Eng. V. K. NECHAEV.

CALCULATION OF LEAF SPRINGS FOR PARACHUTE DEVICES.

Foreword.

Questions concerning the calculation of leaf springs, which are widely used in the parachute mechanisms of hoisting cages, have been very poorly covered not only in Russian but also in foreign literature. The absence of sufficiently developed instructions on calculation methodology is the cause of a great and unproductive expenditure of time in designing springs.

The present work, carried out on assignment from the Scientific-Research Coal Institute, has the aim of filling this gap and providing a method for calculating the leaf springs of parachute devices that is as accurate as possible and convenient for application in design bureaus.

The expanded calculation scheme given at the end of the work should free the designer from the necessity, each time when designing a spring, of again searching in the text for the final calculation formulas and reviewing their derivations.

Six nomograms compiled by the author are appended to the work; they can considerably reduce the time spent on calculating a spring.

Despite the somewhat specific nature of the problem set before the author, the calculation method presented may be almost fully used in designing springs for automobiles and railway rolling stock.

The present work, together with the nomograms, even before being submitted for publication, was adopted for calculations in the Tomsk Design Sector of Shakhtstroy.

At the beginning of the work, A. P. Bushkovsky, a postgraduate student of the Department of Applied Mechanics of the Siberian Mechanical-Machine-Building Institute, took part, with the active cooperation of the entire staff of the department.

A number of suggestions and instructions were given to the author by the head of the Department of Applied Mechanics of SMMI, Associate Professor N. P. Shubin, who reviewed the work in manuscript.

Eng. V. Nechaev.

1. CALCULATION OF A LEAF SPRING FOR STRENGTH.

Every leaf spring intended for operation in the parachute device of a hoisting cage must satisfy a whole series of requirements, of which the following may be regarded as the principal ones:

1. The lowest possible weight of the spring, with sufficient strength;

2. A sufficiently large (not only elastic) deflection of the spring under the action of the design load;

3. No divergence of the ends of the spring leaves during its deformation.

The first and third of the indicated requirements are obvious. The second requirement is due primarily to the design features of the parachute mechanism and, in addition, follows from the need to obtain the smallest possible dependence of the reliability of parachute operation on the degree of wear of the guides. With an excessively stiff spring, even slight wear of the guides may cause a considerable decrease in the pressure exerted on them by the braking elements of the parachute and, consequently, a decrease in the braking force.

When designing the spring, its deformation is determined in advance by design considerations and by the scheme of the entire parachute device.

Let us note that similar requirements are imposed on a leaf spring when it is used in automobile, railway transport, etc. Here the flexibility of the spring determines the softness of the spring suspension of the automobile body, and so forth.

For maximum satisfaction of the above-mentioned requirements, leaf springs are designed in a form close to that of a body of equal resistance to bending. This is the basis of the usual method for calculating a leaf spring as a beam of equal resistance.

1. Elementary case.

From the course in Strength of Materials it is known that a cantilever beam of rectangular cross-section with constant height \(h\), having in plan the form of an isosceles triangle, represents a bar of equal resistance to bending when a bending force \(Q\) acts at its end (Fig. 1).

The maximum normal stress in the extreme (by height) fibers of such a beam is the same along its entire length and is determined by the following expression:

\[ \sigma=\frac{6Ql}{Bh^2}, \]

where \(B\) is the width of the beam at the place of its clamping.

Under the action of the force \(Q\), the beam bends along the arc of a circle and receives at its end a deflection:

\[ f=\frac{Ql^{3}}{2EJ_{\max}}, \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \quad (1) \]

where:

\[ J_{\max}=\frac{Bh^{3}}{12}. \]

This value \(f\) is one and a half times greater than the deflection of a prismatic beam with constant cross-section \(B \times h\), having, obviously, the same strength as the beam of uniform resistance under consideration.

The indicated beam of uniform resistance is the starting point for the construction of a spring. Let us cut this beam into \(2n\) strips of equal width

\[ \frac{b}{2}=\frac{B}{2n}, \]

join them in pairs and stack them one upon another, as shown in Fig. 1. We obtain a leaf spring with \(n\) leaves. The thickness of all leaves is the same and equal to \(h\).

Neglecting friction between the individual leaves and assuming that all these strips will bend in the same way as the corresponding parts of the whole triangular beam (an analysis of this assumption will be made below), it may be considered that all the strips will bend along the arc of one circle. With small leaf thickness and a large radius of curvature of the spring, it may likewise be considered with sufficient accuracy that all the leaves bend along arcs described from one center. Consequently, when the spring bends, its leaves will not separate.

Fig. 1.

2. Leaf spring on straight shackles.

Ordinary leaf springs are composed of separate strips of corresponding length, cut cold from strip spring steel, of rectangular cross-section \(b \times h\). These strips are bent along arcs of a circle, hardened, and stacked one upon another. All leaves are clamped in the middle by a band put on hot or tightened

with the aid of a wedge. To prevent possible lateral displacement of individual leaves under the influence of accidental side shocks, grooved steel (OST—28) is usually used for leaf springs.

To obtain a beam of equal resistance to bending, the ends of the spring leaves should be cut to a triangular shape. In practice, however, they are more often cut to a trapezoidal shape. The effect of this deviation from the theoretically required form will be considered below.

The fitting of the clamp increases the stiffness of the spring in its middle part. On this basis, some authors¹ propose considering half of the spring as a cantilever beam, fixed in the clamp, and taking as the calculated length of this beam the value \(l-\dfrac{a}{2}\) (Fig. 2).

Fig. 2.

However, taking into account the difficulty of achieving a fully rigid and tight fit of the clamp, it is more expedient to neglect its reinforcing effect and to take as the calculated length simply half the chord of the spring, \(l\).

Under the action of an external load, the initial curvature of the spring decreases and becomes comparatively small. Therefore, with some approximation, one may regard half of the spring as a straight cantilever beam of equal resistance, of length \(l\), loaded at the end by the force \(\dfrac{P}{2}\) (Fig. 2).

Here \(l\) is half the chord of the spring in its deformed state, \(P\) is the external load applied at the middle of the length of the spring.

The maximum bending moment at the point of fixing is:

\[ M_{\max}=\frac{Pl}{2}. \]

The section modulus of the dangerous section of the spring is:

\[ w=n\frac{bh^2}{6}, \]

where \(n\) is the number of leaves of the spring,
\(b \times h\) is the cross-section of each leaf.

¹ Vanderjucht and Korotkevich. Fundamentals of Car Building. 1930.

Strength equation:

\[ \frac{Pl}{2}=n\,\frac{bh^2}{6}\,\sigma_{\max} \]

and hence the bending stress:

\[ \sigma_{\max}=\frac{3Pl}{nbh^2}\ . . . . . . . . . . . . . . . . . (2) \]

The deflection of the spring \(f\) is determined by formula (1), if in it \(P/2\) is substituted for \(Q\):

\[ f=\frac{Pl^3}{4EJ_{\max}}. \]

Substituting the value for the maximum moment of inertia

\[ J_{\max}=n\,\frac{bh^3}{12} \]

we obtain:

\[ f=\frac{3Pl^3}{Enbh^3}\ . . . . . . . . . . . . . . . . . (3) \]

or

\[ f=\frac{3Pl}{nbh^2}\cdot\frac{l^2}{h}. \]

Introducing here the value for \(\sigma_{\max}\) from formula (2), we have:

\[ f=\frac{l^2}{Eh}\,\sigma_{\max}\ . . . . . . . . . . . . . . . . . (4) \]

Thus, for one and the same permissible bending stress \(R_b\), the deflection of the spring is inversely proportional to the thickness of the leaves of which it is composed and directly proportional to the square of the length of the spring. The greater the flexibility required of the spring, the thinner the leaves chosen for its manufacture must be.

3. Leaf spring on inclined shackles.

This case is practically the most important. As will be shown below, by changing the angle of inclination of the spring shackles it is possible to vary, within certain limits, the flexibility of an already finished spring.

Let us introduce the notation (Fig. 3):

\(\lambda\)—deflection of the spring after deformation,
\(2l\)—chord of the spring after deformation,
\(P\)—external load on the spring, applied at its middle,
\(\alpha\)—angle of inclination of the spring shackles to the vertical,
\(f\)—elastic deformation (deflection) of the spring under the action of the force \(P\).

a) Determination of stresses.

Under the action of the force \(P\), forces \(Z\) arise in the spring suspension shackles, directed obliquely downward. Their magnitude is determined from the obvious relation:

Fig. 3.

\[ 2Z\cos\alpha = P \]

or

\[ Z=\frac{P}{2\cos\alpha}. \ . \ (5) \]

The force \(Z\) may be resolved into vertical and horizontal directions.

The vertical force is equal to:

\[ Q=Z\cos\alpha=\frac{P}{2}, \]

—and the horizontal force

\[ H=Z\sin\alpha=\frac{P}{2}\tg\alpha. \]

The vertical force \(Q\) produces bending of the spring, while the horizontal force \(H\), in addition to bending, also causes tension of the main spring leaf. The bending moment in the middle section of the spring is:

\[ M_{max}=Ql+H\lambda=\frac{P}{2}l+\frac{P}{2}\lambda\tg\alpha \]

or

\[ M_{max}=\frac{P}{2}(l+\lambda\tg\alpha). \]

The strength equation for bending:

\[ \frac{P}{2}(l+\lambda\tg\alpha)=n\frac{bh^2}{6}\sigma_{max}\ . . . . . . . . \ (6) \]

and hence the bending stress in the middle section of the spring is:

\[ \sigma_{max}=\frac{3P(l+\lambda\tg\alpha)}{nbh^2}. \]

The tensile stress in the main leaf is:

\[ \sigma=\frac{H}{bh}=\frac{P\tg\alpha}{2bh}\ . . . . . . . . \ (7) \]

and the total stress in it:

\[ \sigma'_{\max}=\sigma_{\max}+\sigma=\frac{3P(l+\lambda \tg\alpha)}{nbh^2}+\frac{P\tg\alpha}{2bh}\ldots\ldots(8) \]

Comparison of equations (2) and (6) shows that, in calculating a spring on oblique hangers, its curvature can no longer be neglected and the formulas obtained for straight bars cannot be applied.

b) Determination of deformations.

α) Approximate method.

To determine the magnitude of the deflection of the spring, we shall again use Castigliano’s theorem. The work of the bending deformations of one half of the spring under the forces \(Q\) and \(H\) is determined by the general expression:

\[ U=\frac{1}{2E}\int_0^l \frac{M_x^2\,ds}{J_x} \]

where \(ds\) is an element of the length of the spring arc.

