Fa \]

The segment \(Fb=\rho\) gives us the force taken up by the legs \(EA\) and \(ID\). We transfer its point of application to \(G\) and resolve it along the directions \(EG\) and \(GD\); the segments \(Gc\) and \(Gd\) give us the forces directed along these legs, so that:

Fig. 64.

\[ \sigma_2=-Gc \quad \text{and} \quad \sigma_1=-Gd. \]

For the proof of what has been stated, let us denote by \(\delta\) the angle of inclination of \(FG\) to the vertical line, and form the ratios:

\[ \frac{\sigma_3}{\sin(\varepsilon-\delta)} = \frac{\rho}{\sin(\zeta-\delta)} = \frac{T}{\sin[180-(\zeta-\delta)]}, \]

\[ \frac{\sigma_1}{\sin(\xi-\delta)} = \frac{\sigma_2}{\sin(\eta+\delta)} = \frac{\rho}{\sin[180-(\eta+\xi)]}; \]

whence it is easy to obtain, changing the sign of \(T\):

\[ \sigma_3 = -T\frac{\sin(\varepsilon-\delta)}{\sin(\zeta-\delta)}, \]

\[ \sigma_2 = -T\frac{\sin(\zeta-\varepsilon)}{\sin(\eta+\xi)} \cdot \frac{\sin(\eta+\delta)}{\sin(\zeta-\delta)}, \tag{64} \]

\[ \sigma_1 = -T\frac{\sin(\zeta-\varepsilon)}{\sin(\eta+\xi)} \cdot \frac{\sin(\xi-\delta)}{\sin(\zeta-\delta)}. \]

According to the drawing,

\[ \operatorname{tg}\delta=\frac{GM}{FM}; \]

and since

\[ GM = NB - I_0B -- OI,\quad \text{and} \]

\[ MF = FN - GO, \]

it remains to determine these lengths as functions of the known quantities.

And so \(AF = BF\cos\zeta,\ NB = BF\sin\zeta\); to determine \(BF\) we have the equality:

\[ \frac{BE}{\sin(90+\varepsilon)} = \frac{l_2}{\sin(\zeta-\varepsilon)}; \]

in view of this—

\[ AF = l_2\frac{\cos\zeta\cos\varepsilon}{\sin(\zeta-\varepsilon)}, \quad NB = l_2\frac{\sin\eta\cos\varepsilon}{\sin(\zeta-\varepsilon)}; \]

further:

\[ GO = IG\cos\xi,\quad OD = IG\sin\xi, \]

while—

\[ \frac{IG}{\sin(90-\eta)} = \frac{l}{\sin(\eta+\xi)}, \quad \text{so that} \]

\[ GO = l\frac{\cos\eta\cos\xi}{\sin(\eta+\xi)} \quad \text{and} \quad OI = l\frac{\cos\eta\sin\xi}{\sin(\eta+\xi)}. \]

Finally, we have:

\[ \operatorname{tg}\delta = \left[ l_2 \frac{\sin \zeta \cos \varepsilon}{\sin(\zeta-\varepsilon)} - l_0 - l \frac{\sin \xi \cos \eta}{\sin(\eta+\xi)} \right] : \left[ l_2 \frac{\cos \zeta \cos \varepsilon}{\sin(\zeta-\varepsilon)} - l \frac{\cos \xi \cos \eta}{\sin(\eta+\xi)} \right]. \]

It remains to transform the expression for $\sigma_1$, $\sigma_2$ and $\sigma_3$ so that $\operatorname{tg}\delta$ enters into them instead of $\cos\delta$ and $\sin\delta$, which is very easy to do; substituting the expression found above, we obtain formulas identical with (62).

The graphical method presented gives us the possibility of selecting the dimensions of the headframe so that all the stresses in the legs are caused by compressive forces; namely, the point of intersection of the direction of the force $T$ with the leg $BC$ (or, more precisely, with its prolongation) must lie between $F$ and $e$, for only in this case will the direction $GF$ be inside the angle $EGD$. In any other position of the point $F$ it will go outside, and then the middle or the front legs will be subjected to tension.

If the front and middle legs are placed vertically, then the general course of the construction remains the same. The force $T$ at the point $F$ (Fig. 65) is resolved along the direction of the inclined leg and the vertical; the latter, in turn, is resolved into directions parallel to it along the legs. The condition of compression must be taken to be the circumstance that $F$ should not go outside the space bounded by the prolongations of the vertical legs.

Fig. 65.

Sometimes structures are encountered in which the inclined leg supports the sheave beams in the space between the middle legs (Fig. 66). Of course, the scheme given in the drawing under consideration differs from actual structures in that the middle and rear legs do not lie in one plane. Let us suppose that at the point of their intersection in the drawing there is no connection whatsoever, so that the legs can freely slide one past the other. To determine the stresses in the legs we shall form equations of moments with respect to the upper extreme joints. According to

Fig. 66.

with the notation shown in the drawing, we shall have:

\[ \sigma_2 l\cos\eta+\sigma_3(l_1+l_0)\cos\zeta+T l_1\cos\varepsilon=0, \]

\[ \sigma_1 l\cos\xi+\sigma_3(l_2-l_0)\cos\zeta+T l_2\cos\varepsilon=0. \]

The equations of static equilibrium of the support reactions will be:

\[ \sigma_1\cos\eta+\sigma_2\cos\xi+\sigma_3\cos\zeta+T\cos\varepsilon=0, \]

\[ -\sigma_1\sin\eta+\sigma_2\sin\xi+\sigma_3\sin\zeta+T\sin\varepsilon=0. \]

The first of these equations follows directly from the equation of moments of the forces, so that we have only three equations. For the middle joints, and also with respect to the supports, we do not write equations of moments, since they cannot differ from the preceding ones. The equations of static equilibrium, after eliminating \(\sigma_1\), give the equality—

\[ \sigma_2\sin(\xi+\eta)+\sigma_3\sin(\zeta+\eta)+T\sin(\varepsilon+\eta)=0, \]

which, together with the first equation of moments, enables us to determine \(\sigma_2\) and \(\sigma_3\); after substitution into the equation of static equilibrium of the horizontal forces, we also find \(\sigma_1\). Thus we have:

\[ \sigma_3=-T\frac{l_1\cos\varepsilon\sin(\xi+\eta)-l\cos\eta\sin(\varepsilon+\eta)} {(l_1+l_0)\sin(\xi+\eta)\cos\zeta-l\cos\eta\sin(\zeta+\eta)}, \]

\[ \sigma_2=T\frac{l_1\cos\varepsilon\sin(\zeta+\eta)-(l_1+l_0)\cos\zeta\sin(\varepsilon+\eta)} {(l_1+l_0)\sin(\xi+\eta)\cos\zeta-l\cos\eta\sin(\zeta+\eta)}; \tag{66} \]

\[ \sigma_1=T\frac{l_1\cos\varepsilon\sin(\xi-\zeta)+(l_1+l_0)\cos\zeta\sin(\varepsilon-\xi)+l\cos\eta\sin(\zeta-\varepsilon)} {(l_1+l_0)\sin(\xi+\eta)\cos\zeta-l\cos\eta\sin(\zeta+\eta)}. \]

Most often the supporting legs are placed under the point of application of the active force; if, moreover, the legs are set vertically, then

\[ \eta=\xi=0 \quad \text{and} \quad l_0=0, \]

and then—

\[ \sigma_1=-T\frac{l_2\sin(\zeta-\varepsilon)}{\sin\zeta}, \]

\[ \sigma_2=-T\frac{l_1\sin(\zeta-\varepsilon)}{l\sin\zeta}, \tag{67} \]

$$ \sigma_3=-T\frac{\sin\varepsilon}{\sin\zeta}. $$

The following graphical method may be proposed for determining the forces acting along the legs.

We extend the direction of the force $T$ until it intersects the continuation of the support leg $CD$ at $F$; we transfer its point of application to $F$ and resolve $T$ into forces directed along $FD$ and $FG$, that is, along the direction of the leg and the straight line connecting $F$ with the point of intersection of the extensions of the front and middle legs. We transfer the point of application of the force $Fb=\rho$ to $G$ and resolve it into two forces directed along $KG$ and $EG$. The forces compressing the legs will be given by the segments $Fa$, $Gc$, $Gd$, namely:

Fig. 67.

$$ \sigma_3=-Fa,\quad \sigma_2=-Gc\quad \text{and}\quad \sigma_1=-Gd. $$

The condition for compression in all the legs is that the point $F$ should not lie outside the angle $AGB$. The proof of the validity of the indicated construction is no different from those given above. From the parallelograms of forces the equalities are obtained:

$$ \frac{\rho}{\sin(\zeta-\varepsilon)} = \frac{\sigma_3}{\sin(\varepsilon+\beta)} = \frac{T}{\sin(\zeta+\beta)}, $$

$$ \frac{\sigma_2}{\sin(\eta-\beta)} = \frac{\rho}{\sin(\eta+\xi)} = \frac{\sigma_1}{\sin(\xi+\beta)}, $$

whence:

$$ \sigma_3=-T\frac{\sin(\zeta-\varepsilon)}{\sin(\zeta+\beta)}, $$

$$ \sigma_2=-T\frac{\sin(\zeta-\varepsilon)}{\sin(\eta+\xi)}\cdot \frac{\sin(\eta-\beta)}{\sin(\zeta+\beta)}, \tag{68} $$

$$ \sigma_1=-T\frac{\sin(\zeta-\varepsilon)}{\sin(\eta+\xi)}\cdot \frac{\sin(\xi-\beta)}{\sin(\zeta+\beta)}. $$

These formulas are convenient for taking logarithms; the magnitude of the angle \(\beta\) may be determined from the formula:

\[ \tg \beta = \left[ \frac{l_0 \cos \zeta \sin \varepsilon}{\sin(\zeta-\varepsilon)} + l_2 - l \frac{\cos \eta \sin \xi}{\sin(\eta+\xi)} \right] : \left[ \frac{l \cos \eta \cos \xi}{\sin(\eta+\xi)} - l_0 \frac{\cos \zeta \cos \varepsilon}{\sin(\zeta-\varepsilon)} \right]. \]

Of course, formulas (68) may be reduced to the form of equations (66).

On the upper part of headframes.

Very often, for pyramidal headframes, as well as for prismatic ones, the following construction of the upper part is used.

The upper parts of both the front and the rear legs are covered by beams, which, in turn, are connected by longitudinal beams. On the transverse beams are placed the sheave beams, parallel to the connecting beams (Fig. 16). The transmission of forces from the axes of the sheaves to the legs takes place by means of the transverse beams, in view of which we must first consider the distribution of forces. The question of the transmission of vertical pressures to the transverse beams was considered by us earlier, namely—in considering the beam frame of tower structures; the present case does not differ from it in any way, and therefore we may use formulas (29).

As for the horizontal forces, here we may distinguish the following two cases: 1) if both ends of the sheave beam are immovably fastened to the transverse beams, and 2) if one of the ends can slide along them. In this second case the entire horizontal force is transmitted to the fixed end, or, more precisely,—the entire horizontal force minus the force produced by friction from the vertical pressure at the free end.

Fig. 68.

In Fig. 68 (a) the end \(A\) is fixed, the part of the beam \(AO\) is stretched and takes the horizontal tension \(U\), while the end \(B\)—the force \(U'\)

\[ U' = U - Pf\frac{l_1}{l} \quad \text{and} \quad U'' = Pf\frac{l_1}{l} \]

In case \(b\), the fixed end \(B\) is subjected to the pressure \(U''\), and the end \(A\) to \(U'\):

\[ U'' = U - Pf \frac{l_2}{l} \quad \text{and} \quad U' = Pf \frac{l_2}{l}; \]
the part \(O_1B_1\) is compressed.

In case \(c\), the part of the beam \(A_2O_2\) is stretched, and the part \(O_2B_2\) is compressed; to determine \(U'\) and \(U''\), it is necessary to consider the deformations. If by \(i_1\) and \(i_2\) we denote, respectively, the elongations and compressions per unit area, by \(E_1\) and \(E_2\) the moduli of elasticity, and, finally, by \(\omega\) the cross-section of the beam, then we shall have:

\[ U' = E_1 \omega i_1 \quad \text{and} \quad U'' = E_2 \omega i_2. \]

The total shortening of the length, or elongation of the beam, may be expressed by—

\[ \Delta l = l_2 i_2 - l_1 i_1. \]

But since our transverse beams are connected along the sides, and, moreover, there is also the resistance of the legs, in reality \(\Delta l\) is small.