In view of the usually small curvature of the spring in the deformed state, it may be approximately assumed that \(ds\) is equal to an element of the length of the spring chord \(dx\), i.e.

\[ ds \cong dx, \]

then

\[ U=\frac{1}{2E}\int_0^l \frac{M_x^2\,dx}{J_x}. \]

The deflection of the middle of the spring is now determined on the basis of Castigliano’s theorem:

\[ f=\frac{\partial U}{\partial Q}=\frac{1}{2E}\int_0^l \frac{2M_x\cdot \dfrac{\partial M_x}{\partial Q}}{J_x}\,dx. \]

The bending moment in section \(x\) (Fig. 3):

\[ M_x=Qx+Hy. \]

Making again the assumption that one half of the spring is deformed in the same way as a cantilever beam, curved in the shape of the spring and having a triangular form in plan, one may write:

\[ J_x=J_{\max}\cdot \frac{x}{l}; \]

where

\[ J_{\max}=n\frac{bh^3}{12}. \]

Owing to the small curvature of the elastic curve of the spring and the small magnitude of the deflection \(\lambda\) in comparison with \(l\), it may be approximately assumed that the ordinate \(y\) varies proportionally to \(x\).

This assumption is equivalent to replacing half of the arc of the spring by a straight line. Consequently, approximately:

\[ y=\lambda\frac{x}{l}. \]

Now the bending moment:

\[ M_x=Qx+Hy=Qx+H\lambda\frac{x}{l} \]

and hence

\[ \frac{\partial M_x}{\partial Q}=x. \]

Substitute the obtained values for \(J_x\), \(M_x\), and \(\dfrac{\partial M_x}{\partial Q}\) into the expression for \(f\):

\[ f=\frac{1}{2E}\int_0^l \frac{2\left(Qx+H\lambda\frac{x}{l}\right)x} {J_{\max}\frac{x}{l}}\,dx= \]

\[ =\frac{l}{EJ_{\max}}\int_0^l \left(Qx+H\lambda\frac{x}{l}\right)\,dx = \frac{l^2}{2EJ_{\max}}(Ql+H\lambda)= \]

\[ =\frac{l^3}{2EJ_{\max}}\left(Q+\frac{\lambda}{l}H\right). \]

Introduce here the values for \(Q\) and \(H\):

\[ f=\frac{l^3}{2EJ_{\max}} \left(\frac{P}{2}+\frac{\lambda}{l}\frac{P\tg\alpha}{2}\right) = \frac{Pl^3}{4EJ_{\max}} \left(1+\frac{\lambda}{l}\tg\alpha\right). \]

Finally, substituting the value for the moment of inertia:

\[ J_{\max}=n\frac{bh^3}{12}, \]

we obtain the final formula.

\[ f=\frac{3Pl^3}{Enbh^3} \left(1+\frac{\lambda}{l}\tg\alpha\right) \tag{10} \]

It may be written in the form:

\[ f=\frac{3Pl^3}{Enbh^3}\cdot \varphi \tag{11} \]

or else

\[ f=\frac{Pl^3}{4EJ_{max}}\cdot\varphi \tag{12} \]

Here the first part of the expression for \(f\):

\[ \frac{3Pl^3}{Enbh^3}\quad \text{or}\quad \frac{Pl^3}{4EJ_{max}} \]

obviously gives the deflection of a straight cantilever beam of uniform strength, of length \(l\), loaded at the end by a force \(P/2\), and having, at the point of fixing, a moment of inertia equal to the moment of inertia of the middle section of the spring.

The coefficient

\[ \varphi=\left(1+\frac{\lambda}{l}\tg\alpha\right) \tag{13} \]

takes into account the curvature of the spring and the tensile action of the inclined shackles.

From this expression it is seen that increasing the angle of inclination of the shackles to the vertical increases the flexibility of the spring.

Formula (10), written in the form:

\[ f=\frac{6l^2Q}{Enbh^3}\,(l+\lambda\tg\alpha), \]

where \(Q=\dfrac{P}{2}\)—the vertical load at the end of the spring, is given in Hütte’s handbook 1).

β) A more exact method.

The method considered above for determining the deformations of a spring on inclined shackles is based on a whole series of assumptions.

Half of the spring arc was taken as a straight line; in calculating the work of deformation, the length of the arc was taken equal to the length of the chord \((ds=dx)\), and the work of stretching of the master leaf by the horizontal thrust of the shackles was not taken into account.

Let us solve the problem of determining the deformation of the spring by a more exact method. To this end we consider half of the spring as a cantilever plate of thickness \(h\), bent along an arc of radius \(R\) and having, in the deformed state, in plan the form of a triangle (Fig. 4).

Let us denote by \(\varphi_0\) one-half of the angle between the extreme radii of the bent spring.

The bending moment in section \(x\) is equal to:

\[ M_x=Qx+Hy, \tag{14} \]

1) Hütte, vol. I, p. 728, 13th edition.

where, according to the drawing:

\[ \left. \begin{aligned} x &= l - R\operatorname{Sin}\varphi\\ y &= R\operatorname{Cos}\varphi - a \end{aligned} \right\} \tag{14a} \]

The work of deformation in bending a bar of small curvature, as is known, is determined by the formula¹):

\[ U=\int_0^s \left(\frac{M_x^2}{2EJ_x}+\frac{N^2}{2EF}\right)ds. \tag{15} \]

Here:

\(ds\)—element of the arc of the spring,
\(J_x\)—moment of inertia of the cross-section,
\(F\)—area of the cross-section of the bar,
\(N\)—normal force stretching the given section.

Since only one root leaf is subjected to stretching by the force \(H\), we have:

\[ F=bh. \]

The normal force stretching the given section varies along the length of the arc and is obviously determined by the expression:

\[ N=H\operatorname{Cos}\varphi. \]

Fig. 4.

The deflection of the spring \(f\) is determined from expression (15):

\[ f=\frac{\partial U}{\partial Q} = \int_0^s \frac{2M_x\frac{\partial M_x}{\partial Q}}{2EJ_x}\,ds. \]

(The second term of the work \(T\) gives a derivative with respect to \(Q\) equal to zero, since it does not depend on the magnitude \(Q\).)

Substituting here the values:

\[ M_x=Qx+Hy, \]

\[ \frac{\partial M_x}{\partial Q}=x, \]

\[ ds=R\,d\varphi, \]

\[ J_x=J_{\max}\frac{x}{l}, \]

¹) S. P. Timoshenko. Course of Strength of Materials, 1931, p. 408.

we have:

\[ f=\int_{0}^{\varphi_0}\frac{2M_x\cdot x\cdot R}{2EJ_{\max}\cdot \dfrac{x}{l}}\,d\varphi = \frac{Rl}{EJ_{\max}}\int_{0}^{\varphi_0} M_x\,d\varphi= \]

\[ = \frac{Rl}{EJ_{\max}}\int_{0}^{\varphi_0}(Qx+Hy)\,d\varphi. \]

On the basis of formulas (14-a):

\[ f=\frac{Rl}{EJ_{\max}}\int_{0}^{\varphi_0} \left[Q(l-R\sin\varphi)+H(R\cos\varphi-a)\right]\,d\varphi, \]

whence, after integration:

\[ f=\frac{Rl}{EJ_{\max}} \left[Q(l\varphi_0+R\cos\varphi_0-R)+H(R\sin\varphi_0-a\varphi_0)\right]. \]

According to Fig. 4:

\[ R\cos\varphi_0=a, \]

\[ R\sin\varphi_0=l, \]

and

\[ f=\frac{Rl}{EJ_{\max}} \left[Q(l\varphi_0+a-R)+H(l-a\varphi_0)\right]= \]

\[ =\frac{Rl}{EJ_{\max}} \left[Q(l\varphi_0-\lambda)+H(l-a\varphi_0)\right], \tag{16} \]

since

\[ R-a=\lambda. \]

The half-chord of the spring \(l\) may be regarded as the altitude of a right triangle inscribed in a circle of radius \(2R\) and resting on its diameter. The altitude \(l\) divides the base of the triangle into two segments¹):

\[ 2R-\lambda \quad \text{and} \quad \lambda. \]

On the basis of the well-known geometrical theorem:

\[ l^2=(2R-\lambda)\lambda. \]

Hence:

\[ R=\frac{l^2+\lambda^2}{2\lambda}, \]

\[ a=R-\lambda=\frac{l^2-\lambda^2}{2\lambda}. \]

¹ The same considerations underlie the derivation of formula (30), see Fig. 8.

Moreover:

\[ \operatorname{Sin}\varphi_0=\frac{l}{R}=\frac{2\lambda l}{l^2+\lambda^2}, \]

\[ \varphi_0=\arcsin\frac{2\lambda l}{l^2+\lambda^2}. \]

Now expression (16) can be rewritten in the following form:

\[ f=\frac{l}{EJ_{\max}}\left(\frac{l^2+\lambda^2}{2\lambda}\right) \left[ Q\left(l\arcsin\frac{2\lambda l}{l^2+\lambda^2}-\lambda\right)+ H\left(l-\frac{l^2-\lambda^2}{2\lambda}\arcsin\frac{2\lambda l}{l^2+\lambda^2}\right) \right]. \]

Substituting here for \(Q\) and \(H\) their values in terms of \(P\):

\[ Q=\frac{P}{2}, \]

\[ H=\frac{P}{2}\tg\alpha, \]

after simple algebraic transformations we finally obtain:

\[ f=\frac{Pl^3}{4EJ_{\max}}\cdot\varphi, \tag{17} \]

where:

\[ \varphi=A+B\tg\alpha, \tag{18} \]

with:

\[ A=\left(k+\frac{1}{k}\right)\left[\arcsin\frac{2}{k+\frac{1}{k}}-k\right], \]

\[ B=\left(k+\frac{1}{k}\right)\left[1-\left(\frac{1}{2k}-\frac{k}{2}\right)\arcsin\frac{2}{k+\frac{1}{k}}\right], \tag{19} \]

\[ k=\frac{\lambda}{l}. \tag{20} \]

Here again the dependence of the magnitude of the deflection of the leaf spring on the angle of inclination \(\alpha\) of the spring shackles is confirmed. An increase in \(\alpha\) gives an increase in the flexibility of the leaf spring.

The example of calculation of a leaf spring given below shows that the difference between the results obtained by the simplified and the more exact methods, for small curvature of the leaf spring, is insignifi-

is significant. On this basis, in the subsequent derivations and calculations we shall use the shorter formulas obtained by the simplified method.

4. Design of the Leaf Spring.

Above, in the analysis of the stresses and deformations of the spring on oblique hangers, the following relations were obtained:

1) The strength equation of the spring in bending:

\[ \frac{P}{2}(l+\lambda \operatorname{tg}\alpha)=n\frac{bh^2}{6}\sigma_{max}, \tag{16} \]

which can now be written in the form:

\[ n\frac{bh^2}{6}\sigma_{max}=\frac{Pl}{2}\varphi . \tag{21} \]

2) Deflection of the spring:

\[ f=\frac{3Pl^3}{Enbh^3}\varphi . \tag{17} \]

3) The coefficient taking into account the curvature of the spring and the angle of inclination of its hangers:

\[ \varphi=\left(1+\frac{\lambda}{l}\operatorname{tg}\alpha\right). \tag{13} \]

Calling the flexibility of the spring \(\Delta\) the magnitude of its deflection in centimeters under the action of a load equal to \(1000\ \mathrm{kg}\), one may write:

\[ \Delta=\frac{f}{P/1000}, \tag{22} \]

or

\[ f=\frac{P}{1000}\Delta, \tag{23} \]

where the load \(P\) is measured in kilograms.