Let us suppose that

\[ \Delta l = 0; \]

this we are entitled to do—in view of the fundamental proposition on the geometrical invariability of the form of the headframe. Moreover, since

\[ U = U' + U'', \]

it is easy to find that

\[ \begin{aligned} U' &= \frac{E_1 l_2}{E_1 l_2 + E_2 l_1}\, U,\\ U'' &= \frac{E_2 l_1}{E_2 l_1 + E_1 l_2}\, U. \end{aligned} \tag{69} \]

The horizontal forces taken up by the joints are, after the foregoing, very easy to compute. Let us denote the distance of the pulley axes from the outer connecting beams by \(\lambda\), and the distance between the pulleys by \(\lambda_1\) (Fig. 16). If the corresponding horizontal forces are \(U_1''\) and \(U_2''\), then, denoting by \(U_B\) and \(U_C\) the opposing forces in the joints, we shall have the moments:

\[ U_1''(\lambda + \lambda_1) + U_2'' \lambda - U_B(2\lambda + \lambda_1) = 0, \]

\[ U_2''(\lambda+\lambda_1)+U_1''\lambda-U_C(2\lambda+\lambda_1)=0, \]

whence:

\[ \begin{gathered} U_B=\frac{U_1''(\lambda+\lambda_1)+U_2''\lambda}{2\lambda+\lambda_1},\\[6pt] U_C=\frac{U_2''(\lambda+\lambda_1)+U_1''\lambda}{2\lambda+\lambda_1}. \end{gathered} \tag{70} \]

We have taken the simplest example; but it is clear that even for a more complex relation between the distances the method of calculation will be the same.

Calculation of Shaft Headframes

It remains for us now to apply the information gathered in the preceding chapters to the calculation of typical structures. We shall confine ourselves to only a few examples of wooden and iron headframes.

The calculation is initially performed according to a schematic, simplified drawing, with the weight of the parts of the structure being taken approximately, in accordance with existing arrangements. If it subsequently turns out that the designed parts have obtained dimensions that differ considerably from those originally adopted, corrections must be introduced. In the absence of suitable practical data, the preliminary calculation may be performed without taking into account the weight of the structure, and it may then be included in the secondary, final calculation on the basis of the dimensions obtained by the preliminary calculation. Graphical methods may be recommended both for the preliminary layout of the structure and for checking the numerical results of the calculation. Individual trusses may, of course, be calculated by generally accepted graphical methods.

The calculation of iron and wooden headframes in general differs little from one another. The difference consists only in the fact that, in small wooden structures, certain simplifications are possible: very often, for example, the pressure of the wind and lateral overturning forces are not taken into account, which cannot be allowed in solid iron headframes. In addition, iron structures are much more complicated, and the transmission of pressure to the legs and supports is not as simple as in wooden headframes; finally, here it is much more significant to avoid members subjected to tension.

At the present time, in most cases iron headframes are built that belong to the prismatic type, with rear—inclined and front—vertical legs, the so-called two-legged headframes, and combined, prismatic-pyramidal headframes—with vertical legs. For completeness of presentation we shall also consider older types of headframes.

For the transition from the expressions derived in the chapter “Theory of Shaft Headframes” to those used for calculations, let us adopt:

\[ \Pi = T \cos \varepsilon = R \cos \frac{\alpha}{2} + G_0 + p_0 = P + G_0 + p_0, \]

\[ \Omega = T \sin \varepsilon = R \sin \frac{\alpha}{2} + v_0 = U + v_0. \tag{71} \]

In these formulas \(T\) denotes the resultant of all the forces applied to the headframe, after their points of application have been transferred to the axis of the sheave, as we saw above; the angle \(\varepsilon\) is its inclination to the vertical, as in Chapter III. \(R\) is the resultant of the rope tensions, \(\frac{\alpha}{2}\) is its inclination to the vertical, as in Chapter I; \(P\) and \(U\) have exactly the same meanings. Finally, \(G_0\) is the weight of the structure applied to the axis of the sheave, while \(p_0\) and \(v_0\) are the vertical and horizontal components, respectively, of the wind pressure referred to the same point. From formula (71) it follows that

\[ \tg \varepsilon = \frac{\Omega}{\Pi} = \frac{U + v_0}{P + G_0 + p_0}. \tag{72} \]

It is self-evident that for \(\operatorname{Max}\Pi\) and \(\operatorname{Max}\Omega\) we shall have the expressions:

\[ \operatorname{Max}\Pi = \operatorname{Max}P + G_0 + p_0, \]

\[ \operatorname{Max}\Omega = \operatorname{Max}U + v_0, \]

in which \(\operatorname{Max}P\) and \(\operatorname{Max}U\) are determined according to equations (25) and (26). Let us expand expressions (37) and (38) and substitute into them our new notation; then the prismatic headframes in which the force \(T\) remains in the space between the legs are subjected only to compression; the forces directed in the planes of these legs parallel to the axial plane of the headframe will be:

\[ t_1 = -\sigma_1 = \frac{\Pi \sin \xi - \Omega \cos \xi}{\sin(\xi+\eta)}, \]

\[ t_2 = -\sigma_2 = \frac{\Pi \sin \eta + \Omega \cos \eta}{\sin(\xi+\eta)}. \tag{73} \]

In this case \(t_1\) and \(t_2\) denote not stresses, but the forces into which \(T\) is resolved.

If the resultant passes on the outside of the front legs, then

\[ t_1=-\sigma_1=-\frac{\Pi \sin \xi+\Omega \cos \xi}{\sin(\eta-\xi)}, \]

\[ t_2=-\sigma_2=\frac{\Pi \sin \eta-\Omega \cos \eta}{\sin(\eta-\xi)}, \tag{74} \]

and, moreover, both legs lie to the left of the vertical. Formulas (73), for the case when the rear legs are vertical, take a simpler form; namely, putting \(\xi=0\), we find:

\[ t_1=-\frac{\Omega}{\sin \eta}, \]

\[ t_2=\Pi-\frac{\Omega}{\operatorname{tg}\eta}; \tag{75} \]

and conversely, when the front legs are vertical, then, putting \(\eta=0\), we find:

\[ t_1=\Pi-\frac{\Omega}{\operatorname{tg}\xi}, \]

\[ t_2=-\frac{\Omega}{\sin \xi}. \tag{76} \]

We shall have occasion to use these last formulas for the calculation of an English three-legged headframe, similar to the headframe shown in Fig. 26. A headframe of this type, in which the pulley beams are placed horizontally in the direction of the axis of the whole structure, must be constructed so that the prolongations of the planes of the front vertical legs and of the rear inclined legs intersect along a straight line coinciding with the axis of the pulleys. If this condition is not observed, and the upper part of the headframe consists of beams laid horizontally one upon another, as shown in Fig. 16, then although such a headframe, strictly speaking, cannot be called pyramidal, since its legs, when prolonged, do not intersect at a single point, nevertheless its calculation is identical with the calculation of pyramidal headframes, and therefore we shall not consider it separately. The three-legged English headframe is shown schematically in Figs. 69 and 70 (p. 92). According to the theory, the vertical forces are taken up

vertical legs, while the horizontal ones are resolved into reverse verticals of the first kind and into forces acting in the plane of the inclined legs. The magnitudes of these and other forces are given by formula (76). It is clear that

Fig. 69.

\[ \operatorname{Max} t=\operatorname{Max} \Pi-\frac{\Omega}{\operatorname{tg}\xi}, \]

\[ \operatorname{Max} t_{1}=\frac{\operatorname{Max}\Omega}{\sin \xi}; \]

and since \(\operatorname{Max}\Pi\) corresponds to \(\operatorname{Max}P\), it follows that the vertical posts must be calculated for the vertical pressure occurring when the lower rope breaks, and for the horizontal force \(U\) corresponding to this case according to formula (26); the inclined posts, however, must be calculated for the horizontal force when the upper rope breaks.

Let us now consider the front and rear legs separately. The truss of the inclined legs is shown schematically in Fig. 70; in its plane, during normal hoisting, forces \(t_{2}'\) and \(t_{2}''\) act, parallel to the axis of this truss; their magnitude varies constantly so that

\[ t_{2}' \gtrless t_{2}''. \]

Fig. 70.

These forces are transmitted to the legs by means of the beam \(ABC\). The calculation of this beam is performed chiefly for bending, and for simplicity it is usually assumed that it consists of two separate parts \(AB\) and \(CB\). Thus, the outer posts must resist the forces \(\frac{1}{2}t_{2}'\) and \(\frac{1}{2}t_{2}''\), while the middle one must resist the force \(\frac{1}{2}(t_{2}'+t_{2}'')\); since the outer posts are inclined at a certain angle \(\alpha\) to the axis of the truss, their compressive stresses are equal to:

\[ t_{2}'\frac{1}{2\cos\alpha}\quad \text{and}\quad t_{2}''\frac{1}{2\cos\alpha}. \]

The horizontal beams \(AC\) and \(CB\), however, are compressed by the forces—

\[ t_{2}'\frac{\operatorname{tg}\alpha}{2}\quad \text{and}\quad t_{2}''\frac{\operatorname{tg}\alpha}{2}. \]

Since \(\operatorname{Max}t_{2}\) occurs when the upper rope at the sheaves breaks, therefore the post corresponding to this sheave should be…

it would be necessary to make the other one shorter, but an economy of material of this kind is hardly appropriate; moreover, a given sheave cannot be assigned once and for all to the lower or the upper rope, and therefore the force compressing the outermost posts is found from the formula:

\[ s_1=\operatorname{Max} t_2' \frac{1}{2\cos\alpha}, \]

whereas the middle post is calculated for the force—

\[ s_2=\frac{1}{2}\left(\operatorname{Max} t_2' + t_2''\right). \]

\(t_2''\) is found from formula (76) for the case of the greatest normal load of the sheave of the lower rope, that is, when the loaded cage is at the lowest point of the shaft. The transverse beam \(ABC\) is checked in compression from the force—

\[ s=\operatorname{Max} t_2' \frac{\operatorname{tg}\alpha}{2}. \]

Since, from the force—

\[ s_0=\left(\operatorname{Max} t_2' - t_2''\right)\frac{\operatorname{tg}\alpha}{2} \]

racking is possible, the rear legs are tied together by diagonal rods. The stresses arising from the force \(s_0\) in the entire truss could be calculated exactly by means of formulas (48) of Chapter III, but usually, in view of the enormous margin of strength, the corresponding corrections are not introduced into the calculation; likewise, wind pressure is not taken into account.

The calculation of a truss composed of vertical legs is based on the calculation just given; it differs only in that rupture of the lower rope at the sheaves and the full load of the upper one are assumed. If the legs are set vertically, \(\frac{\alpha}{2}=0\), and there are two of them, not three, then the compression of each of them is determined by the force:

\[ s_1'=\operatorname{Max} t_1''. \]

It remains for us to determine \(P\) and \(\Omega\) for the given headframe. Since, in small wooden structures, we neglect

by wind pressure, then for the present case it remains only to indicate the value—

\[ G_0 = G + \frac{1}{4}(G_1 + G_2), \]

where \(G\) is the weight of the sheave and of half the beams covering the upper part of the headframe, \(G_1\) is the weight of the vertical truss, and \(G_2\) that of the inclined truss. We assign one half of the weight of each of them to the upper joint, and the other half to the supports; thus, one quarter of the total weight falls on each sheave axle. Since the lower parts of the legs carry the full weight of the truss, in view of the foregoing, the forces compressing them are somewhat greater than those indicated by the formulas, and therefore we must introduce corrections for this second half of the weight; then the force compressing the inclined legs downward will be determined by the formula:

\[ S_1 = s_1 + \frac{1}{8}\frac{G_2}{\cos \xi}, \]

while the force falling on the middle parts of the legs will be—

\[ S_2 = s_2 + \frac{1}{4}\frac{G_2}{\cos \xi}. \]

Strictly speaking, the inclination of the inclined outer legs differs somewhat from the angle \(\xi\), but, since in the calculation presented by us we admit a whole series of simplifications, we find it possible to take this angle as equal to \(\xi\). The vertical legs (if there are two of them) will be compressed downward by the force—

\[ S_1' = s_1' + \frac{1}{4}G_1. \]

As for the support reactions, it is clear that their magnitudes must be equal and opposite to \(S_1\), \(S_2\), \(S_1'\), and the masonry abutments must be designed accordingly. If there are wooden sills against which the legs bear, then they must be designed according to the method of fastening; for example, for tension and shear if a portion at the vertical post is fastened; the forces causing these stresses are, evidently, then equal to:

\[ S_1 \sin \xi \quad \text{or} \quad S_2 \sin \xi. \]

Let us now consider the English wooden two-legged headframe (Fig. 25). The point of application of the forces \(R_1\) and \(R_2\) lies somewhat in

from the inclined legs; if we transfer it perpendicularly to the leg onto its axis, then we obtain the moment \(Re\); here \(e\) is the distance of the axis of the sheave from the axis of the leg. This moment tends to turn the bearing or part of the leg into the plane of its contact with the bearing. Since \(e\) is small, we neglect this moment, although it must be noted that in general it is undesirable, and therefore structurally the axes of the sheaves should be placed in the plane of the axes of the legs; of course, in those cases where this is not associated with great inconveniences.