To determine the flexibility \(\Delta\), substitute into (17) the value \(P=1000\ \mathrm{kg}\):

\[ \Delta=\frac{3000l^3}{Enbh^3}\varphi\frac{cm}{tn}. \tag{24} \]

Let us represent expression (24) in the following form:

\[ \Delta=\frac{500l^3}{En\,\dfrac{bh^2}{6}}\varphi \]

or

\[ n \frac{bh^2}{6}=\frac{500l^3}{E\Delta h}\varphi \]

and substitute it into the strength equation (21):

\[ \frac{500l^3}{E\Delta h}\varphi\,\sigma_{max}=\frac{Fl}{2}\varphi, \]

whence finally¹

\[ P=\frac{1000l^2}{E\Delta h}\sigma_{max}. \tag{25} \]

From (24) we shall also determine the number of leaves of the spring:

\[ n=\frac{3000l^3}{E\Delta h b^3}\varphi. \tag{26} \]

Equations (13), (23), (25), and (26) are fundamental in calculating a spring. Their application is shown below on a specific example (p. 20) and is given in the general calculation scheme for a leaf spring.

For a given load \(P\) and selected:

1) leaf profile \(b \times h\),
2) allowable bending stress \(R_b\),
3) flexibility of the spring \(\Delta\) or desired deflection \(f\),
4) angle of inclination of the shackles \(\alpha\),
5) camber of the deformed spring \(\lambda\),

the indicated equations make it possible to determine the required number of spring leaves \(n\) and its chord in the deformed state \(2l\).

To establish the relation between the geometric elements of the spring in the free and deformed states (Fig. 5), let us denote:

\(R\)—the manufacturing radius of curvature of the main leaf,
\(2L\)—the manufacturing chord of the spring,
\(F\)—the manufacturing camber of the spring, measured from the line connecting the centers of the eyes to the first, main leaf,
\(S\)—the length of the straightened main leaf,
\(2\varphi_0\)—the angle formed by the extreme radii of the free spring,
\(r\)—the mean radius of the eyes of the main leaf.

\[ P=\frac{1000l^2}{E\Delta h}R_b \tag{25-a} \]

fabrication dimensions, i.e., the dimensions of the spring in the free state (without load), are necessary for its manufacture and must be indicated on the working drawing.

Under the action of the load \(P\), the fabrication camber \(F\) decreases to the value \(\lambda\), called above the camber of the spring in the deformed state.

The elastic deformation (deflection) of the spring is:

\[ f = F - \lambda . \]

The values of \(f\) and \(\lambda\) adopted in the above calculation make it possible to determine the fabrication camber:

\[ F = f + \lambda . \tag{27} \]

Fig. 5.

The main leaf in the deformed state is an arc of a circle having chord \(AB = 2l\) (between the vertical axes passing through the centers of the eyes) and camber \(\lambda - r\), where \(r\) is the mean radius of the eyes (Fig. 5).

For a known chord \(2l\) and camber \(\lambda - r\), the length of the straightened main leaf (the length of the arc \(AB\)) can, with accuracy sufficient for practice, be determined by the approximate formula¹ (Hütte, vol. I, p. 35):

\[ S = \sqrt{(2l)^2 + \frac{16}{3}(\lambda - r)^2} \]

or

\[ S = 2 \sqrt{l^2 + \frac{4}{3}(\lambda - r)^2} \tag{28} \]

During deformation of the spring, the length \(S\) of the main leaf remains unchanged, i.e., the fabrication camber and chord are constantly related by the relation:

\[ S = 2 \sqrt{L^2 + \frac{4}{3}(F - r)^2} \]

or, on the basis of formula (28):

\[ L^2+\frac{4}{3}(F-r)^2=l^2+\frac{4}{3}(\lambda-r)^2, \]

\[ L^2=l^2+\frac{4}{3}(\lambda-r)^2-\frac{4}{3}(F-r)^2. \]

Whence we obtain the final expression for determining the shop chord of the spring:

\[ 2L=2\sqrt{\,l^2+\frac{4}{3}(\lambda-r)^2-\frac{4}{3}(F-r)^2\,}. \tag{29} \]

Fig. 6.

For the right triangle \(ABC\) (Fig. 6), resting on the diameter \(2R\) and having height \(BD=L\), on the basis of a known geometrical theorem one may write:

\[ \frac{F-r}{L}=\frac{L}{2R-(F-r)}, \]

whence:

\[ R=\frac{L^2+(F-r)^2}{2(F-r)}. \]

Substitute here the value of \(L\) from (29):

\[ R= \frac{ l^2+\frac{4}{3}(\lambda-r)^2-\frac{4}{3}(F-r)^2+(F-r)^2 }{ 2(F-r) } = \]

\[ = \frac{ l^2+\frac{4}{3}(\lambda-r)^2-\frac{1}{3}(F-r)^2 }{ 2(F-r) }. \]

But according to (28):

\[ l^2+\frac{4}{3}(\lambda-r)^2=\frac{S^2}{4} \]

and

\[ R= \frac{ \frac{S^2}{4}-\frac{(F-r)^2}{3} }{ 2(F-r) }. \]

or

\[ R=\frac{3S^2-4(F-r)^2}{24(F-r)}. \tag{30} \]

The last expression serves to determine the shop radius of curvature of the main leaf of the spring.

Finally, the angle between the extreme radii of the spring in the free state is determined from the obvious relation:

\[ 2\varphi_0=\frac{S}{R}\frac{360}{2\pi}=57.4\frac{S}{R}. \tag{31} \]

(the angle \(\varphi_0\)—in degrees).

5. Nomograms for calculating springs.

To simplify the calculations for determining the dimensions of leaf springs, the author has compiled nomograms based on the calculation equations (25) and (26) obtained above.

At the end of the work six nomograms are appended, for the six cross-section sizes of spring steel according to OST—28 that are most commonly used in practice. Each of these nomograms is valid only for those dimensions of the cross section \(b \times h\) of the spring steel that are indicated in large type in its upper right corner.

The method of using the nomograms is set forth in detail in the general scheme for calculating a leaf spring (p. 44). In addition, an approximate calculation path is shown on each of them by dotted lines and arrows.

For constructing these nomograms, equations (25) and (26) were reduced to linear form by taking logarithms.

Considering it unnecessary to give here a description of the method of their construction and the tables of all the constructions carried out in doing so, we shall note only the following.

For spring hardened steel the modulus of elasticity \(E\) varies within the limits\(^{1}\):

\[ E=1.9\cdot10^6 \div 2.2\cdot10^6\ kg/cm^2, \]

in the handbook “Hütte” the following value is given for spring steel:

\[ E=2.2\cdot10^6\ kg/cm^2. \]

In calculating the nomograms we adopted:

\[ E=2.1\cdot10^6\ kg/cm^2, \]

which may be regarded as a sufficiently reliable value.

\(^{1}\) Vandergucht and Korotkevich. Fundamentals of Car Building. 1930.

Nevertheless, for a certain selected value of \(E\), the nomograms do not lose their applicability for any other values of Young’s modulus.

Let us suppose that the material available at the plant for designing the spring has a modulus of elasticity \(E_1\), not equal to the value adopted by us above, \(E = 2.1 \cdot 10^6\ kg/cm^2\). Using the appended nomograms, in the usual way, from the specified load \(P\) and the selected \(R_b\), \(\Delta\), and \(\varphi\), we determine the half-chord of the deformed spring \(l\) and the required number of leaves \(n\). With modulus \(E_1\), the spring will retain the required values \(R_b\), \(\varphi\), and \(\Delta\), if its half-chord length \(l_1\) and number of leaves \(n_1\) are made equal to the following values:

\[ \left. \begin{aligned} l_1 &= l\sqrt{\frac{E_1}{E}}\\ n_1 &= n\frac{E}{E_1} \end{aligned} \right\} \tag{32} \]

These relations are easily obtained from equations (25) and (26).

6. Springs with a built-up main leaf.

To reinforce the leaves of the spring to which the shackles are attached, several (usually not more than two) additional leaves with rectangular ends are sometimes laid on the upper main leaf. The first added, sub-main leaf is usually made so long that it partially embraces the eyes. In this way a combined spring, or a spring with a built-up main leaf, is obtained (Fig. 7). The expediency of such a device is clarified by the following considerations. Fracture of the main leaf, upon which the entire load acts directly, is obviously considerably more dangerous than fracture of any of the intermediate leaves. Meanwhile, in addition to the bending moments distributed among all the leaves, the main leaf bears entirely the tensile forces of the oblique shackles and, in the section between the eye and the first sub-main leaf, is subjected to considerable shearing forces.

Fig. 7.

Moreover, bending the eyes, which is an operation more difficult than bending the leaf to a template, requires more intense heating of its ends. Increased heating, however, does not pass without trace for the structure of the metal. Finally, under accidental lateral loads on the spring, the main leaf is subjected to twisting.

Both (or three) main leaves usually have the same cross-section, lie one upon the other, and bend together as a single leaf, whose moment of inertia is equal to the sum of the moments of inertia of the component leaves.

The addition of new leaves, of course, reduces the flexibility of the spring. In order to preserve the validity of all the preceding calculation, it is necessary to choose the thickness of the main leaf and of the additional leaves so that the flexibility of the entire spring remains as before.

The moment of inertia of the cross-section of a composite main leaf must be equal to the moment of inertia of the section of the ordinary main leaf of a spring of normal construction.

Let us consider two cases:

1) The main leaf is made up of two leaves; the thickness of each of them is \(h_1\).

Condition of equal flexibility:

\[ 2\frac{bh_1^3}{12}=\frac{bh^3}{12}, \]

where \(h\) is the thickness of all the remaining leaves of the spring.

Hence:

\[ h_1=\frac{h}{\sqrt[3]{2}}=0.794\,h, \]

i.e., each of the two leaves must have a thickness:

\[ h_1 \simeq \frac{4}{5}h. \]

The section modulus of the composite main leaf:

\[ 2\frac{bh_1^2}{6}=2\frac{b\cdot(0.8h)^2}{6}=1.28\frac{bh^2}{6}; \]

i.e., the composite main leaf is 28% stronger than a simple one.

2) The main leaf is made up of three leaves, of thickness \(h_2\).

Condition of equal flexibility:

\[ 3\frac{bh_2^3}{12}=\frac{bh^3}{12}, \]

\[ h_2=\frac{h}{\sqrt[3]{3}}=0.693\,h\simeq 0.7\,h, \]

and the total section modulus:

\[ 3\cdot\frac{bh_2^2}{6}=3\frac{b\cdot(0.7h)^2}{6}\simeq 1.47\frac{bh^2}{6}. \]

It follows from this that the combined spring can be calculated in the same way as a spring of the usual normal construction. After the calculation, it is only necessary to take spring steel of thickness \(0.8h\) for a double main leaf, and \(0.7h\) for a triple one, whereby the strength of the leaf increases, in the first case by \(28\%\), and in the second by \(47\%\), with the flexibility of the entire spring unchanged.

7. Example of spring calculation.

We shall show, by a concrete example, the application of the calculation formulas derived above for determining the dimensions of a leaf spring.

Let us calculate a spring for a specified load:

\[ P = 4\ tn, \]

and the required flexibility:

\[ \Delta = 1\ cm/tn. \]

We choose the allowable bending stress:

\[ R_b = 5000\ kg/cm^2 \]

and the dimensions of the cross-section of the leaves (for trough-shaped spring steel, according to OST—28):

\[ b = 100\ mm,\quad h = 13\ mm. \]

1) Half the chord of the spring in the deformed state. From formula (25-a), with \(E = 2.1\cdot 10^6\ kg/cm^2\):

\[ l = \sqrt{\frac{PE\Delta h}{1000 R_b}} = \sqrt{\frac{4000\cdot 2.1\cdot 10^6\cdot 1\cdot 1.3}{1000\cdot 5000}} = 46.7\ cm. \]

2) Coefficient \(\varphi\). We take the deflection of the spring in the deformed state:

\[ \lambda = 5\ cm. \]

and the angle of inclination of the shackles:

\[ \alpha = 45^\circ,\quad tg\alpha = 1. \]

According to (13):

\[ \varphi = 1 + \frac{\lambda}{l}tg\alpha = 1 + \frac{5}{46.7} = 1.107. \]

3) Number of leaves. By formula (26)

\[ n = \frac{3000\,l^3\varphi}{E\Delta b h^3} = \frac{3000\cdot 46.7^3\cdot 1.107}{2.1\cdot 10^6\cdot 1\cdot 10\cdot 1.3^3} = 7.3. \]

Rounding off:

\[ n = 7. \]

4) Actual flexibility of the spring. From formula (26):

\[ \Delta=\frac{3000\,l^{3}\varphi}{Enbh^{3}} =\frac{3000\cdot 46{,}7^{3}\cdot 1{,}107}{2{,}1\cdot 10^{6}\cdot 7\cdot 10\cdot 1{,}3^{3}} =1{,}045\ \mathrm{cm/tn}. \]

5) Elastic deflection:

\[ f=P\Delta=4\cdot 1{,}045=4{,}18\ \mathrm{cm}. \]

6) Manufacturing camber of the spring:

\[ F=f+\lambda=4{,}18+5{,}0=9{,}18\ \mathrm{cm}\simeq 9{,}2\ \mathrm{cm}. \]

7) Actual bending stress according to formula (25):

\[ \sigma_{max}=\frac{PE\Delta h}{1000\,l^{2}} =\frac{4000\cdot 2{,}1\cdot 10^{6}\cdot 1{,}045\cdot 1{,}3}{1000\cdot 46{,}7^{2}} =5220\ \frac{kg}{cm^{2}}. \]

8) Force stretching the main leaf (see p. 6):

\[ H=\frac{P}{2}\tg\alpha=\frac{4000}{2}=2000\ kg. \]

9) Tensile stress in the main leaf:

\[ \sigma=\frac{H}{bh}=\frac{2000}{10\cdot 1{,}3}=154\ kg/cm^{2}. \]

10) Total normal stress in the main leaf:

\[ \sigma'_{max}=5220+154=5374\ kg/cm^{2}, \]

which may still be considered permissible.

11) Length of the straightened leaf (between the axes passing through the eyes), according to formula (25), taking the mean radius of the eyes as

\[ r=2{,}5\ \mathrm{cm} \]

\[ S=2\sqrt{l^{2}+\frac{4}{3}(\lambda-r)^{2}} =2\sqrt{46{,}7^{2}+\frac{4}{3}(5-2{,}5)^{2}} =93{,}6\ \mathrm{cm}^{1)}. \]

12) Manufacturing chord of the spring according to formula (29):

\[ 2L=2\sqrt{l^{2}+\frac{4}{3}(\lambda-r)^{2}-\frac{4}{3}(F-r)^{2}}= \]

\[ =2\sqrt{46{,}7^{2}+\frac{4}{3}(5-2{,}5)^{2}-\frac{4}{3}(9{,}2-2{,}5)^{2}} =92{,}3\ \mathrm{cm}. \]

1) The full construction length of the main leaf will be greater than the calculated value \(S\) by the amount necessary for bending the eyes.

13) The factory radius of curvature; according to (30):

\[ R=\frac{3S^{2}-4(F-r)^{2}}{24(F-r)} =\frac{3\cdot 93{,}6^{2}-4(9{,}2-2{,}5)^{2}}{24\cdot(9{,}2-2{,}5)} =162\ \text{cm}. \]

14) The angle between the extreme radii of the free spring, according to (31):

\[ 2\varphi_{0}=57{,}4\,\frac{S}{R} =57{,}4\cdot\frac{93{,}6}{162} =33{,}1^\circ. \]

15) Lengths of the individual leaves in the straightened state. From the obvious relations:

\[ S_{1}=S\cdot\frac{n-1}{n}, \]

\[ S_{2}=S\cdot\frac{n-2}{n}, \]

\[ \ldots\ldots \]

\[ S_{n-1}=S\cdot\frac{1}{n}, \]

where \(S_{1}\) is the length of the first underspring leaf,
\(S_{n-1}\) is the length of the smallest leaf. In our case:

\[ S_{1}=93{,}6\cdot\frac{7-1}{7}=80{,}4\ \text{cm}, \]

\[ S_{2}=93{,}6\cdot\frac{7-2}{7}=66{,}9\ \text{cm}, \]

\[ \ldots\ldots\ldots \]

\[ S_{6}=93{,}6\cdot\frac{1}{7}=13{,}4\ \text{cm}. \]

Let us check the deformation of the calculated spring by more exact methods. For this purpose we shall use formulas (17)—(20), derived above, on p. 12, and finally apply the method proposed by U. Tanaka.

a) Determination of the deformation of the spring by formulas (17)—(20).

The elastic deflection of the spring is determined by the general expression (17):

\[ f=\frac{Pl^{3}}{4EJ_{\max}}\varphi. \]

where the coefficient \(\varphi\), taking into account the curvature of the spring and the inclination of the shackles, can be determined with great accuracy by formula (18):

\[ \varphi=A+B\,\operatorname{tg}\alpha, \]

where

\[ A=\left(k+\frac{1}{k}\right)\left[\arcsin\frac{2}{k+\frac{1}{k}}-k\right], \]

\[ B=\left(k+\frac{1}{k}\right)\left[1-\left(\frac{1}{2k}-\frac{k}{2}\right)\arcsin\frac{2}{k+\frac{1}{k}}\right], \]

\[ k=\frac{\lambda}{l}. \]

In our case:

\[ k=\frac{5}{46.7}=0.107, \]

\[ A=\left(0.107+\frac{1}{0.107}\right)\left[\arcsin\frac{2}{0.107+\frac{1}{0.107}}-0.107\right]=1.002, \]

\[ B=\left(0.107+\frac{1}{0.107}\right)\left[1-\left(\frac{1}{2\cdot0.107}-\frac{0.107}{2}\right)\arcsin\frac{2}{0.107+\frac{1}{0.107}}\right]= \]

\[ =0.150, \]

\[ \varphi=1.002+0.150\cdot\operatorname{tg}45^\circ=1.152. \]

The moment of inertia of the average middle section of the spring:

\[ J_{max}=n\frac{bh^3}{12}=7\cdot\frac{10\cdot1.3^3}{12}=12.82\ \mathrm{cm}^4. \]

Deflection:

\[ f=\frac{4000\cdot46.7^3}{4\cdot2.1\cdot10^6\cdot12.82}\cdot1.152=4.35\ \mathrm{cm}, \]

which differs from the value of \(f\) obtained in the preceding calculation by 4%.

Determination of the deformation of a spring by the method of Y. Tanaka.

Y. Tanaka, in the article “Allgemeine Theorie der Blattfeder”¹, proposed a sufficiently deeply substantiated method for determining de—

deformations of a leaf spring, making it possible to take into account its curvature, pressure, and friction between the leaves, and also suitable when the individual leaves have different thicknesses and shapes.

Fig. 8.

Assuming that all the spring leaves, initially made with different radii of curvature and then assembled together and compressed in the middle by a clamp, touch one another only at their ends and at the middle (under the clamp), Tanaka considers the deformation of each of the leaves under the action of the following four forces (Fig. 8).

\(X_m\)—the force transmitted from the leaf above,

\(X_{m-1}\)—the force transmitted to the leaf below,

\[ \mu (X_m+p_0) \quad \text{and} \quad \mu (X_{m-1}+p_0) \]

the friction forces between the leaf under consideration and, respectively, the upper and lower leaves.

Here \(p_0\) is the initial pressure between the ends of the leaves, which arose when the clamp was tightened.

Without presenting here all of his derivations and reasoning, and referring the reader to the original for details, we shall give only the final expressions needed for determining the deformation of the spring.

At the same time, we note that the expressions written below have, in Tanaka’s cited work, a somewhat different form, since the notation adopted by him does not coincide with ours.

The vertical deflection of the \(n\)-th, main leaf (or, what is the same, the elastic deflection of the spring \(f\)), under the action of the inclined force \(Z\) applied at its end, is determined by the expression:

\[ \begin{aligned} f={1\over 6EJ_n}\Bigg\{& \left[\left[2l^3\cos\alpha+\left(3e+\frac{5}{2}\lambda\right)l^2\sin\alpha\right]\cdot Z\right] \\ &-(3l-a_{n-1})\cdot a_{n-1}^2 X_{n-1} \\ &-\frac{1}{2}\mu\left[3h(2l-a_{n-1})\cdot a_{n-1} +\lambda\frac{a_{n-1}^3}{l^2}(4l-a_{n-1})\right](X_{n-1}+p_0) \Bigg\} \end{aligned} \tag{33} \]

Here (Figs. 3 and 8):

\(f\)—the elastic deflection of the spring,
\(Z\)—the force in the inclined shackle,
\(X_{n-1}\)—the vertical force acting between the main leaf and the end of the second leaf,

\(p_0\)—the initial pressure between the ends of the leaves,
\(\lambda\)—the camber of the spring in the deformed state,
\(2l\)—the chord of the spring in the deformed state,
\(2a_{n-1}\)—the chord of the main leaf, under the same conditions,
\(h\)—the thickness of each leaf,

\[ J_n=\frac{bh^3}{12} \]

—the moment of inertia of the cross section of the leaf,

\(e\)—the shortest distance of the point of application of the force \(Z\) from the neutral line of the main leaf, measured at the end of the leaf,
\(n\)—the number of leaves of the spring,
\(\mu\)—the coefficient of friction between the leaves.