Drawing 71.

Thus, for the calculation of the headframe we shall assume that the point of application of the force lies in the axial plane of the inclined legs. To determine the forces in the planes of the front and rear legs, into which this force is resolved, we shall use formula (43), assuming in them \(T_1 = 0\), so that

\[ T_2 \sin \varepsilon_2 = \Omega \quad \text{and} \quad T_2 \cos \varepsilon_2 = \Pi, \]

and then:

\[ t_1 = -\sigma_1 = \frac{H_1}{H}\, \frac{\Pi \sin \xi - \Omega \cos \xi}{\sin(\eta+\xi)}, \tag{78} \]

\[ t_2 = -\sigma_2 = -\frac{H_1}{H}\, \frac{\Pi \sin \xi - \Omega \cos \xi}{\sin(\eta+\xi)} \cdot \frac{\cos \eta}{\cos \xi} + \frac{\Pi}{\cos \xi}. \]

For our case these expressions are further simplified because \(\eta = 0\); therefore—

\[ t_1 = \frac{H_1}{H} \left( \Pi - \frac{\Omega}{\operatorname{tg}\xi} \right), \tag{79} \]

\[ t_2 = \left( 1-\frac{H_1}{H} \right) \frac{\Pi}{\cos \xi} - \frac{\Omega}{\sin \xi}; \]

Formulas (78) and (79) may also be used for the exact determination of forces directed parallel to the axes of the frames and caused by the wind pressure and by the weight of the frames. For the present example, however, we shall calculate the stresses caused by the weight according to the method adopted in the first calculation; so that if \(G_0\) has the same value as before, but without the weight of the sheaves and sheave beams, then

the total component in the vertical truss will, for our case, be:

\[ t_1 = t_0 + G_0. \]

The vertical truss experiences the greatest stress when the lower rope breaks; this stress is expressed by the formula:

\[ \operatorname{Max} t_1 = G_0 + \frac{H_1}{H}\left(\operatorname{Max}\Pi - \frac{\Omega}{\operatorname{tg}\xi}\right), \]

whereas the inclined truss does so when the upper one breaks. The magnitude of the stress is determined by the equality:

\[ \operatorname{Max} t_2 = \left(1 - \frac{H_1}{H}\right)\frac{\Pi}{\cos \xi} - \frac{\operatorname{Max}\Omega}{\sin \xi} \]

We now have to consider both trusses separately. In the inclined truss (see the diagram in Fig. 72), the forces \(\Pi_1\), \(\Pi_2\) and \(\Omega_1\), \(\Omega_2\) are applied to the pulley axes. These forces may be resolved into components normal and parallel to the pulley beams; in this case—

\[ N = \Pi \sin \xi - \Omega \cos \xi, \]

\[ T = \Pi \cos \xi + \Omega \sin \xi. \]

Thus, each pulley beam must be calculated for bending by the formula:

\[ \operatorname{Max} N = \frac{1}{2}\left(\operatorname{Max}. \Pi \sin \xi - \Omega \xi\right), \]

since the pulleys are located at the middle of the interval between the beams. It is clear that all the beams are made of the same thickness.

Fig. 72

The horizontal transverse beam \(CD\) takes one half of the normal pressures on the pulley beams, as well as part of the longitudinal pressure on them. From the resolution of the normal and longitudinal pressure, the forces \(t_1\) and \(t_2\), which we determined above, are finally obtained. Since the original forces are applied at the pulleys, the force \(t\) may be regarded as applied at the same point, and consequently the parts of it taken up by the beams \(CD\) and \(KH\) are determined by formula (69). In view of the fact that the pulleys are placed approxi-

approximately at the middle of the beams, we may assume that half of the forces is transmitted to beam \(CD\), and the other half to beam \(KH\). In view of the fact that for iron the modulus of elasticity in compression differs little from the modulus of elasticity in tension, such an assumption will be sufficiently close to reality for iron headframes. It is also admissible for small wooden headframes, since in this case the errors of calculation may, without harm to the structure, be somewhat greater. If the distance \(Cd = bD\) is denoted by \(l_1\), and \(db\) by \(l_2\), then the pressures at the joints \(C\) and \(D\) are determined by the expressions:

\[ \text{pressure at joint } D = \frac{l_1 t_2' + (l_1 + l_2)t_2''}{2(2l_1 + l_2)}, \]

\[ \text{” \quad ” \quad ” } C = \frac{l_1 t_2'' + (l_1 + l_2)t_2'}{2(2l_1 + l_2)} . \]

If the inclination of the legs to the axis of the truss under consideration is denoted by \(\alpha\), then the forces transmitted to the legs will be

\[ s_2 = \frac{l_1 t_2' + (l_1 + l_2)t_2''}{2(2l_1 + l_2)\cos\alpha}, \]

\[ s_1 = \frac{l_1 t_2'' + (l_1 + l_2)t_2'}{2(2l_1 + l_2)\cos\alpha}. \]

If the inclination of the braces \(LK'\) and \(K'G\) to the axis of the truss is given as the angle \(\beta\), then the forces taken up by them will be:

\[ t_2' \frac{1}{2\cos\beta} \quad \text{and} \quad t_2'' \frac{1}{2\cos\beta}. \]

These forces are transmitted to the legs and to the diagonal ties; but since, in order to have a clear passage in the lower part of the headframe, it is often preferred to use a horizontal tie, then, resolving the pressure from the braces in the directions of the leg and the horizontal, we obtain from the parallelogram of forces:

\[ s_1 = t_2' \frac{1}{2\cos\alpha} \quad \text{and} \quad s_2'' = t_2'' \frac{1}{2\cos\alpha}. \]

In the horizontal tie the forces causing tension will be:

\[ t_2'\,\frac{\sin(\beta-\alpha)}{2\cos\alpha\cos\beta} \quad \text{and} \quad t_2''\,\frac{\sin(\beta-\alpha)}{2\cos\alpha\cos\beta}; \]

the transom \(H'K'\) is compressed by the forces:

\[ t_2'\,\frac{\operatorname{tg}\beta}{2} \quad \text{and} \quad t_2''\,\frac{\operatorname{tg}\beta}{2}, \]

and the upper beam—by the forces:

\[ \frac{l_1 t_2' + (l_1+l_2)t_2''}{2(2l_1+l_2)}\,\operatorname{tg}\alpha \quad \text{and} \quad \frac{l_1 t_2'' + (l_1+l_2)t_2'}{2(2l_1+l_2)}\,\operatorname{tg}\alpha; \]

the compression of the legs in the lower parts is determined by the formulas:

\[ -\sigma_2' = s_1' + s_1'' + \frac{G}{2\cos\xi}, \]

\[ -\sigma_2'' = s_2' + s_2'' + \frac{G}{2\cos\xi}; \]

it must, however, be noted that these formulas give only an approximate value for \(-\sigma_2'\) and \(\sigma_2''\), since the inclination of the legs is not exactly equal to \(\xi\). If in \(\sigma_2'\) and \(\sigma_2''\) we substitute the expressions for \(s_1'\), \(s_1''\), \(s_2'\), \(s_2''\), the following formulas are obtained:

\[ -\sigma_2' = \frac{(3l_1+2l_2)t_2' + l_1t_2''}{2(2l_1+l_2)\cos\alpha} + \frac{G}{2\cos\xi}; \]

\[ -\sigma_2'' = \frac{(3l_1+2l_2)t_2'' + l_1t_2'}{2(2l_1+l_2)\cos\alpha} + \frac{G}{2\cos\xi}. \tag{80} \]

If \(\operatorname{Max.} t_2'' > \operatorname{Max.} t_2'\), which occurs when \(t_2''\) corresponds to the sheave of the upper rope and \(t_2'\) to the lower one, then the absolute value

\[ \operatorname{Max}\sigma_2'' > \operatorname{Max}\sigma_2'; \]

this is easily proved by comparing expressions (80), and therefore we have to calculate the legs, and in general the entire inclined frame, for the case of rupture of the upper rope.

Thus, we shall consider the lower part of the legs compressed by a force determined by the expression—

\[ -\sigma_2 = \frac{(3l_1+2l_2)\operatorname{Max} t_2'' + l_1 t_2'} {2(2l_1+l_2)\cos\alpha} +\frac{G}{\cos\xi}; \tag{81} \]

the horizontal tie is stretched by the force—

\[ \frac{\sin(\beta-\alpha)}{2\cos\alpha\cos\beta}\operatorname{Max} t_2'', \tag{82} \]

the crossbar \(K'H'\)—

\[ \frac{\operatorname{tg}\beta}{2}\operatorname{Max} t_2'' \tag{20} \]

and the upper beam by the force—

\[ \frac{\operatorname{tg}\alpha}{2}\, \frac{l_1 t_2' + (l_1+l_2)\operatorname{Max} t_2''} {(2l_1+l_2)} \tag{84} \]

The course of the calculation of the vertical truss does not differ in general from that just indicated. The points of application of the forces acting in the plane of this truss are \(a\) and \(b\), so that the joints receive and transmit directly to the pressure legs:

\[ \frac{l_1 t_1'' + (l_1+l_2)t_1'}{(2l_1+l_2)} \quad\text{and}\quad \frac{l_1 t_1' + (l_1+l_2)t_1''}{(2l_1+l_2)}. \]

But since in this case

\[ \operatorname{Max} t_1' > \operatorname{Max} t_1'', \]

that is, the greatest vertical pressure arises when the lower rope breaks, the calculation of the lower part of the legs is carried out for compression by the force

\[ -\sigma_1 = \frac{l_1 t_1'' + (l_1+l_2)\operatorname{Max} t_1'} {(2l_1+l_2)} +\frac{G}{2}. \]

It remains also to calculate the horizontal beam \(CD\); it is subjected to bending from two forces \(\rho_1\) and \(\rho_2\), applied at the points \(a\) and \(b\), and from its own weight. The magnitude of these forces is determined by the expressions:

\[ \rho_1 = \sqrt{t^2+t''^2-\cos\xi\, t't''} \quad\text{and}\quad \rho_2 = \sqrt{t_1'^2+t_1''^2-\cos\xi\, t_1't_1''}. \]

These forces do not lie in one plane, and therefore the beam is subjected to compound bending. For simplicity of calculation we shall bring them into one plane and shall calculate the beam for the force \(\rho_1 + \operatorname{Max}\rho_2\) or \(\operatorname{Max}\rho_1 + \rho_2\), and shall use the results of that calculation which gives the larger values. It is clear that the maximum values of \(\rho\) are determined by the largest values of \(t_2\) and \(t_1\).