To compute the deflection \(f\) by this formula it is necessary to know the pressure between the leaves \(X_{n-1}\). The quantity \(X_{n-1}\) is determined from the simultaneous solution of the following system of linear equations:

\[ \left. \begin{aligned} \alpha_n Z-\beta_n X_{n-1}+\gamma_n X_{n-2}&=-\xi_n p_0,\\ \alpha_{n-1}X_{n-1}-\beta_{n-1}X_{n-2}+\gamma_{n-1}X_{n-3}&=-\xi_{n-1}p_0,\\ &\ldots\\ &\ldots\\ \alpha_2 X_2-\beta_2 X_1&=-\xi_2 p_0 \end{aligned} \right\} \tag{34} \]

In ordinary spring designs all leaves have the same thickness and shape (except for the main leaf), and each leaf is shorter than the preceding one by the same amount \(2a\) (Fig. 8).

In this case, for the main leaf:

\[ \left. \begin{aligned} \alpha_n&=(2n+1)(n-1)^2\cos\alpha+\frac{1}{a}\left\{3e+\frac{\lambda}{2n^2}\left[6n^2-(n-1)^2\right]\right\}\\ &\quad\cdot(n-1)^2\sin^2\alpha,\\ \beta_n&=4(n-1)^3+\frac{3}{2}\mu\,\frac{\lambda (n-1)^4}{a n^2}+6F,\\ \gamma_n&=(2n-1)(n-2)^2+\frac{3}{2}\mu\,\frac{h}{a}\,n(n-2),\\ \xi_n&=-\frac{3}{2}\frac{\mu}{a}\left[\frac{\lambda (n-1)^4}{n^2}-hn(n-2)\right]+6N, \end{aligned} \right\} \tag{35} \]

and for all the remaining leaves:

\[ \left. \begin{aligned} \alpha_m&=(2m+1)(m-1)-\frac{3}{2}\mu\,\frac{h}{a}(m-1)^2,\\ \beta_m&=4(m-1)^3+6F,\\ \gamma_m&=(2m-1)(m-2)^2+\frac{3}{2}\mu\,\frac{h}{a}\,m(m-2),\\ \xi_m&=-\frac{3}{2}\mu\,\frac{h}{a}+6N. \end{aligned} \right\} \tag{36} \]

Here \(m\) is the number of each leaf, the smallest leaf of the spring being taken as the first \((m=1)\).

Coefficients:

\[ \left. \begin{aligned} F&=-\frac{1}{(1-\lambda')^{2}} \left[\frac{\lambda'}{(1-\lambda')}+\frac{1}{2}\mu\frac{h}{a}\right] \left[\lambda'\ln\lambda'+(1-\lambda')-\frac{1}{2}(1-\lambda')^{2}\right] +\frac{1}{6},\\[6pt] N&=\frac{1}{2}\mu\frac{h}{a}\frac{1}{(1-\lambda')^{2}} \left[\lambda'\ln\lambda'+(1-\lambda')-\frac{1}{2}(1-\lambda')^{2}\right]. \end{aligned} \right\} \tag{37} \]

In these expressions the coefficient \(\lambda'\) takes into account the shape of the ends of the spring leaves. According to Fig. 9:

\[ \lambda'=\frac{b'}{b}. \tag{38} \]

Fig. 9.

In our approximate calculation the force in the shackle is:

\[ Z=\frac{P}{2\operatorname{Cos}\alpha}=\frac{4000}{2\operatorname{Cos}45^\circ}=2830\ \mathrm{kg}. \]

number of leaves:

\[ n=7, \]

deflection of the spring after deformation:

\[ \lambda=5\ \mathrm{cm}, \]

thickness of the leaves:

\[ h=1.3\ \mathrm{cm}, \]

moment of inertia of the section of each leaf:

\[ J_n=\frac{bh^{3}}{12}=\frac{10\cdot 1.3^{3}}{12}=1.83\ \mathrm{cm}^{4}. \]

Moreover:

\[ e\simeq \frac{h}{2}=0.7\ \mathrm{cm}, \]

\[ a=\frac{l}{n}=\frac{46.7}{7}=6.67\ \mathrm{cm}. \]

To ascertain the influence of the shape of the ends of the spring leaves on the magnitude of its deformation, let us consider two limiting cases: 1) the leaves are tapered to a triangle:

\[ b'=0,\quad \lambda'=0. \]

2) the plates are cut at a right angle:

\[ b' = b,\quad \lambda' = 1. \]

In this case, to simplify the calculations, we shall neglect the friction between the plates, i.e., assume

\[ \mu = 0. \]

First case \(\lambda' = 0\).

From (37):

\[ F = \frac{1}{6}, \]

\[ N = 0. \]

For the root plate, by formulas (35):

\[ \alpha_7=(2\cdot 7+1)(7-1)^2 \operatorname{Cos}45^\circ+ \]

\[ +\frac{1}{6{,}67}\left[3\cdot 0{,}7+\frac{5}{2\cdot 7^2}(6\cdot 7^2-6^2)\right](7-1)^2\operatorname{Sin}45^\circ=440{,}4, \]

\[ \beta_7=4(7-1)^3+6\cdot\frac{1}{6}=865, \]

\[ \gamma_7=(2\cdot 7-1)(7-2)^2=325, \]

\[ \xi_7=0. \]

For the sub-root plate \((m=6)\), by formulas (36):

\[ \alpha_6=(2\cdot 6+1)(6-1)^2=325, \]

\[ \beta_6=4(6-1)^3+6\cdot\frac{1}{6}=501, \]

\[ \gamma_6=(2\cdot 6-1)(6-2)^2=176, \]

\[ \xi_6=0. \]

Similarly, for all the remaining plates \((m=5 \div 2)\), we obtain:

\[ \begin{aligned} &\alpha_5=176, && \beta_5=257, && \gamma_5=81,\\ &\alpha_4=81, && \beta_4=109, && \gamma_4=28,\\ &\alpha_3=28, && \beta_3=33, && \gamma_3=5,\\ &\alpha_2=5, && \beta_2=5, && \gamma_1=0, \end{aligned} \]

\[ \xi_5=\xi_4=\xi_3=\xi_2=0. \]

We form the system of equations (34):

\[ \begin{aligned} 440{,}4\,Z - 865\,X_6 + 325\,X_5 &= 0,\\ 325\,X_6 - 501\,X_5 + 176\,X_4 &= 0,\\ 176\,X_5 - 257\,X_4 + 81\,X_3 &= 0,\\ 81\,X_4 - 109\,X_3 + 28\,X_2 &= 0,\\ 28\,X_3 - 33\,X_2 + 5\,X_1 &= 0,\\ 5\,X_2 - 5\,X_1 &= 0. \end{aligned} \]

The joint solution of these equations gives:

\[ X_6=0.816\quad Z=2310\ kg. \]

Half the chord of the root leaf:

\[ a_{n-1}=a_6=l-a=46.7-6.67\cong 40.0\ cm. \]

The elastic deformation of the spring according to formula (45):

\[ \begin{aligned} f={}&\frac{1}{6\cdot 2.1\cdot 10^6\cdot 1.83} \Bigl\{ \bigl[ 2\cdot 46.7^3\operatorname{Cos}45^\circ \\ &+\left(3\cdot 0.7+\frac{5}{2}\cdot 5\right)\cdot 46.7^2\cdot \operatorname{Sin}45^\circ \bigr]\cdot 2830 \\ &-\left(3\cdot 46.7-40.0\right)\cdot 40.0^2\cdot 2310 \Bigr\} =4.40\ cm. \end{aligned} \]

Second case \(\lambda'=1\).

All coefficients \(\alpha\) and \(\gamma\), as independent of \(\lambda'\), retain their previous values, calculated above, in the first case \((\lambda'=0)\). The coefficient \(F\), determined by formula (37), for \(\lambda'=1\) receives an indeterminate expression:

\[ F=\frac{0}{0}. \]

Resolving the indeterminacy by the usual method of differential calculus gives the actual value:

\[ F=0. \]

Now, according to formulas (35) and (36),

\[ \begin{aligned} \beta_7&=864,\\ \beta_6&=500,\\ \beta_5&=256,\\ \beta_4&=108,\\ \beta_3&=32,\\ \beta_2&=4. \end{aligned} \]

Moreover, as in the preceding case:

\[ N=0, \]

\[ \xi_7=\xi_6=\xi_5=\xi_4=\xi_3=\xi_2=0. \]

Let us form the system of equations (34):

\[ \begin{aligned} 440.0\,Z-864\,X_6+325\,X_5&=0,\\ 325\,X_6-500\,X_5+176\,X_4&=0,\\ 176\,X_5-256\,X_4+81\,X_3&=0,\\ 81\,X_4-108\,X_3+28\,X_2&=0,\\ 28\,X_3-32\,X_2+5\,X_1&=0,\\ 5\,X_2-4\,X_1&=0. \end{aligned} \]

Hence:

\[ X_6 = 0.826 \quad Z = 2340\ \mathrm{kg}. \]

The elastic deformation of the spring according to formula (33):

\[ f = \frac{1}{6 \cdot 2.1 \cdot 10^6 \cdot 1.83} \left\{ \left[ 2 \cdot 46.7^3 \cos 45^\circ + \left(3 \cdot 0.7 + \frac{5.5}{2}\right) \cdot 46.7^2 \sin 45^\circ \right] \cdot 2830 - (3 \cdot 46.7 - 40.0) \cdot 40.0^2 \cdot 2340 \right\} = 4.20\ \mathrm{cm}. \]

To assess the degree of accuracy of the above methods for determining the deformation of a leaf spring, let us compare the values of the elastic deflection \(f\) calculated in the example. On the basis of the approximate formulas (12) and (13), it was obtained

\[ f = 4.18\ \mathrm{cm}. \]

By the more exact formulas (17)—(20):

\[ f = 4.35\ \mathrm{cm}. \]

Finally, by Tanaka’s method, for sharp ends of the leaves (cut according to a triangle):

\[ f = 4.40\ \mathrm{cm} \]

and for rectangular ends of the leaves:

\[ f = 4.20\ \mathrm{cm}. \]

The actual value of the deflection will be intermediate between \(4.20\) and \(4.40\ \mathrm{cm}\), since in practice the ends of the leaves are usually cut according to a trapezoid.

On the basis of the summary given, the following conclusions may be drawn:

1) The shape of the ends of the spring leaves does not exert a substantial influence on the magnitude of the elastic deformation.

2) The approximate formulas (12) and (13) possess accuracy sufficient for ordinary practical cases.

3) If it is necessary to obtain a more exact value for \(f\), formulas (17)—(20) may be used.

4) Tanaka’s method, which requires lengthy computations, has no special advantages for practical purposes.