As for the support reactions, they are very easy to calculate, namely: the pressure on the supports of the vertical posts is equal to \(\sigma_1\), while the inclined posts press in the direction of their axes and, in addition, transmit part of the horizontal force from the resolution of \(T\) into the horizontal direction and the direction parallel to the axis of the inclined king-post. The magnitude of this force is determined by the formula—

\[ \operatorname{Max}\sigma_0 = (P\tg \xi + \operatorname{Max} U)\frac{H-H_1}{H}, \]

and consequently the resultant of \(\sigma_2\) and \(\sigma_0\), that is, the actual pressure taken by the support, will be—

\[ \rho_0 = \sqrt{\sigma_2^2 + \sigma_0^2 + \sigma_2\sigma_0\sin\xi}; \]

the angle of its inclination to the horizon—

\[ \sin\gamma = \cos\xi \frac{\sigma_0}{\rho_0}. \]

It is clear that, for the calculation of \(\rho\) and \(\gamma\), the largest values of \(\sigma_0\) and \(\sigma_2\), obtained at \(\operatorname{Max} U\), are introduced.

In the calculation presented by us there are simplifications and even some inaccuracies. Their appropriateness is explained by their application only to small hoisting devices; more exact methods we shall apply in the calculation of large iron headframes.

Pyramidal headframes.

Let us now take up the application of our theory to pyramidal headframes. As an example, let us consider the simple wooden structure described earlier (Fig. 36). From the theory of beam-type sheave stands we know that both vertical and horizontal forces are transmitted to the joints through the transverse beams. Depending on how the sheave beams are fastened, this transmission may take place in different ways, but in the final result it is always horizon-

vertical forces act along the connecting beams, and the difference in the design is reduced only to whether these beams are in tension or compression. If the sheave beams are attached quite immovably, then the horizontal forces are transmitted both to the front and to the rear joints. We can calculate the vertical forces by formula (29), and the horizontal forces by formula (69).

Thus, on the basis of theoretical considerations, we conclude that for the calculation of a headframe in general the method of fastening the beams is immaterial; the whole difference is reduced only to the fact that in one case these beams have to be calculated for compression, and in the other—for tension.

On the basis of all that has been said, the magnitude of the forces acting at each joint may be considered known. Let us first consider the vertical forces. Suppose that at joint \(A\) a vertical force \(p_A\) is applied. Through the inclined leg and this force we draw a plane, which intersects the plane of the connecting beams along the straight line \(CA\). If we regard the force \(p_A\) as the resultant of forces directed along the leg and the straight line \(CA\), and call the angle made by the leg with the vertical straight line \(\gamma\), then the force directed along the leg will be equal to

Fig. 73-a.

\[ s_1=\frac{p_A}{\cos \gamma}, \]

while the horizontal component along \(AC\) will be

\[ p_A \tg \gamma . \]

If the angles made by the direction \(AC\) with the connecting and transverse beams are denoted by \(\delta\) and \(o\), then, resolving this force in the directions \(DA\) and \(BA\), we obtain

\[ p_A \tg \gamma \cos \delta \quad \text{and} \quad p_A \tg \gamma \cos o . \]

If we take the continuation \(AE\) (see Fig. 73-a), \(AE_1\), to be the diagonal of the parallelepiped whose side directions are \(DA\), \(BA\), and the continuation \(p\), then, denoting the angle between \(DA\) and \(AE\) by \(\alpha\),

between \(BA\) and \(AE\) through \(\beta\), on the basis of the geometrical properties of the parallelepiped we have:

\[ \cos \delta=\frac{\cos(180-\alpha)}{\sin\gamma} \quad\text{and}\quad \cos o=\frac{\cos(180-\beta)}{\sin\gamma}, \]

and consequently the force acting in the direction \(AD\) will be

\[ s_2=-p_A\frac{\cos\alpha}{\sin\gamma}, \]

and along \(AB\)—

\[ s_2=-p_A\frac{\cos\alpha}{\sin\gamma}. \]

Since the angles \(\beta\) and \(\alpha\) are obtuse, \(s_2\) and \(s_3\) have positive values. The forces determined by us produce opposite stresses in the members of the joint; therefore, if we denote these stresses by \(s_1'\), \(s_2'\), \(s_3'\), then we shall have the equalities:

\[ s_1=-s_1',\quad s_2=-s_2'\quad\text{and}\quad s_3=-s_3'. \]

If we write the equilibrium equation for this joint with respect to a system of three mutually perpendicular axes \(X-(AD)\), \(Y-(AB)\), and \(Z-(p)\), then, of course, we shall be convinced of the correctness of the resolution of the force \(p\). These equations will be

\[ p_A+s_1'\cos\alpha=0, \]

\[ p_A\frac{\cos\alpha}{\cos\gamma}+s_2'=0, \tag{85} \]

\[ p_A\frac{\cos\beta}{\cos\gamma}+s_3'=0. \]

All three stresses are caused by compressive forces. In exactly the same way we determine the forces and stresses acting along the rods in the other three joints. If in \(p_A\), \(p_B\), etc. we assume included the weight of all the pulleys, transverse and connecting beams, and the entire corresponding part of the inclined leg with all the ties and braces, then for the forces directed along the connecting beams \(AB\) and \(CD\), we may write the following expressions:

\[ \Omega_1=S_{AB}=\omega_1-p_A\frac{\cos\alpha_1}{\cos\gamma_1}+p_B\frac{\cos\alpha'}{\cos\gamma'}, \tag{86} \]

\[ \Omega_2 = S_{CD} = \omega_2 - p_C \frac{\cos \alpha_1}{\cos \gamma_1} + p_D \frac{\cos \alpha'}{\cos \gamma'}; \tag{86} \]

along the transverse beams \(AC\) and \(BD\) there act the forces:

\[ \Omega' = S_{AC} = \omega' - p_A \frac{\cos \beta_1}{\cos \gamma_1} + p_C \frac{\cos \beta_1}{\cos \gamma_1}, \]

\[ \Omega'' = S_{BD} = \omega'' - p_B \frac{\cos \beta'}{\cos \gamma'} + p_D \frac{\cos \beta'}{\cos \gamma'}. \tag{87} \]

These overturning forces \(\Omega'\) and \(\Omega''\), in the calculation of small pyramidal headframes, are either not taken into account at all, or, if they are sometimes taken into account, then only for checking the transverse beams in compression, and, of course, the check is made for the larger of them. However, in order not to return any more to headframes of this type, we shall investigate the stresses caused by all the overturning forces.

In view of the fact that in a pyramidal headframe the front legs as well as the rear ones, taken together, form trusses having the form of a trapezoid, very often without diagonal members, inclined somewhat toward one another, we shall have to turn to formula (48). Transforming them by the general method and replacing the stresses by the forces which cause them, namely—

\[ t_1 = -\sigma_1 \quad \text{and} \quad t_2 = -\sigma_2, \]

we find that

\[ t_2 = \frac{(l_1 + H \operatorname{tg}\eta)\Pi + H\Omega} {(l + H \operatorname{tg}\eta + H \operatorname{tg}\xi)\cos \xi}, \]

\[ t_1 = \frac{(l_2 + H \operatorname{tg}\xi)\Pi + H\Omega} {(l + H \operatorname{tg}\eta + H \operatorname{tg}\xi)\cos \eta}; \tag{88} \]

but since we have already distributed the vertical pressures in the direction of the legs and of the longitudinal and transverse beams, and it is desirable to introduce only the action of the transverse horizontal forces, for this case one should assume

\[ \Pi = 0, \]

and then:

\[ t'_{1-2} = -\frac{H\Omega}{L\cos\eta}, \]

\[ t''_{1\cdot 2} = -\frac{H\Omega}{L\cos\xi}. \tag{89} \]

Substituting into these expressions \(Q'\) and \(Q''\), as well as the corresponding values for the angles of inclination of the legs to the axis of the headframe, we determine the forces \(t_1'\), \(t_1''\), \(t_2'\), \(t_2''\), which, when summed with the previous ones (85), allow us to compute the stresses in the legs by the formulas:

\[ \sigma_1 = - (s_1 - t_1);\quad \sigma_2 = - (s + t). \]

It remains for us to use the expressions obtained in order to be able to judge the role of each of the active forces taken into account, which, owing to their complexity, I shall not present.

For determining the horizontal forces \(Q_1\) and \(Q_2\) we have expressions (69).

Formulas (29), indicating how the vertical pressures are distributed, enable us to compute the vertical forces acting at the nodes:

\[ p_A = \frac{l_2}{l}\frac{P_1(\lambda+\lambda_1)+P_2\lambda}{(2\lambda+\lambda_1)},\quad p_B = \frac{l_1}{l}\frac{P_1(\lambda+\lambda_1)+P_2\lambda}{(2\lambda+\lambda_1)}, \]

\[ p_C = \frac{l_2}{l}\frac{P_2(\lambda+\lambda_1)+P_1\lambda}{(2\lambda+\lambda_1)},\quad p_D = \frac{l_1}{l}\frac{P_2(\lambda+\lambda_1)+P_1\lambda}{(2\lambda+\lambda_1)}. \]

If, for simplicity, we denote

\[ \frac{P_1(\lambda+\lambda_1)+P_2\lambda}{(2\lambda+\lambda_1)} \quad \text{by } \Pi_1 \quad \text{and} \quad \frac{P_2(\lambda_2+\lambda_1)+P_1\lambda}{(2\lambda+\lambda_1)} \quad \text{by } \Pi_2, \tag{91} \]

then we shall have:

\[ P_A + \frac{l_2}{l}\Pi_1,\quad P_B = \frac{l_1}{l}\Pi_1,\quad P_C = \frac{l_2}{l}\Pi_2,\quad P_D = \frac{l_1}{l}\Pi_2, \]

where \(\Pi_1\) and \(\Pi_2\) are the forces by which, on the connecting beams \(BA\) and \(CD\), we may replace the vertical pressures \(P_1\) and \(P_2\) at the axes of the sheaves.

The method of calculation presented is, in its technique, similar to those applied in practice, but without those liberties that are usually allowed. The rear legs are calculated for the force

\[ \frac{l_1 \operatorname{Max}\Pi_1}{l\cos\gamma'} \quad \text{or} \quad \frac{l_1 \operatorname{Max}\Pi_2}{l\cos\gamma'}, \]

and then the whole structure is checked for overturning from the forces \(Q_1\) and \(Q_2\).

As is evident, such a calculation, strictly speaking, has no basis. We shall subsequently make use of the formulas derived for the distribution of forces over the joints when calculating pyramidal headframes in which the legs are connected by diagonal rods.

For the case where the invariability of the geometrical form of the headframe is ensured only by horizontal ties, we shall use the method of the ideal truss; that is, we shall replace the pressures transmitted to the transverse beams by the forces \(P\) and \(\Omega\), by forces directed in the planes of the front and rear legs and applied at the middle of the distance between the transverse beams supporting each of the sheaves. If the inclination of the truss of the front legs to the vertical plane is \(\eta\), and that of the rear legs is \(\xi\), then, substituting in equation (88), respectively, \(P_1, \Omega_1\) and \(P_2, \Omega_2\), we find two groups of forces:

\[ t_2',\ t_1' \quad \text{and} \quad t_2'',\ t_1''. \]

Thus in the front inclined truss, in its plane, the forces (Fig. 73) \(t_1'\) and \(t_1''\) act, and in the rear one—\(t_2'\) and \(t_2''\).

Let us consider the rear truss; the inclination of the legs in it is the same; let the angle made by the leg with the horizontal transverse beam be \(\beta'\); consequently their inclination to the axis of the truss is \(\beta' - 90^\circ\). The inclined height of this truss is

\[ H_2=\frac{\cos \xi}{H}. \]

The stresses in the legs of this truss are caused by the forces \(t_2'\) and \(t_2''\) and by the lateral wind pressure, as well as by the force caused by the deviation of the rope \(v_2\); these stresses are expressed by the formulas:

Fig. 73.

\[ \sigma_B= \frac{[\lambda+H_2\operatorname{tg}(\beta'-90^\circ)]t_2'} {L_2\cos(\beta'-90^\circ)} - \frac{[(\lambda+\lambda_1)+H_2\operatorname{tg}(\beta'-90^\circ)]t_2''} {L_2\cos(\beta'-90^\circ)} \pm \frac{H_2 v_2} {L_2\cos(\beta'-90^\circ)}, \tag{92} \]

\[ \sigma_D= \frac{[(\lambda+\lambda_1)+H_2\operatorname{tg}(\beta'-90^\circ)]t_2'} {L_2\cos(\beta'-90^\circ)} + \frac{[\lambda+H_2\operatorname{tg}(\beta'-90^\circ)]t_2''} {L_2\cos(\beta'-90^\circ)} \mp \frac{H_2 v_2} {L_2\cos(\beta'-90^\circ)}. \]

The double sign for the force of wind pressure and lateral deviations of the rope is placed because the wind may be directed both from one side and from the other; however, this term is often neglected, since the longitudinal wind pressure has already entered into \(\Omega_1\) and \(\Omega_2\):

\[ L_2 = 2\lambda + \lambda_1 + 2H_2 \operatorname{tg}(\beta' - 90^\circ). \]

It is clear that if

\[ t_2' > l_2'', \]

then \(\operatorname{Max}\ \sigma_p\) will correspond to \(\operatorname{Max} t_1'\), and conversely, if

\[ t_2' > t_2'' \]

then the legs should be calculated according to \(\operatorname{Max} t_2''\), namely, \(\sigma_A\) should be taken into account for the above-mentioned value of \(t_2''\). Thus, we must determine which of the inequalities

\[ \operatorname{Max} t_2' \gtrless \operatorname{Max} t_2'' \]

takes place. But since, in general, the greatest stresses arise when the ropes break, in our case the following forces may occur for each rope (sheave):

\[ (\operatorname{Max} t_2)_1 = \frac{(l + H\operatorname{tg}\eta)\operatorname{Max}\Pi + H\Omega} {L\cos \xi} \]

\[ (\operatorname{Max} t_2)_2 = \frac{(l_1 + H\operatorname{tg}\eta)\Pi + H\operatorname{Max}\Omega} {L\cos \xi}. \]

Let us first consider the stresses occurring at the sheave of the upper rope, and, for simplicity, we shall take into account only the tensions caused by hoisting. When the rope breaks at the sheave, the magnitude of the force \(t_1'\), according to equation (25) of Chapter I, will be—

\[ \frac{(l_1 + H\operatorname{tg}\eta)\sigma(1 + \cos\alpha_1') + H\sigma\sin\alpha_1'} {L\cos \xi}; \]

if the rope breaks when the loaded cage is in the lower part of the shaft, then this force is determined by the expression

\[ \frac{(l_1 + H\operatorname{tg}\eta)\sigma(1 + \cos\alpha_1'') + H\sigma\sin\alpha_1''} {L\cos \xi}. \]

The difference of these quantities is positive or negative, according as which of the inequalities written below holds:

\[ (l_1+H\operatorname{tg}\eta)\left(\cos \alpha_1' - \cos \alpha_1''\right) +H\left(\sin \alpha_1' - \sin \alpha_1''\right)\gtreqless 0. \]

Let us write it as follows:

\[ \frac{l_1}{H}+\operatorname{tg}\eta+M\gtreqless 0, \]

where

\[ M=\frac{\sin \alpha_1' - \sin \alpha_1''}{\cos \alpha_1' - \cos \alpha_1''}. \]

To determine \(M\), we shall use formula (7), which gives us (on replacing \(L\) by \(D\)) the following equalities:

\[ \sin \alpha=\frac{1}{H^2+D^2}\left[D\sqrt{H^2+D^2-(\rho-\nu)^2}+H(\rho-\nu)\right], \]

\[ \cos \alpha=\frac{1}{H^2+D^2}\left[H^2\sqrt{H^2+D^2-(\rho-\nu)^2}-D(\rho-\nu)\right]. \]

If, in these expressions, we substitute for \(\rho\) its limiting values \(\rho_1\) and \(\nu\), then in the first case we determine \(\sin \alpha_1'\) and \(\cos \alpha_1'\), and in the second—\(\sin \alpha_1''\) and \(\cos \alpha_1''\); thus:

\[ M= \frac{ L\sqrt{H^2+D^2-(\rho_1-\nu)^2}+H(\rho_1-\nu)-D\sqrt{H^2+D^2} }{ H\sqrt{H^2+D^2-(\rho_1-\nu)^2}-D(\rho_1-\nu)-H\sqrt{H^2+D^2} }, \]

and, with accuracy sufficient for our reasoning,

\[ M=-\frac{H}{D}; \]

therefore, if

\[ \frac{l_1}{H}+\operatorname{tg}\eta-\frac{H}{D}>0, \tag{93a} \]

then \(\operatorname{Max} t_1\) occurs at \(\operatorname{Max} U\). Conversely, if

\[ \frac{l_1}{H}+\operatorname{tg}\eta-\frac{H}{D}<0, \tag{93b} \]

then the greatest value of \(t_1\) is attained at \(\operatorname{Max} P\). Reasoning in exactly the same way with respect to the stresses at the pulley of the lower rope,

we shall arrive at literally the same conclusion, and consequently the rear legs should be designed for \(\operatorname{Max}\Omega\), when inequality \(93a\) holds, and for \(\operatorname{Max}\Pi\) (equation 25) when inequality \(93b\) holds.

Let us now consider the truss of the front legs. The stresses in them are determined by expressions analogous to (92), with the corresponding change of notation; these stresses will be:

\[ \sigma_A= \frac{[\lambda+H_1\operatorname{tg}(\beta-90^0)t_1'}{L_1\cos(\beta-90^0)} + \frac{[(\lambda+\lambda_1)+H_1\operatorname{tg}(\beta-90^0)]t_1''}{L_1\cos(\beta-90^0)} \pm \frac{H_1v_1}{L_1\cos(\beta-90^0)}; \tag{94} \]

\[ \sigma_c= \frac{[(\lambda+\lambda_1)+H_1\operatorname{tg}(\beta-90^0)]t_1'}{L_1\cos(\beta-90^0)} + \frac{[\lambda+H_1\operatorname{tg}(\beta-90^0)]t_1''}{L_1\cos(\beta-90^0)} \pm \frac{H_1v_1}{L_1\cos(\beta-90^0)}. \]

In order that both stresses be compressive, it is necessary that \(t_1'\) and \(t_1''\) be positive quantities. The condition necessary for this is expressed in equation (53); if in it \(\operatorname{tg}\varepsilon\) is replaced by \(\operatorname{tg}\alpha\), which can be calculated as indicated above, then

\[ \operatorname{tg}\alpha= \frac{D\sqrt{H^2D^2-(\rho-v)^2}+H(\rho-v)} {H\sqrt{H^2D^2-(\rho-v)^2}-D(\rho-v)}, \]

but, limiting ourselves to a rather rough approximation, we may take that

\[ \operatorname{tg}\alpha=\frac{D}{H} \]

and then we obtain the following:

\[ \operatorname{tg}\xi>\frac{D-l_2}{H}. \]

However, in the theory this case is considered more precisely. Our inequality clearly shows that the type of pyramidal headframes, especially wooden ones, is suitable only for small machines, which, owing to the small length of the drums, can be installed directly at the shafts.

Thus, if condition (53) is satisfied, then it is clear that

\[ \operatorname{Max}t_1= \frac{(l_2+H\operatorname{tg}\xi)\operatorname{Max}\Pi-\Omega H} {L\cos\eta} \]

\[ \operatorname{Min}t_1= \frac{(l_2+H\operatorname{tg}\xi)\Pi-H\operatorname{Max}\Omega} {L\cos\eta}; \]

on the contrary, if the condition is not satisfied, and the front legs have to be calculated for tension, then the preceding minimum must be taken as the basis of the calculation; in the present case it will be the maximum tension. The remaining considerations are the same as in the calculation of the rear legs, so that it is unnecessary to repeat them.

Let us now pass to the case when the front and rear legs are connected by diagonal rods. In this case the headframe may be regarded as a structure consisting of two longitudinal lattice trusses connected with one another. The calculation is carried out in the following manner: all acting forces are distributed over the joints, and the vertical ones are resolved into horizontal forces directed along the legs, according to the method indicated at the beginning of the chapter on pyramidal headframes. Since the legs consist of continuous timbers, the forces directed along them are temporarily considered nonexistent, and the further calculation consists in resolving the horizontal forces in the direction of the legs and of the ties. Thus, in the present case it is by no means immaterial how the sheave beams are attached to the transverse ones, for on the method of this attachment depends the circumstance to what degree the horizontal force is transmitted to the front and rear legs (see the drawing). In the case where the beams are connected rigidly, then, combining formulas (69) with (70), we obtain the equalities:

\[ \omega_A = K_1 \omega_1,\quad \omega_b = K_2 \omega_1,\quad \omega_D = K_1 \omega_2 \ \text{and}\ \omega_C = K_2 \omega_2, \]

in which

\[ K_1=\frac{E_1 l_2}{E_1 l_2+E_2 l_1}\quad \text{and}\quad K_2=\frac{E_2 l_1}{E_1 l_2+E_2 l_1}. \]

And so, the horizontal forces acting in the direction of the headframe axis will be:

\[ \begin{aligned} \Omega_A &= K_1\omega_1-\frac{l_2}{l}\Pi_1\frac{\cos\alpha_1}{\cos\gamma_1},\quad \Omega_b = K_2\omega_1+\frac{l_1}{l}\Pi_1\frac{\cos\alpha'}{\cos\gamma'},\\ \Omega_C &= K_2\omega_2-\frac{l_2}{l}\Pi_2\frac{\cos\alpha_1}{\cos\gamma_1},\quad \Omega_D = K_2\omega_1+\frac{l_1}{l}\Pi_2\frac{\cos\alpha'}{\cos\gamma'} ; \end{aligned} \tag{95} \]

while the forces acting perpendicular to the headframe axis are—

\[ \begin{aligned} \Omega'_A &= \omega'_1-\frac{l_2}{l}\Pi_1\frac{\cos\beta_1}{\cos\gamma_1},\quad \Omega'_b = \omega'_2-\frac{l_1}{l}\Pi_1\frac{\cos\beta'}{\cos\gamma'},\\ \Omega'_C &= \omega'_1+\frac{l_2}{l}\Pi_2\frac{\cos\beta_1}{\cos\gamma_1},\quad \Omega'_D = \omega'_2+\frac{l_1}{l}\Pi_2\frac{\cos\beta'}{\cos\gamma'}. \end{aligned} \tag{96} \]

In the second group of expressions the signs are set as though the force \(\Omega'\) and its components \(\omega'_1\) and \(\omega'_2\) are directed from \(A\) to \(C\); one should not forget, however, that it may change its direction, so that for a completely exact calculation the structure should also be checked for this latter combination of forces. According to the method of fastening the sheave beams, it is necessary to modify the design of the side trusses if it is desired that they be statically determinate. They are most often calculated by the method of resolving forces along the members, which in the present case is convenient to apply; of course, one could successfully use one of the general methods of Ritter, Cremona, etc., but in most cases the method we indicate creates no difficulties. Thus, if the sheave beams are fixed immovably to the transverse beam at the front legs, then the active forces, as we know, are transmitted mainly to the front joints; we may take for this case

Fig. 74.

\[ \Omega_A = \Omega_{AA'}, \]

\[ \Omega_C = \Omega^{*}_{CD}, \]

where, of course, it is assumed that

\[ \Omega_B = \Omega_D = 0. \]

The truss corresponding to this case is shown in drawing 74. The notation is clear without explanation. We resolve the forces, beginning at joint \(A\), in the direction of the leg \(AA_1\) and the member \(Ab_1\):

\[ s'_1 = \Omega_A \frac{\sin \delta_1}{\sin(\alpha_1 - \delta_1)};\qquad s_1 = \Omega_A \frac{\sin \alpha_1}{\sin(\alpha_1 - \delta_1)}. \]

Then we transfer the point of application of the force \(\rho_1\) to point \(b_1\) and resolve it in the direction of the tie \(a_1b_1\) and the leg \(BB_1\):

\[ s'_2 = \rho_1 \frac{\sin \delta'_1}{\sin \alpha'};\quad \Omega_2 = \rho_1 \frac{\sin(\alpha' - \delta'_1)}{\sin \alpha'}. \]

We now transfer \(\Omega_2\) to \(a_1\) and resolve it as was done above:

\[ s''_1 = \Omega_2 \frac{\sin \delta_2}{\sin(\alpha_1 - \delta_2)};\qquad s_2 = \Omega_2 \frac{\sin \alpha_1}{\sin(\alpha_1 - \delta_2)}. \]

then we make the resolution of forces at point \(b\):

\[ s_2''=\rho_2\frac{\sin\delta_2}{\sin\alpha'} \ \text{and}\ \Omega_3=\rho_2\frac{\sin(\alpha'+\delta_2)}{\sin\alpha'} \]

and so on; the forces \(\sigma_1'\), \(\Omega_2\), \(\Omega_3\) are tensile, while \(\sigma_2'\), \(\rho_1\), \(\rho_2\), \(\rho_3\) are compressive. The total stress in the lower parts of the legs is determined by the forces:

\[ S_1-\Sigma s_1^i \ \text{and}\ S_2+\Sigma s_2^i, \]

where \(n\) is the number of belts of the truss. Of course, in iron structures it is desirable, and in wooden ones necessary, that

\[ S_1-\Sigma s_1^i>0. \]

If the rear ends of the sheave beams are fixed, while the front ones may slide, then

\[ \Omega_b=\Omega_{AB}; \ \text{and}\ \Omega_D=\Omega_{CD}, \]

where, of course,

\[ \Omega_A=\Omega_C=0. \]

The corresponding structure is shown in Fig. 75. The procedure of calculation, obviously, is no different from that indicated above.