8. Friction in the spring.

In the example calculation set forth above, the influence of frictional forces between the leaves was not taken into account at all. In Tanaka’s formulas, for

for the simplification of the calculations, the terms containing the friction coefficient \(\mu\) were discarded.

In the absence of friction, the elastic deformation of the spring should vary in proportion to each change in the load acting on it. In reality, however, owing to friction between the leaves, the magnitude of the deformation depends not only on the magnitude of the load, but also on the direction of change of the latter.

When the load is gradually increased from \(0\) to \(P\), part of the force \(P\) is balanced by frictional forces, and the spring obtains a deformation \(f_1\) smaller than the value \(f\) determined above. The ratio

\[ \varepsilon = \frac{f - f_1}{f}, \]

may be called the coefficient of frictional resistance in the spring. Hence:

\[ f_1 = f(1 - \varepsilon). \]

Conversely, when the load is gradually decreased from some value \(P_{\max}\) to \(P\), part of the internal elastic forces of the deformed spring is balanced by friction between the leaves, and the elastic deflection arrow is:

\[ f_1 = f(1 + \varepsilon). \]

The coefficient \(\varepsilon\) increases with an increase in the coefficient of friction \(\mu\), the initial pressure between the leaves \(p_0\), the thickness \(h\), and the number of leaves \(n\), and decreases with an increase in the load \(P\) and the spring length \(2l\).

U. Tanaka gives the following simplified formula:

\[ \varepsilon = \frac{\mu}{2a}\left[h + \lambda\left(1 - \frac{8}{3n}\right)\right]\left(1 + \frac{p_0}{P}\right), \]

and also indicates a less exact expression obtained by G. Marié:

\[ \varepsilon = \frac{\mu (n - 1)h}{l}. \]

To calculate the coefficient of frictional resistance \(\mu\) by these formulas, it is necessary to know the value of the coefficient of friction \(\mu\) between the leaves.

Depending on the quality of the surface of the spring leaves and the lubrication between them, the coefficient \(\mu\) varies within very wide limits. Thus, for example, according to Herdner’s data¹ one may consider—

take the coefficient of friction as \(\mu = 0.80\) for new springs and \(\mu = 0.20\) for springs with lubrication between the leaves. In his approximate calculations, U. Tanaka assumes:

\[ \mu = 0.30 \div 0.45. \]

The impossibility of a sufficiently accurate estimate of the value of \(\mu\) and of the initial pressure \(p_0\) reduces the reliability of all calculations that take into account the friction forces in the spring.

When calculating the springs of parachute devices, the picture becomes still more complicated, since, in addition to the friction between the leaves, the frictional resistances in all the other links of the parachute mechanism must be taken into account. An analytical calculation of these resistances is very complicated and does not guarantee even minimal accuracy.

Only further experimental investigations in this direction can yield reliable results characterizing the magnitude of the harmful resistances in the parachute mechanism.

II. DETERMINATION OF THE SPRING RESPONSE TIME.

A description of the designs and operating principles of the various types of parachute devices used in practice is not part of the task of our work. We shall note only the most essential points necessary for the subsequent exposition.

The action of most parachute mechanisms is based on the use of the potential energy of a deformed spring or leaf spring.

During normal operation of the hoisting device, the spring is constantly loaded by a force \(P\), as a rule considerably smaller than the full weight of the load being lifted. This force, required in calculating the spring for strength, is usually taken as equal to from 0.8 to 0.9 of the weight of the empty cage \(^{1}\).

The magnitude of the load \(P\) is limited by special stop devices fastened in the frame of the cage and not allowing deformations of the spring greater than the calculated elastic deflection \(f\).

A force equal to the difference between the weight of the loaded cage and the force \(P\) is transmitted to the hoisting rope already bypassing the spring.

At the moment of rupture of the hoisting rope, the spring is relieved of the load, begins to move into its free position, and sets the parachute mechanism in motion.

After a certain interval of time, necessary for the corresponding displacement of all the links of the parachute mechanism, the brake shoes press against the guides. Only from this

\(^{1}\) See, for example: Prof. Ternigorev. Mine Haulage. 1929.

from that moment a braking process will begin, consisting in the conversion of the kinetic energy of the falling cage into the work of friction and cutting of the material of the guides.

The magnitude of the time interval $\tau$, elapsing from the moment of rupture of the rope to the moment of contact of the knives of the parachute device with the guides, is called the response time of the spring.

The determination of the response time should be considered an essential point in designing the parachute mechanism, since its magnitude determines the duration of the free, uniformly accelerated fall of the broken-away cage and the distance traversed by it before braking.

As the response time increases, the velocity and the live force possessed by the cage at the beginning of braking increase, and the dynamic load caused by the cage being retarded in the guides and crossheads also increases.

1. Period of natural oscillations.

It was noted above that at the moment of rupture of the rope, the deformed spring begins to “straighten,” and its potential energy is expended in setting the parachute mechanism in motion.

Let us note that the term “straighten” cannot be applied literally in the case of a spring of ordinary normal construction, since when the load is decreased such a spring does not straighten, but bends still more. Its curvature increases as the load decreases.

Nevertheless, the term “straighten” describes the phenomenon much more simply and clearly than indicating a change in the magnitude of the radius of curvature.

If the spring were completely free and not connected with the parachute mechanism, then, approaching with gradually increasing velocity its mean position, owing to the inertia of its mass it would pass beyond this mean, free position and would begin to bend in the opposite direction. Having reached the maximum curvature, the spring would again begin to straighten, and so on. The spring would perform free harmonic oscillations.

In the complete absence of any resistances, the amplitude of the oscillations would remain unchanged and equal to $f$.

In reality, however, the spring connected with the parachute mechanism will not be able to make even one complete oscillation, since, before yet reaching its mean position, it will press the braking knives against the guides.

In determining the magnitude of the interval of time elapsed from the moment of rupture of the rope to the pressing of the knives against the guides (during this time the spring will perform part of one complete oscillation),

it is necessary to compute the period of free oscillations of the spring and of the parachute mechanism connected with it.

To derive the basic equations of the oscillatory motions of the leaf spring, we shall replace it by a simple helical spring carrying, at its free end, a certain mass \(m\) (Fig. 10). Such a replacement will not introduce errors into the subsequent conclusions if the values of the elastic characteristics of the spring and the mass \(m\) are chosen in accordance with their actual values in the spring.

Fig. 10.

The flexibility of the helical spring must be equal to the flexibility \(\Delta\) of the leaf spring. The mass \(m\), called the reduced mass, must take into account not only the magnitude of the masses of all moving links of the parachute mechanism, but also the degree of their participation in the oscillatory motion. The method for calculating it will be given below.

In Fig. 10 the mean, free position of the mass \(m\) is denoted by the line \(0—0\). Under the action of the applied force \(P\), the spring will be compressed by the amount \(f\), and the mass \(m\) will occupy the position \(a—a\).¹

In this position the external load \(P\) is balanced by the elastic force of the spring, equal to \(cf\), where \(c\) is the so-called stiffness of the spring, i.e., the force required to deform it by 1 centimeter.

Obviously:

\[ c=\frac{1000}{\Delta}\ kg/cm, \tag{39} \]

if \(\Delta\), as above, is measured in centimeters per ton.

Upon the sudden cessation of the action of the force \(P\), the mass \(m\), under the action of the straightening spring, will begin to move downward. The deformation of the spring decreases and at some instant of time \(t\) is equal to \(x\). The quantity \(x\) simultaneously represents the vertical displacement of the mass from its equilibrium position \(0—0\).

At the instant of time \(t\), the mass \(m\) moves with velocity:

\[ v=\frac{dx}{dt}, \]

and with acceleration:

\[ j=\frac{d^2x}{dt^2}. \]

The inertial force of the mass is equal to:

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

where the minus sign indicates that the direction of the force is always opposite to the direction of the acceleration.

To set up the differential equation of motion, we shall apply d’Alembert’s principle.

On the basis of this principle, the equation of motion may be obtained analogously to the equations of statics. It is only necessary, in addition to the external and elastic forces, to take into account the inertial forces of the moving masses. In our case:

\[ m \frac{d^2 x}{dt^2} + cx = 0, \tag{40} \]

or

\[ \frac{d^2 x}{dt^2} + \frac{c}{m} x = 0. \]

The solution of this differential equation gives:

\[ x = A \operatorname{Sin} \omega t + B \operatorname{Cos} \omega t, \tag{41} \]

where:

\[ \omega = \sqrt{\frac{c}{m}}. \tag{42} \]

The arbitrary constants \(A\) and \(B\) must be determined in such a way that the initial conditions of the motion are satisfied.

Taking as the initial instant of time the instant of rupture of the rope, we have:

\[ \text{for } t = 0,\quad x = f,\quad \frac{dx}{dt} = 0. \]

Under these conditions, from equation (53):

\[ A = 0,\quad B = f. \]

Substituting these values of the constants into the general integral (41):

\[ x = f \operatorname{Cos} \omega t, \tag{43} \]

or

\[ x = f \operatorname{Cos} \sqrt{\frac{c}{m}}\, t. \tag{44} \]

From expression (43) it is evident that from adding to \(t\) the quantity:

\[ T = \frac{2\pi}{\omega}, \tag{45} \]

the values of \(x\) do not change. Over the time interval \(T\) the mass \(m\) executes one complete oscillation, returns to its initial position, and acquires its initial velocity \(v=0\). The quantity \(T\) is called the period of the system’s natural oscillations. According to (42) and (45):

\[ T=2\pi\sqrt{\frac{m}{c}}, \tag{46} \]

or, on the basis of definition (51):

\[ T=2\pi\sqrt{\frac{m\Delta}{1000}}. \tag{47} \]

The last expressions show that the period of oscillation (in the absence of resistance) does not depend on the amplitude of oscillation \(f\) and is fully determined by the magnitude of the mass \(m\) and by the elastic properties of the spring.

Let us imagine a vector \(OK\) (Fig. 10), having a length equal to the amplitude of oscillation \(f\) and rotating with constant angular velocity \(\omega\) about point \(0\).

During the time \(t\) the vector turns through the angle \(\omega t\) from the vertical. From equation (43) it is evident that each instantaneous position of the oscillating mass \(m\) is represented by the projection of the end of the vector \(OK\) onto the vertical. In the absence of harmful resistances, the free harmonic oscillation of the mass can be represented by a sinusoid in the coordinates \(x\) and \(t\).

The quantity \(\omega\), determined by expression (42), is called the angular velocity of harmonic oscillation.

2. Reduction of the masses of the parachute mechanism.

To compute the period of oscillation \(T\) of the spring by formula (47), it is necessary to know the magnitude of the mass \(m\) participating in the oscillatory motion. In the case of a complex oscillating system, such as we have in the parachute mechanism, the masses of all moving parts should be replaced by one reduced mass, dynamically equivalent.

Fig. 11.

Using Rayleigh’s method, we shall call the reduced mass \(m\) of the entire parachute mechanism such a fictitious mass, applied at the middle of the spring, whose kinetic energy at every instant of time is equal to the kinetic energy of the actual oscillating masses¹.

We shall show the general method for determining the magnitude of the reduced mass on a concrete example. Let us consider one of the designs of parachute mechanisms, shown schematically in Fig. 11. Assuming that the arrangement of the mechanism is sufficiently clear from the drawing, we shall note only the following.

The ends of the spring \(A\) and the supports of the shafts \(E\) are fixed in the car body. Under the action of the tension of the hoisting rope, attached to the spring yoke, the latter bends somewhat upward. In this case the brake shoes \(F\) move away from the guides.

When the rope breaks, the spring “straightens,” the yoke moves downward, and the shoes \(F\), turning together with the shafts \(E\), cut into the guides. The car begins gradually to stop.

In our case (Fig. 11) the total reduced oscillating mass is composed of the following parts:

1) The reduced mass of the spring itself—\(m_{A\,red.}\)
2) The actual mass of the spring yoke and the crossbar connected with it \(B\)—\(m_B\).
3) The actual mass of two vertical earrings \(C\)—\(2m_C\).
4) The reduced mass of two rotating levers \(D\)—\(2m_{D\,red.}\)
5) The reduced mass of two shafts \(E\)—\(2m_{E\,red.}\)
6) The reduced mass of four shoes \(F\)—\(4m_{F\,red.}\)

Let us consider all these points separately.

1) Determination of the reduced mass of the spring \(m_{A\,red.}\)

By the reduced or transformed mass of the spring we mean such a fictitious mass \(m_{A\,red.}\), applied at the middle of the spring, whose live force in oscillatory motion is equal to the live force (kinetic energy) of the actual spring.

We consider the spring as a straight beam of constant height \(h\) and consisting in plan of two triangles (Fig. 12). The maximum width of the beam:

\[ B = nb, \]

where, according to the notation adopted above, \(b\) is the width of each spring leaf, \(n\) is the number of leaves.

Fig. 12.

At the middle of the beam let us apply an arbitrary vertical force \(Q\). In this case the beam will bend according to some elastic curve, whose equation may be written in the general form:

\[ y=f(x), \]

For an arbitrary cross section \(x\) we have:

\[ M_x=\frac{Q}{2}x, \]

\[ J_x=B\frac{x}{l}\frac{h^3}{12}=J_{\max}\frac{x}{l}, \]

where the maximum moment of inertia (of the middle section) is:

\[ J_{\max}=\frac{Bh^3}{12}. \]

The differential equation of the elastic curve:

\[ EJ_xy''=M_x=\frac{Q}{2}x, \]

\[ EJ_{\max}\cdot\frac{x}{l}y''=\frac{Q}{2}x, \]

or:

\[ EJ_{\max}\cdot y''=\frac{Ql}{2}. \]

Integrating twice:

\[ EJ_{\max}\cdot y'=\frac{Ql}{2}x+C_1, \]

\[ EJ_{\max}\cdot y=\frac{Qlx^2}{4}+C_1x+C_2, \]

with:

\[ x=0,\ y=0,\ C_2=0; \]

\[ x=l,\ y'=0,\ C_1=-\frac{Ql^2}{2}. \]

Now:

\[ EJ_{\max}\cdot y=\frac{Qlx^2}{4}-\frac{Ql^2x}{2}, \]

or, changing signs (taking the downward direction as positive):

\[ EJ_{\max}\cdot y=\frac{Ql^2x}{2}-\frac{Qlx^2}{4}. \tag{48} \]

The maximum deflection at the middle, for \(x=l\):

\[ EJ_{max}\, y_{max}=\frac{Ql^3}{2}-\frac{Ql^3}{4}=\frac{Ql^3}{4}; \]

\[ y_{max}=\frac{Ql^3}{4EJ_{max}}. \]

From (48) the general equation of the elastic curve is:

\[ y=\frac{Q\frac{l^2x}{2}-Q\frac{lx^2}{4}}{EJ_{max}}, \]

or

\[ y=y_{max}\frac{\left(Q\frac{l^2x}{2}-Q\frac{lx^2}{4}\right)\cdot 4EJ_{max}}{EJ_{max}\cdot Ql^3}. \]

Hence, after cancellations:

\[ y=y_{max}\frac{2lx-x^2}{l^2}. \tag{49} \]

We assume that the deflection curve of the beam during vibration has the same form as the elastic curve under static loading\(^1\). Then the velocities of any two points of the spring, at each given instant, are related to one another in the same way as the corresponding ordinates \(y\).

If \(v_{max}\) is the instantaneous vibration velocity of the middle section of the spring, then the vibration velocity of a section located at a distance \(x\) from the support is equal to:

\[ v=v_{max}\frac{y}{y_{max}}=v_{max}\frac{2lx-x^2}{l^2}. \tag{50} \]

By two infinitely close transverse sections \(x\) and \(x+dx\), we cut out from the spring an elementary volume \(dV\). Obviously:

\[ dV=B\frac{x}{l}h\,dx. \]

The separated elementary volume has mass:

\[ dm=\frac{\gamma}{g}\,dV=\frac{\gamma}{g}B\frac{x}{l}h\,dx, \]

\(^1\) S. P. Timoshenko, Theory of Vibrations in Engineering, 1931.

and kinetic energy:

\[ dm \cdot \frac{v^2}{2}=B\,\frac{h}{l}\,\frac{\gamma}{g}\,\frac{v^2}{2}\,x\,dx. \]

Here \(\gamma\) is the specific weight of the material, \(g\) is the acceleration due to gravity.

The kinetic energy of the entire beam must be equal to the kinetic energy of the reduced mass \(m_{A\ \mathrm{red.}}\), applied at the middle of the spring and moving with velocity \(v_{max}\). Therefore:

\[ m_{A\ \mathrm{red.}}\cdot \frac{v^2_{max}}{2} = 2\int_0^l \frac{v^2}{2}\,dm, \]

or

\[ m_{A\ \mathrm{red.}}\cdot \frac{v^2_{max}}{2} = 2\int_0^l \frac{v^2}{2}\,\frac{Bh}{l}\,\frac{\gamma}{g}\,x\,dx. \]

Substituting here the values for \(v\) from formula (50) and taking the constant quantities outside the integral sign, after reducing by \(v^2_{max}\) we obtain:

\[ m_{A\ \mathrm{red.}} = \frac{2Bh\gamma}{l^5 g} \int_0^l (2lx-x^2)^2\cdot x\,dx. \]

Integration and substitution of the limits gives:

\[ m_{A\ \mathrm{red.}}=\frac{11}{15}\,\frac{Bh\gamma l}{g}. \tag{51} \]

But the weight of the entire spring is:

\[ G_A=2\,\frac{Bl}{2}\,h\gamma=Blh\gamma=nbhl\gamma. \tag{52} \]

Consequently:

\[ m_{A\ \mathrm{red.}}=\frac{11}{15}\,\frac{G_A}{g}, \tag{53} \]

i.e. the reduced mass of the spring is \(11/15\) of its actual mass.

2, 3) Determination of the masses \(m_B\) and \(m_C\).

The spring clamp, the crosspiece \(B\), and the earrings \(C\) move during oscillation with the same velocity \(v_{max}\). Their reduced mass is equal to the actual mass:

\[ m_B+m_C=\frac{G_B+2G_C}{g}. \]

The weights of these parts \(G_B\) and \(G_C\) are calculated directly from the dimensions of the drawing or, for an already constructed parachute mechanism, may be determined by weighing.

4) Reduction of the mass of levers \(D\).

Let us denote (Fig. 13):

\(J_D\)—the moment of inertia of the mass of the lever with respect to the axis of rotation \(E—E\),

\(\omega\)—the instantaneous angular velocity of the lever,

\(r_D\)—the radius (length) of the lever.

With the lever \(D\) and shackle \(C\) in a mutually perpendicular position, point \(a\) has a velocity \(v_{max}\) equal to the velocity of the middle point of the spring (the point of reduction).

Fig. 13.

The reduced mass of the lever \(m_{D\,red.}\) is determined from the necessary equality:

\[ m_{D\,red.}\frac{v^2_{max}}{2}=\frac{1}{2}J_D\omega^2, \]

or:

\[ m_{D\,red.}\frac{r_D^2\omega^2}{2}=\frac{1}{2}J_D\omega^2, \]

since the velocity of point \(a\) is:

\[ v_{max}=r_D\omega. \]

Hence:

\[ m_{D\,red}=\frac{J_D}{r_D^2}. \tag{54} \]

The moment of inertia of the mass of the lever \(J_D\) can be determined from the dimensions of the drawing, using the usual formulas for moments of inertia of masses\(^1\). For manufactured levers, the quantity \(J_D\) may be found experimentally. We suspend the lever on a prism, on the axis \(EE\), and measure the time \(T\), in seconds, required for one complete (double) oscillation. Then, on the basis of the theory of the physical pendulum:

\[ J_D=\frac{T^2}{4\pi^2}\cdot G_D e\ kg\,cm\,sec^2 \tag{55} \]

where \(G_D\) is the weight of the lever in kg, and \(e\) is the distance of the center of gravity of the shackle from the axis of rotation \(E—E\), in centimeters\(^1\).

5) Reduction of the mass of shaft \(E\).

Each shaft \(E\) is a straight circular cylinder of radius \(R\). The moment of inertia of the shaft with respect to the axis of rotation is:

\[ J_E=\frac{G_E R^2}{2g}, \tag{56} \]

Fig. 14.

\(^1\) Hütte, vol. I, 13th ed., p. 248.
\(^1\) See, for example, G. Dubbel, Handbook of Mechanical Engineering, vol. I.

where \(G_E\) is the weight of the shaft. The reduced mass (analogously to the reduction of the lever mass):

\[ m_{E\,red}=\frac{J_E}{r_D^2}. \tag{57} \]

Here \(r_D\), as above, denotes the length of the lever \(D\).

b) Reduction of the masses of the cams \(F\).

The moment of inertia \(J_F\) of each knife relative to the axis \(E—E\) is determined by calculation from the dimensions in the drawing, or by rocking on a prism, analogously to what was stated above with respect to the levers \(D\).

The reduced mass of each knife (cam):

\[ m_{F\,red}=\frac{J_F}{r_D^2}, \]

where \(r_D\) has the previous meaning.

Having determined the reduced masses of all the links of the parachute mechanism, one can calculate the period of natural oscillations of the spring \(T\) by formula (59):

\[ T=2\pi\sqrt{\frac{m\Delta}{1000}}. \]

For the scheme of the parachute mechanism considered by us (Fig. 11), it is necessary to substitute here, in place of \(m\), the quantity:

\[ m=m_{A\,red}+m_B+2m_C+2m_{D\,red}+2m_{E\,red}+4m_{F\,red}. \]

The flexibility of the spring \(\Delta\) in formula (59) is measured in centimeters per ton.