In the present case the horizontal braces are subjected to compression, and the diagonal rods to tension:

\[ s_2'=\Omega_B\frac{\sin\delta_1}{\sin(\alpha'-\delta_1)} \ \text{and}\ \rho_1=\Omega_B\frac{\sin\alpha'}{\sin(\alpha'-\delta_1)} \]

and so on.

Fig. 75.

For design purposes it is not superfluous to note that, as the angles \(\delta\) increase, \(\sigma_1'\) and \(\sigma_2'\), etc., increase, while \(\rho^1\), \(\rho_2\ldots\) decrease; as for the forces \(\Omega_2\), \(\Omega_3\ldots\), they increase together with \(\delta\). In the case where both ends of the sheave beams are fastened immovably to the transverse beams, so that the horizontal forces are distributed over all the joints, according to formula (95), one should use the truss shown in Fig. 76, with crossing diagonal rods; in this arrangement both the connecting beam \(AB\) and, in general, all horizontal braces are superfluous. The calculation

in general remains the same, although it is sufficiently clear from the drawing. The members \(Ab_1\), \(a_1 b_2\), \(a_2 B_1\) are in compression, while the members \(Ba_1\), \(b_1 a_2\), \(b_2 A_1\) are in tension. If for some reason it is desirable to use, in addition to the diagonal bracing, horizontal ties as well, then for an approximate calculation one may apply the method of resolving the truss into the simplest cases, namely, calculate the forces according to the 1st and 2nd cases and add the results. The horizontal ties will prove to be subjected to tension and compression, and therefore their dimensions must be determined from the difference of these forces. However, most often, even if horizontal ties are installed in such cases, they are not included in the calculation, since they impart to the structure a certain, though indeterminate, excess of strength.

Drawing 76.

Transverse trusses, which consist of the front or rear legs, have to be provided with crossing diagonal members, since they are subjected alternately to overturning forces first from one side and then from the other. If, in the calculation of the longitudinal trusses, the maximum wind pressure has already been taken into account, then there is no need to introduce it a second time in the calculation of the transverse trusses; in this case the force \(\omega'\) is determined according to formula (20), on the basis of the lateral deflection of the ropes.

As regards the question of which of the maximum forces, vertical or horizontal, should be made the basis for calculating a headframe of the type under consideration, one should use the indications given above and formula (93).

For verification, one may also, using the method of the principal ideal truss with diagonal bracing, determine the support reactions.

Prismatic-pyramidal headframes (six-legged).

Equations (57) and (58), derived for the basic form of this headframe and transformed for the calculation of forces, will be:

\[ t_1=\frac{l_2 \Pi}{l\cos\eta}, \]

\[ t_2=\frac{(l_1\cos\eta\sin\zeta-l_2\sin\eta\cos\zeta)\Pi-l\cos\eta\cos\zeta\,\Omega} {l\cos\eta\sin(\zeta-\xi)}, \tag{97} \]

\[ t_3=\frac{(-l_1\cos\eta\sin\xi+l_2\sin\eta\cos\xi)\Pi+l\cos\eta\cos\xi\,\Omega} {l\cos\eta\sin(\zeta-\xi)}. \]

In iron structures the front and middle legs are most often made vertical; since in this case

\[ \cos\eta=\cos\xi=1 \quad \text{and} \quad \sin\eta=\sin\xi=0, \]

then

\[ t_1=\frac{l_2}{l}\Pi, \]

\[ t_2=\frac{l_1\Pi-l\Omega}{l\,\operatorname{tg}\zeta} =\left(\frac{l_1}{l}\Pi-\Omega\right)\cos\zeta, \tag{98} \]

\[ t_3=\frac{\Pi}{\sin\zeta}. \]

In order that compression should always occur in the middle legs, we have derived a conditional inequality. It is clear that the front legs must be calculated for the maximum vertical pressure. Likewise, if the conditional inequality is satisfied even under the maximum horizontal pressure, then the middle legs must be calculated for compression from the vertical pressure. Conversely, the rear legs must be calculated for the greatest horizontal force. However, we shall have occasion to return to this question later.

The analysis of structures of the type under consideration may be carried out in two ways, namely: 1) all active forces may be distributed by resolving them at the upper joints; in this case the vertical forces will be applied at all joints, while the horizontal ones only at the middle joints. 2) The vertical forces at the front joints may be resolved into forces directed along the legs, and the two horizontal forces—along the front transverse and horizontal side connecting beams. Then the point of application of these new horizontal forces is to be transferred to the middle joints, upon which the sum of the indicated horizontal forces plus the vertical forces previously determined will act. These vertical forces we resolve into horizontal forces directed along the transverse beams and into others, the direction of which is given by the straight line of intersection of the vertical plane passing through the transverse beam and the plane drawn through the middle and rear leg. It now remains to resolve this force, as well as the previously found horizontal ones, in the direction of the middle and rear leg. To determine the stresses caused by the side horizontal forces just found, in conjunction with the forces arising from the deviation of the hoisting ropes from the axial direction and from the lateral pressure of the wind, we consider the rear legs, which, together with the horizontal transverse beam, form a trapezoidal truss, and in which the distribution of forces is determined by the arrangement of the transverse and diagonal members. These forces are determined according to one of the methods indicated above. As regards the analogous middle truss, the question is how to distribute over it the horizontal forces directed along the transverse beam. In the present method of analysis this question is solved approximately; for example, it is assumed that their magnitude is proportional to the magnitude of the forces directed along the legs; this is quite close to the truth. It is simpler, but less exact, to divide the horizontal force equally and assume that each of these halves is taken up by the leg trusses separately. The stresses in the legs, calculated in considering the transverse trusses, are added to those determined earlier. As to which maximum stresses should be adopted in the calculation, the instructions have been given by us above.

Of course, the magnitude of the support reactions is equal to the forces in the lower parts of the legs.

The method of calculation we have presented is rather complicated; therefore we must resort to the general method of resolving the active forces along the directions of the members of an ideal main truss, of which ...

in the given case there are two, entirely analogous to the same in pyramidal headframes. The decomposition may be carried out by means of formulas (97), or graphically by the method indicated earlier (Fig. 61).

Thus, in the plane of each of the transverse trusses there will act the forces \(s_1\) and \(s'_1\), \(s_2\) and \(s'_2\), \(s_3\) and \(s'_3\), which are active forces. In accordance with them we carry out the calculation of the truss, taking into account, of course, its construction. Most often, division into several panels with intersecting diagonal members is used, so that they have to be calculated by the method of distributing the forces among the joints and then by successive decomposition along the members. Of course, Ritter’s method, or some other, may be applied. Often, in order to have free access to the shaft mouth, one has to be satisfied only with intersecting rods and horizontal ties in the upper parts of the truss, while below the legs remain without bracing; in this case we may use formula (48). The correction for the lateral wind pressure and the force caused by the deflection of the rope may be introduced separately, distributing this pressure among the joints, and at the rear joint, moreover, among two trusses, as indicated above. From all that has been set forth it follows that any lateral ties connecting the front legs with the middle ones, or these latter with the rear ones, have no strictly definite significance and, if they are provided with a knee brace, as is shown in Figs. 37 and 38, this is done to eliminate lateral bending of the legs, so that these parts of the structure are not amenable to calculation; nor is there any need for it.

For the case of headframes with set-back supporting legs, the calculation in general remains the same, but, of course, for determining the forces one should use formulas which, in expanded form, will be:

\[ t_1 = \frac{l\Omega - (l_1 \operatorname{tg}\zeta + l_2 \operatorname{tg}\xi)\Pi} {\left[l(\operatorname{tg}\xi - \operatorname{tg}\zeta) + l_0 \operatorname{tg}\xi\right]\cos\eta}, \]

\[ t_2 = \frac{-(l + l_0)\Omega + \left[(l + l_1)\operatorname{tg}\zeta - l_2 \operatorname{tg}\eta\right]\Pi} {\left[l(\operatorname{tg}\xi - \operatorname{tg}\zeta) + l_0(\operatorname{tg}\eta + \operatorname{tg}\xi)\right]\cos\xi}, \tag{99} \]

\[ t_3 = \frac{l\Omega - \left[(l + l_1)\operatorname{tg}\xi + l_1 \operatorname{tg}\eta\right]\Pi} {\left[l(\operatorname{tg}\xi - \operatorname{tg}\zeta) + l_0(\operatorname{tg}\eta + \operatorname{tg}\xi)\right]\cos\zeta}. \]

For the case most often employed in practice, when the front and middle legs are placed vertically, these formulas

are simplified in view of the fact that \(\operatorname{tg}\eta=\operatorname{tg}\xi=0\); so that

\[ t_1=\frac{-\,l\Omega+l_1\Pi\operatorname{tg}\zeta}{l\operatorname{tg}\zeta}, \]

\[ t_2=\frac{(l+l_0)\Omega-(l+l_1)\Pi\operatorname{tg}\zeta}{l\operatorname{tg}\zeta}, \tag{100} \]

\[ t_3=-\,\frac{\Omega}{\sin\zeta}. \]

For headframes in which the supporting legs are brought directly under the sheave beams, the general expressions for the forces acting in the transverse trusses may be written as follows:

\[ t_1=-\frac{ \left[l_1\sin(\xi-\zeta)-(l_1+l_0)\cos\zeta\sin\xi+l\cos\eta\sin\zeta\right]\Pi +\left[(l_1+l_0)\cos\zeta\cos\xi-l\cos\eta\cos\zeta\right]\Omega }{ (l_1+l_0)\sin(\xi+\eta)\cos\zeta-l\cos\eta\sin(\zeta+\eta) }, \]

\[ t_2=-\frac{ \left[l_1\sin(\xi+\eta)-(l_1+l_0)\cos\zeta\sin\eta\right]\Pi -(l_1+l_0)\cos\zeta\cos\eta\,\Omega }{ (l_1+l_0)\sin(\xi-\eta)\cos\zeta-l\cos\eta\sin(\zeta+\eta) }, \tag{101} \]

\[ t_3=\frac{ \left[l_1\sin(\xi+\eta)-l\cos\eta\sin\eta\right]\Pi -l\cos^2\eta\,\Omega }{ (l_1+l_0)\sin(\xi+\eta)\cos\zeta-l\cos\eta\sin(\zeta+\eta) }; \]

these formulas are considerably simplified if the supporting legs are brought directly under the sheaves, and if the rear and middle legs are inclined identically. If the middle and rear legs are placed vertically, then these formulas take the form—

\[ t_1=l_2\,\frac{\Pi\sin\zeta-\Omega\cos\zeta}{l\sin\zeta} =\frac{l_2}{l}\left(\Pi-\frac{\Omega}{\operatorname{tg}\zeta}\right), \]

\[ t_2=l_1\,\frac{\Pi\sin\zeta-\Omega\cos\zeta}{l\sin\zeta} =\frac{l_1}{l}\left(\Pi-\frac{\Omega}{\operatorname{tg}\zeta}\right), \tag{102} \]

\[ t_3=\frac{\Omega}{\operatorname{tg}\zeta}. \]

Let us now consider the construction of the headframe shown in drawing 34. This headframe, in its most simplified form, may be represented by the diagram shown in drawing 77. At the center of the sheaves act the horizontal forces \(U_1, U_2\) and the vertical forces \(P_1, P_2\). Obviously, the outermost

the cushions take the forces \(\frac{1}{2}U_1\) and \(\frac{1}{2}U_2\), \(\frac{1}{2}P_1\) and \(\frac{1}{2}P_2\), and the middle one—\(\frac{1}{2}(U_1+U_2)\) and \(\frac{1}{2}(P_1+P_2)\). These forces, in turn, are transmitted by means of the beams \(CD\) and \(EF\) to the joints, so that vertical forces \(\frac{1}{8}(3P_1+P_2)\) will act at \(C\) and \(F\), and along the beam \(DF\) a horizontal force \(\frac{1}{4}(3U_1+U_2)\). On the other side of the head-frame there will act analogously: \(\frac{1}{8}(P_1+3P_2)\) and \(\frac{1}{4}(U_1+3U_2)\). The further calculation may be carried out either by resolving the vertical forces along the legs and the horizontal ones along the directions \(CF\), \(CD\), \(DE\), and \(EF\), or else, with the aid of an ideal truss constructed analogously to the side trusses of the head-frame, if the legs are joined together unchangeably by means of horizontal or inclined ties not reaching the supports. In this case the vertical planes are passed through \(CFDE\). If the planes of the front and rear legs are inclined to the vertical at the angles \(\eta\) and \(\xi\), then, putting

\[ \Omega'=\frac{1}{4}(3U_1+U_2);\quad \Pi'=\frac{1}{4}(P_1+3P_2)\quad \text{and}\quad l_1=l_2=\frac{1}{2}l, \]

by means of formulas (48) we determine \(t'\) and \(t'_1\), acting in the planes \(BCDA\) and \(GFEK\) at the joints \(C\) and \(F\). In exactly the same way we calculate the forces \(t''\) and \(t''_1\) at the joints \(D\) and \(E\). The further calculation reduces to the consideration of the trusses formed by the front and rear legs, similarly to what we saw in the case of pyramidal head-frames.

Let us now consider a head-frame whose simplified diagram may be represented by Fig. 78; its description was given with Fig. 28; the rear legs have been replaced by flexible ties, i.e. by ropes, so that the active forces act at the apex of the triangular truss. The case under consideration is covered by formulas (36), for here all the angles are negative.

Fig. 77.

Fig. 78.

Since it is desirable for us to obtain not the stresses at once, but the forces acting in the planes of the trusses, then, changing the signs

and expanding these expressions, we obtain:

\[ t_1=\frac{-\Pi \sin \xi+\Omega \cos \xi}{\sin(\eta+\xi)}, \]

\[ t_2=\frac{\Pi \sin \eta-\Omega \cos \eta}{\sin(\eta-\xi)}, \tag{103} \]

In the vertical sections drawn through the pulleys, the forces \(t_1'\) and \(t_1''\) will act, stretching the truss of the flexible ties, and \(t_2'\) and \(t_2''\), compressing the legs.

Let us find the forces acting on the ties, resolving the indicated forces in the directions of the ties and in the transverse direction; according to the drawing we shall have:

\[ \sigma_1'=\frac{t'l_1+t''l_2}{l\cos\alpha};\quad \sigma_1''=\frac{t'l_1+t''l_2}{l\cos\alpha}; \]

the lateral overturning force arising from the action of the forces \(t'\) and \(t''\) will be

\[ \omega=(t'-t'')\frac{l_2-l_1}{l}\operatorname{tg}\alpha, \]

where \(l_2-l_1=l_0\) is nothing other than the distance between the pulleys, or the axes of the hoisting compartments, from one another. Thus, with an increase of \(\alpha\), both the tensile stresses in the ties \(\sigma_1'\) and \(\sigma_1''\) and the overturning force \(\omega\), which can be taken up only by the legs, increase. It is desirable to make \(\alpha\) as small as possible. If the ties are directed parallel to the plane of symmetry of the headframe, that is, \(\alpha=0\), then, of course, \(\omega=0\) and

\[ \sigma_1'=\frac{t'l_2+t''l_1}{l},\quad \sigma_1''=\frac{t'l_1+t''l_2}{l}. \]

The truss of the legs is subjected to the action of the compressive forces \(t_2'\) and \(t_2''\), directed along its axis. In addition, it must resist all lateral forces, namely: the \(\omega\) just considered, the wind force, whose pressure for the given case is most dangerous in the direction perpendicular to the axis of the headframe, and also the forces arising from the deflection of the ropes. The tension in the ties and the compression in the legs are determined by the inequalities:

\[ -\Pi \sin \xi+\Omega \cos \xi<0 \quad \text{and} \quad \Pi \sin \eta-\Omega \cos \eta>0, \]

which in turn reduces to the inequalities:

\[ \varepsilon < \xi \quad \text{and} \quad \xi < \eta; \]

however, the first inequality predetermines the second, since \(\eta > \xi\). From these inequalities, moreover, it follows that, with the present arrangement of the headframe relative to the machine, both the greatest amount of tension and the greatest amount of compression will occur at Max. \(\Pi\); consequently, the headframe is calculated for rupture of the lower rope at the sheaves. Both ties, as well as the legs, are, of course, made of equal dimensions, corresponding to the greater of the stresses. If there is a middle tie, then the calculation in general does not change, but the tensile forces have to be distributed among these three directions. The calculation of the leg truss does not differ from other calculations of this kind, and therefore we shall not examine it here.

Instead of widening the distance between the legs at the base, headframes of this type are sometimes provided with side ties \(OD\) and \(O'E\); their purpose is to counteract the lateral overturning forces. These ties can easily be calculated on the assumption that the whole lateral pressure is taken by one tie and one leg. If this pressure is \(U\), then the force stretching the tie \(OD\) will be

\[ \sigma'=\frac{U}{\sin \delta}, \]

and the force compressing the leg—

\[ \sigma=\frac{U}{\operatorname{tg}\delta}. \]

The type of headframes considered may be used exclusively for small hoists.

If, with the same construction of the headframe, the machine were placed not on the left but on the right side (as in the drawing), then the equations for the forces acting in both trusses would take the form:

\[ t_1=-\frac{\Pi\sin \xi+\Omega\cos \xi}{\sin(\eta-\xi)}, \]

\[ t_2=\frac{\Pi\sin \eta+\Omega\cos \eta}{\sin(\eta-\xi)}, \]

for \(\varepsilon\) is a positive quantity. These expressions show that in the ties there will be tension, and in the legs compression, under

for all possible values of \(P\) and \(U\), even when the machine is placed at the shaft itself, so that \(U = 0\). Moreover, these same formulas show that with an increase in \(P\) or \(U\), both \(t_1\) and \(t_2\) increase.

Let us compare the magnitudes of the forces \(t\) in the rupture of the upper and lower rope (absolute):

\[ -t'_1 = \frac{\sigma(1+\cos\alpha_1)\sin\xi+\sigma\sin\alpha_1\cos\xi} {\sin(\eta-\xi)} \quad \text{(for the upper one),} \]

\[ -t''_1 = \frac{\sigma(1+\cos\alpha_2)\sin\xi+\sigma\sin\alpha_2\cos\xi} {\sin(\eta-\xi)} \quad \text{(for the lower one).} \]

In order to determine which of them is greater, subtract one from the other; in this case we shall have:

\[ (\cos\alpha_1-\cos\alpha_2)\sin\xi+ (\sin\alpha_1-\sin\alpha_2)\cos\xi \gtrless 0; \]

we transform this inequality as follows:

\[ -2\sin\frac{\alpha_1+\alpha_2}{2} \sin\frac{\alpha_1-\alpha_2}{2}\sin\xi + 2\cos\frac{\alpha_1+\alpha_2}{2} \sin\frac{\alpha_1-\alpha_2}{2}\cos\xi \gtrless 0, \]

or

\[ 1 \gtrless \operatorname{tg}\frac{\alpha_1+\alpha_2}{2}\operatorname{tg}\xi . \]

In the case of the first inequality, the calculation must be made for rupture of the upper rope, and in the case of the second—for rupture of the lower one.

For the legs we can obtain an analogous expression:

\[ 1 \gtrless \operatorname{tg}\frac{\alpha_1+\alpha_2}{2}\operatorname{tg}\eta . \]

Since \(\eta > \xi\), the lower inequalities follow from one another, but the upper ones may also not coincide. Our reasoning applies to cylindrical drums and is based on formulas (25, 26). It is clear that here \(\frac{\alpha_1+\alpha_2}{2}\) is the angle of the mean inclination of the equilibrants.

It remains for us to make a few further remarks concerning the calculation of large headframes above shafts. These headframes are built in

at the present time chiefly on the pattern of the English two-legged frames or else of combined six-legged frames. In general, the method of calculating these structures has been considered by us earlier; here we shall have to deal with certain constructional features.

If there are horizontal longitudinal ties connecting the inclined legs with the vertical ones (schematic drawing, or the more detailed drawing 33), then, for accuracy of calculation, it is necessary to introduce their weight as well, etc. Let us call the weight of one half of the inclined truss between \(O\) and \(A\) \(g_1\), between \(A\) and \(B\)—\(g_2\), and so on. Let the weight of the horizontal tie \(Aa\) be \(g'\), \(Bb\)—\(g''\), etc. Let us distribute the pressure from the force of gravity over the joints, on the assumption that the centers of gravity of the parts considered are situated at the midpoints of their lengths. We find that at the joints \(O\), \(A\), \(B\), \(C\), and at the support \(D\) there act vertical forces:

Drawing 79.

\[ \frac{g_1}{2},\quad g' + \frac{g_1 + g_2}{2},\quad g'' + \frac{g_2 + g_3}{2},\quad g''' + \frac{g_3 + g_4}{2},\quad \frac{g_4}{2}, \]

which are distributed in the direction of the horizontal ties and along the parts of the inclined legs. The first of these forces will be:

\[ \frac{g_1}{2}\operatorname{tg}\xi,\quad g' + \frac{g_1 + g_2}{2}\operatorname{tg}\xi,\quad g'' + \frac{g_3 + g_4}{2}\operatorname{tg}\xi,\quad \frac{g_4}{2}\operatorname{tg}\xi. \]

The parts of the inclined leg are subjected to the following compressive forces:

\[ \text{Part } OA\text{—force } s_1'=\frac{g_1}{2\cos\xi} \]

\[ \text{Part } AB\quad "\quad s_1''=\frac{2g_1+g'+g_2}{2\cos\xi} \]

\[ \text{Part } BC\quad "\quad s_1'''=\frac{2g_1+2g_2+g'+g''+g_3}{2\cos\xi} \]

and so on.

These same forces, directed perpendicular to the vertical legs, are transmitted partly to the joint \(O\), and partly directly to the supports. If by \(H\) we denote the height of the entire headframe, and by \(h_1\), \(h_2\), \(h_3\),—the heights of the ties above the horizon, then, on the basis of equality of moments-

then the resultant of them at joint \(O\) is determined by the formula:

\[ U_1=\frac{\operatorname{tg}\xi}{2H}\left[g_1h_1+(g'+g_1+g_2)h_2+\ldots+(g'''+g_3+g_4)h_4\right]. \]

This force \(U_1\) (horizontal) can be resolved in the direction of the inclined leg and the vertical as follows:

\[ s_1^0=-\frac{U_1}{\sin\xi},\qquad s_2^0=\frac{U_1}{\operatorname{tg}\xi}, \]

and the stresses caused by these forces can be added to those determined from the active forces, or simply \(U_1\) can be added to the active forces. As regards the forces \(s_1'\), \(s_1''\), etc., the stresses caused by these forces are introduced into the calculation gradually when determining the dimensions of the corresponding parts. If the weights of the parts of the vertical leg \(Oa\), \(ab\), \(bc\), etc., are denoted by \(q_1\), \(q_2\), etc., then the stresses in their parts will be due to the forces:

\[ q_1,\ q_1+q_2+\frac{q'}{2},\ q_1+q_2+q_3+\frac{q'+q''}{2},\ \text{etc.} \]

The stresses caused by wind pressure can be calculated in the same manner; the only difference is that we have to resolve the pressure normal to the legs into a horizontal component and one directed along them. If the wind pressure on the parts \(OA\), \(AB\), \(BC\), etc., is denoted by \(Q_1\), \(Q_2\), \(Q_3\), then at the joints \(O\), \(A\), \(B\ldots\) the horizontal forces will act:

\[ \frac{Q_1}{2\cos\xi},\ \frac{Q_1+Q_2}{2\cos\xi},\ \frac{Q_2+Q_3}{2\cos\xi}; \]

and the forces directed along the legs will be:

\[ \frac{Q_1}{2}\operatorname{tg}\xi;\ \frac{Q_1+Q_2}{2}\operatorname{tg}\xi;\ \frac{Q_2+Q_3}{2}\operatorname{tg}\xi. \]

We introduce these forces into the calculation on the same grounds as the forces arising from the weight of the parts of the headframe.

The calculation of stresses from the hoisting mechanism may be carried out by the general method, namely—to resolve the forces, concentrated at the centers of the sheaves, along the directions obtained by the intersection of the vertical plane drawn through the sheave parallel to the axis

of the headframe, with the planes of the legs. In this way the forces acting along the axes of the inclined and vertical frames are determined. The resultant force of gravity and wind pressure, referred to the top of the headframe, is resolved in the same manner; the points of application of these forces lie at the points of contact of the front and rear legs of the headframe.

However, for a complete calculation it is also necessary to take into account the lateral overturning forces, which consist of the lateral wind pressure \(v_1\) and the tensions \(p\) arising from the deflection of the ropes; thus

\[ \omega = v_1 + p. \]

We refer the force \(\omega\) to the frames of the rear legs. It is unnecessary to mention that the inclined frame must be calculated for the breaking force of the upper rope, and the vertical one for the breaking force of the lower. As an example, let us carry out the algebraic calculation of two typical frames often encountered in existing structures.

It should be noted that it is advisable to calculate in advance which of the wind pressures, axial or lateral, produces the more significant stresses in the legs; in the final calculation, of course, the greater of them should be adopted. We refer the lateral wind pressure, of course, to the top of the headframe, i.e., to the intersection of the legs. It is clear that \(\omega\) may be directed either to one side or to the other. The frames shown in Fig. 80, as is evident even from a simple examination of them, are statically indeterminate, owing to the excessively large number of connecting members. We shall remove, in the first, the horizontal ties, and in the second, the diagonal ones, as well as the struts whose purpose is to support the platforms \(n\). Then we shall obtain the simplified frames of Fig. 81 (p. 124).

Fig. 80.

We shall regard the horizontal force as transferred to the cross-member \(AC\). This is achieved, of course, by recalculating the moment with respect to the supports of those forces which are applied higher. Thus, if the height of application of the center of the lateral wind pressure is \(H_2\), the height

location of the sheaves is \(H_1\), and the height of the headframe to the contact of the legs is \(H\), then

\[ \omega H=\omega_1 H_1+pH_2. \]

In both trusses the upper parts are identical; if \(t_1'\) and \(t_1''\) are the components of the active forces from hoisting, and \(t_1^0\) is that of the other external forces, then at the left bearing block there is a pressure directed along the axis of the truss

\[ \frac{1}{2}t_1' + t_1^0, \]

at the middle bearing blocks—

\[ \frac{1}{2}(t_1' + t_1''), \]

and at the right—

\[ \frac{1}{2}t_1'' + t_1^0. \]

Fig. 81.

The end pressures are resolved in the direction of the legs and the horizontal cross member in such a way that the legs are compressed by the forces:

\[ -\sigma_1'=\frac{t_1' + 2t_1^0}{2\cos\alpha},\qquad -\sigma_2'=\frac{t_1'' + 2t_1^0}{2\cos\alpha}, \]

and along the cross member there is the force

\[ \frac{(t_1' - t_1'')}{2}\operatorname{tg}\alpha. \]

The pressures from the middle bearing blocks are transmitted to joint \(F\); moreover, at this same point we may regard as concentrated all the horizontal forces

\[ \pm \omega + \frac{t_1' - t_1''}{2}\operatorname{tg}\alpha, \]

which, together with the preceding axial forces, are taken up by the struts. The left strut is subjected to the compressive force—

\[ \tau_1= \frac{t_1' + t_1''}{4\cos\beta} - \frac{t_1' - t_1''}{4\sin\beta}\operatorname{tg}\alpha \pm \frac{\omega}{2\sin\beta}. \]

The struts are inclined to the vertical at the angle \(\beta\); the right strut is compressed by the force—

\[ \tau_2= \frac{t_1' + t_1''}{4\cos\beta} + \frac{t_1' - t_1''}{4\sin\beta}\operatorname{tg}\alpha \pm \frac{\omega}{2\sin\beta}. \]

and, consequently, the horizontal force acting along the first tie will be

\[ \Omega=\omega_2-\omega_1, \]

and after substitution—

\[ \Omega=\left(1-\frac{\operatorname{tg}\alpha}{\operatorname{tg}\beta}\right) \left[\pm\omega+(t_1-t_1'')\frac{\operatorname{tg}\alpha}{2}\right]. \]

For the further calculation we shall assume that an active force \(\Omega\) is applied at the joint; the auxiliary sections make it possible for us to compose equations of moments with respect to the poles—the joints:

\[ \sigma_1'' l_2\cos\alpha=-\Omega(H_1-H_2) \quad\text{and}\quad \sigma_2'' l_2\cos\alpha=\Omega(H_1-H_2), \]

\[ \sigma_1''' l_3\cos\alpha=-\Omega(H_1-H_3) \quad\text{and}\quad \sigma_2''' l_3\cos\alpha=\Omega(H_1-H_3), \]

\[ \sigma_1^{\mathrm{IV}} l_4\cos\alpha=-\Omega(H_1-H_4) \quad\text{and}\quad \sigma_2^{\mathrm{IV}} l_4\cos\alpha=\Omega(H_1-H_4) \quad\text{etc.} \]

If the stresses in the ties are denoted by \(\sigma'\), \(\sigma''\), etc., and we begin to pass sections that cut the horizontal ties and the legs, then, by composing moment equations with respect to the joints situated below, we can determine the tensions in the ties. Namely, we shall have the equation:

\[ (\sigma'+\Omega)(H_1-H_2)=\sigma_2' l_2\cos\alpha, \]

whence—

\[ \sigma'=-\Omega+\sigma_2'\frac{l_2\cos\alpha}{H_1-H_2}; \]

further—

\[ \sigma''(H_2-H_1)=\sigma_2'' l_3\cos\alpha-\Omega(H_2-H_3), \]

whence—

\[ \sigma''=-\Omega\frac{H_1-H_3}{H_2-H_2} +\sigma_2''\frac{l_3\cos\alpha}{H_2-H_1}; \]

then—

\[ \sigma'''(H_2-H_3)=\sigma_2''' l_4\cos\alpha-\Omega(H_1-H_4) \quad\text{etc.} \]

To determine the stresses in the lower tie-beam we can use the equation of static equilibrium:

\[ \sigma'+\sigma''+\sigma''+\ldots\sigma^{(n)}+\Omega=0; \]

The further course of the calculation is different for one or the other form, and therefore we shall have to consider them separately. The truss shown in Fig. I we shall calculate by applying the method of successive resolution of forces; for the truss depicted in Fig. II we shall apply Ritter’s method.

Thus, we resolve \(\tau_1\) in the direction of the leg and of the diagonal member, whose inclination to the vertical \(\gamma\) is found from the formula:

\[ \sigma_1''=\tau_1\frac{\sin(\gamma+\beta)}{\sin(\gamma+\alpha)} \quad\text{and}\quad \rho_1=\tau_1\frac{\sin(\beta+\alpha)}{\sin(\gamma+\alpha)}. \]

From the right brace we obtain analogously:

\[ \sigma_2''=\tau_1\frac{\sin(\gamma+\beta)}{\sin(\gamma+\alpha)} \quad\text{and}\quad \rho'=\tau_2\frac{\sin(\beta+\alpha)}{\sin(\gamma+\alpha)}. \]

The stresses \(\sigma_1''\) and \(\sigma_2''\) are compressive, \(\rho_1\) and \(\rho'\) tensile. At the following joints the resolution of the forces transmitted by the members gives us the formulas:

\[ \sigma_1'''=\rho'\frac{\sin(\delta+\gamma)}{\sin(\alpha+\delta)}, \quad \rho_{\mathrm{II}}=\rho'\frac{\sin(\gamma-\alpha)}{\sin(\alpha+\delta)}, \]

and

\[ \sigma_2'''=\rho_1\frac{\sin(\delta+\gamma)}{\sin(\alpha+\delta)}, \quad \rho''=\rho_1\frac{\sin(\gamma-\alpha)}{\sin(\alpha+\delta)}, \]

where the forces \(\sigma_1'''\) and \(\sigma_2'''\) are tensile, \(\rho_{\mathrm{II}}\) and \(\rho''\) are compressive. Thus, the total stresses at the bottom of the legs are determined by the expressions:

\[ \sigma_1=\sigma_1'+\sigma_1''-\sigma_1'''+\sigma_1^{\mathrm{IV}}-\sigma_1^{\mathrm{V}}+\ldots+\sigma_1^{2n}-\sigma_1^{2n+1}, \]

\[ \sigma_2=\sigma_1'+\sigma_2''-\sigma_2'''+\ldots+\sigma_2^{2n}-\sigma_1^{2n+1}. \]

From these expressions it is evident that the members are alternately compressed and stretched—the even-numbered ones are compressed, and the odd-numbered ones are stretched.

Let us now consider the second case—the truss with horizontal ties; we resolve the forces directed along the braces in the directions of the legs and ties:

\[ \sigma_1'=\tau_1\frac{\cos\beta}{\cos\alpha} \quad\text{and}\quad \omega_1=\tau_1\frac{\sin(\beta-\alpha)}{\cos\alpha}, \]

\[ \sigma_2'=\tau_2\frac{\cos\beta}{\cos\alpha} \quad\text{and}\quad \omega_2=\tau_2\frac{\sin(\beta-\alpha)}{\cos\alpha}, \]

if all the \((n-1)\) stresses have already been determined, then the \(n\)-th stress is obtained as the difference between \(\Omega\) and the sum of all the stresses found.

Vertical legs are connected by diagonal and transverse bracings, forming the truss shown in Fig. 82. The course of the calculation, in general, does not differ from the calculation of an inclined truss; however, it is somewhat simpler, namely in that in the formulas derived one should take \(\alpha = 0\). Thus:

Fig. 82.

\[ \sigma'_1=\frac{t' + 2t^0}{2} \quad\text{and}\quad \sigma'_2=\frac{t'' + 2t^0}{2}. \]

The struts are compressed by the forces:

\[ \tau_1=\frac{t' + t''}{4\cos\beta}=\tau_2. \]

The horizontal forces arising from wind pressure and from the deviation of the ropes are taken up by the inclined truss, so that in the present case \(\omega=0\).

In both trusses it is necessary also to introduce a correction relating to the dead weight. In the calculation of the inclined truss it is, of course, necessary to introduce not the whole weight of the members, but the component of it directed parallel to the axis of the truss. The weight of the members is distributed over the joints; the corrections relating to it may be calculated separately and then added to the numerical values obtained in the calculation of the principal forces.

Horizontal bracings often play the role of supporting beams for platforms at unloading levels; in this case they must also be calculated for the load carried by the platforms, as must the struts supporting these beams. These struts are often given the form of brackets.

When the compressive and tensile forces acting along the members of the spatial truss of the headframe have been determined, one proceeds to the computation of their dimensions. These members, again, are lattice trusses, usually statically indeterminate owing to the excessively large number of diagonal and transverse bracings. Usually, without attempting an exact calculation, it is assumed that the forces are transmitted to the main panels, for which the corresponding ...

according to the dimensions, the cross-section of the rolled section. The role of the tie rods is thus reduced to imparting rigidity to the trusses and eliminating longitudinal deflections. Nevertheless, they are still calculated for longitudinal bending—in the event of compression, since their length is usually considerable.

Fig. 42.

Fig. 38.

Fig. 39.

Fig. 40.

Fig. 41.

Fig. 29.

Fig. 30.

Drawing 23.

Drawing 24.

Drawing 25.

Drawing 27.

Drawing 28.

Fig. 43.

Fig. 34.

Fig. 35.