3. Determination of the response time of the spring.

In the free state of the spring, its root leaf is bent along the arc of a circle \(AOC\) (Fig. 15). During normal operation of the cage the root leaf has the form \(ADC\) (if, for simplicity of the drawing, the displacements of the ends of the spring during its deformation are neglected). At the same time the spring has the deflection:

\[ \lambda=F-f \]

and the elastic deformation (sagitta of deflection)

\[ f=P\Delta. \]

Fig. 15.

At the moment the rope breaks, the spring begins to oscillate from position \(ADC\) to \(AEC\), through its zero (free) position \(AOC\).

In the absence of resistance, the amplitude of oscillation is equal to \(f\), i.e.

\[ DO = OE = f. \]

The motion of the spring, which began at the moment when the action of the force \(P\) ceased, will continue until the mid-point \(D\) of the spring reaches a certain point \(F\), lying between \(D\) and \(O\). At that moment the brake shoes will press on the conductors and the motion of the spring will end.

The interval of time necessary for the mid-section of the spring to traverse the path

\[ DF = z, \]

is evidently the required response time \(\tau\) (see p. 32).

In position \(F\), the spring must still have a certain definite tightening \(P_0\), necessary to create sufficient pressure of the brake shoes (cams) on the conductors.

On the basis of the law of proportionality between load and deformation, one may write:

\[ \frac{P_0}{P} = \frac{f - z}{f}. \tag{58} \]

This relation makes it possible to determine the necessary working displacement of the spring \(z\) for the selected tightening \(P_0\), or conversely, the resulting \(P_0\) for the adopted \(z\).

It must always be:

\[ P_0 < P; \quad z < f. \]

The magnitude \(z\), for the selected ratios of the dimensions of the parachute mechanism, determines the shape and wedging angle of the brake cams (shoes).

Above, on p. 35, it was established that each instantaneous position of a harmonically oscillating mass is represented by the projection of the end of the rotating vector \(OK\) onto the vertical. The length of the vector is equal to the amplitude of oscillation (Fig. 16):

\[ OK = f. \]

The angular velocity of rotation of the vector, according to (45):

\[ \omega = \sqrt{\frac{c}{m}} = \sqrt{\frac{1000}{m\Delta}}. \tag{59} \]

During the response time of the spring, the vector will turn through some angle \(\varphi=\omega\tau\), and point \(D\) will arrive at \(F\).

According to the drawing:

\[ DF=z=f-f\cos\varphi. \]

Hence:

\[ \cos\varphi=\frac{f-z}{f}=1-\frac{z}{f}, \]

\[ \varphi=\arccos\left(1-\frac{z}{f}\right). \]

Fig. 3.

The response time of the spring:

\[ \tau=\frac{\varphi}{\omega}=\sqrt{\frac{m}{c}}\arccos\left(1-\frac{z}{f}\right)= \]

\[ =\sqrt{\frac{m\Delta}{1000}}\arccos\left(1-\frac{z}{f}\right). \tag{60} \]

Using expressions (47) and (58), we finally obtain

\[ \tau=\frac{T}{2\pi}\arccos\left(\frac{P_0}{P}\right), \tag{61} \]

or

\[ \tau=\sqrt{\frac{m\Delta}{1000}}\arccos\left(\frac{P_0}{P}\right). \tag{62} \]

From the formulas obtained it can be seen that reducing the working displacement of the spring \(z\) reduces the response time, but makes the operation of the parachute more abrupt.

In conclusion, we note that all the results set forth above were obtained under the assumption that there are no friction forces in the parachute mechanism. It was assumed that, from the moment the cord breaks, the spring moves according to the laws of free harmonic oscillations. In reality, frictional resistance reduces the amplitude of the oscillations. The motion becomes damped. The response time in fact turns out to be somewhat greater than that given by formulas (61) and (62).

Nevertheless, the use of the indicated formulas should be considered expedient, since the calculation of friction forces in the parachute mechanism cannot be carried out with sufficient accuracy.

Appendix.

SCHEME FOR CALCULATING A LEAF SPRING

1. Determination of the dimensions of the spring.

Notation:

$P$—design load on the spring in $kg$.

$2l$—chord of the spring in the deformed state, in $cm$,

$2L$—factory chord of the spring, in $cm$,

$S$—length of the straightened spring, in $cm$,

$f$—elastic deflection of the spring, in $cm$, under the action of the force $P$,

$\lambda$—camber of the spring in the deformed state, in $cm$,

$F = f + \lambda$—factory camber of the spring, in $cm$,

$R$—factory radius of curvature of the main leaf of the spring, in $cm$,

$r$—mean radius of the eyes of the main leaf,

$2\varphi_0$—angle formed by the extreme radii of the spring in the free state,

$n$—number of leaves of the spring,

$b$—width of the spring leaves, in $cm$,

$h$—thickness of the spring leaves, in centimeters,

$R_b$—allowable bending stress, in $kg/cm^2$,

$\alpha$—angle of inclination of the spring shackles to the vertical,

$\Delta$—flexibility of the spring, in $cm/tn$, i.e., its deflection in $cm$ under the action of a force of 1 ton.

For the calculation the following are given:

$P$—force acting on the spring, in $kg$,

$\Delta$—flexibility of the spring, in $cm/tn$ (from the condition of obtaining the necessary magnitude of deformation $f = P\Delta$, selected constructively, in accordance with the entire scheme of the parachute mechanism),

$R_b$—allowable bending stress, in $kg/cm^2$,

$b \times h$—section profile of the spring leaves (according to OST—28), (being guided by the considerations: the greater the flexibility required of the spring, the thinner the leaves chosen for its manufacture should be),

$\alpha$—angle of inclination of the shackles (the greater $\alpha$, the greater the flexibility of the given spring),

$r$—mean radius of the eyes.

A. Determination of $l$ and $n$ (Fig. 17).

1. We take the nomogram for the selected profile $b \times h$ and on the upper scale (loads) mark the design load $P$ (point $a$).

2. Through the obtained point we draw a vertical to its intersection with the line of the selected $R_b$ at point $b$.

3. Through $b$ we draw a horizontal to its intersection with the line of the selected flexibility $\Delta$ at point $c$.

1. Through \(c\) draw the vertical \(cd\) downward, into the second half of the nomogram, as far as the line of the same \(\Delta\).

2. The line \(cd\) intersects the length scale \(l\) at the point \(e\). This point

Fig. 17.

gives the approximate length of half the chord of the deformed [[unclear: word obscured]].

Choosing \(\lambda\) (the deflection of the spring in the deformed state), we calculate the coefficient \(\varphi\):

\[ \varphi=\left(1+\frac{\lambda}{l}\tg\alpha\right). \]

(For greater accuracy, the coefficient \(\varphi\) may be computed from formulas 18–20).

1. Through point \(d\) draw a horizontal line until it intersects the line of the computed \(\varphi\) at point \(f\).

2. The vertical through \(f\) gives, on the lower scale, the required number of leaves of the spring \(n\).

3. Round the obtained number to the nearest integer (upward or downward). We obtain point \(g\).

4. Through \(g\) draw a vertical line until it intersects the preceding line of \(\varphi\) at point \(i\).

5. Through \(i\) draw the horizontal \(ik\) to the preceding \(\Delta\).

6. From \(k\) draw the vertical \(kl\) to the same \(\Delta\).

7. The vertical \(kl\) cuts off on the scale of lengths the actual value of the half-chord of the deformed spring \(l\).

8. Through point \(l\) draw a horizontal line. Its intersection with the vertical \(ab\) gives the actual bending stress \(\sigma_{max}\) in the spring.

9. The maximum stress is obtained in the main leaf of the spring (owing to the additional tension in it from the horizontal component of the reaction of the inclined spring suspension):

\[ \sigma'_{max}=\sigma_{max}+\frac{P\,\operatorname{tg}\alpha}{bh}\;kg/cm^2. \]

1. Knowing \(f\) and \(\lambda\), determine the design camber of the spring:

\[ F=f+\lambda. \]

B. Determination of the design dimensions of the spring.

1. Length of the straightened spring \(^{1}\):

\[ S=2\sqrt{\,l^2+\frac{4}{3}(\lambda-r)^2\,}. \]

1. Design chord of the spring:

\[ 2L=2\sqrt{\,\frac{S^2}{4}-\frac{4}{3}(F-r)^2\,}. \]

1. Design radius of the main leaf of the spring:

\[ R=\frac{3S^2-4(F-r)^2}{24(F-r)}. \]

\(^{1}\) The full design length of the main leaf will be greater than the value \(S\), computed from this formula, by the amount necessary for bending the eyes.

Nomogram for Calculating Leaf Springs

76 × 95

Upper scale: \(P\), t
Lower scale: \(\ell\), cm

\(R_0 = 2000\ \mathrm{kg/cm^2}\)

\(\Delta = 0.5\ \mathrm{cm/t}\)

\(R_0\): 3000, 4000, 5000, 6000, 7000

\(\Delta\): 0.75, 1.00, 1.25, 1.50, 1.75, 2.00, 2.25, 2.50, 2.75, 3.00, 3.50

\(\ell\), cm: 20, 30, 40, 50, 60, 70

\(\Delta = 3.5\ \mathrm{cm/t}\)

\(\Delta\): 3.00, 2.50, 2.25, 2.00, 1.75, 1.50, 1.25, 1.00, 0.75, 0.50

\(q\): 4, 3, 2, 1.0

Bottom scale: 15, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Eng. V. Nechaev, “Calculation of Springs.”

NOMOGRAM

For calculating leaf springs

76 × 11.

\(P\)
t n

Top scale: 15, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

\(R_\mathrm{b}=2000\ \mathrm{kg/cm^2}\)
3000
4000
5000
6000
7000

\(\Delta = 0.5\ \mathrm{cm/t}\)
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.5

\(l\)
cm. 20 30 40 50 60 70

\(\Delta = 3.5\ \mathrm{cm/t}\)
3.0
2.75
2.50
2.25
2.00
1.75
1.50
1.25
1.00
0.75
0.50

\(\varphi = 1.8\)
1.5
1.2
1.0

Bottom scale: \(n\) — 15, 10, 9, 8, 7, 6, 5, 4, 3, 2

Engineer V. Nekrasov, “Calculation of Springs”

Nomogram for Calculating Leaf Springs

76 × 13

Top-left scale: \(P_b\), \(t_n\)

Top scale markings: \(10,\ 9,\ 8,\ 7,\ 6,\ 5,\ 4,\ 3,\ 2,\ 1\)

Upper diagram labels:

• \(f = 2000\ \mathrm{kg/cm^2}\)
• \(3000\)
• \(4000\)
• \(5000\)
• \(6000\)
• \(7000\)
• \(b = 0.5\ \mathrm{cm/kg}\)
• \(0.75\)
• \(1.00\)
• \(1.25\)
• \(1.50\)
• \(1.75\)
• \(2.00\)
• \(2.25\)
• \(2.50\)
• \(2.75\)
• \(3.00\)
• \(3.5\)

Left lower scale: \(l\), cm

Intermediate scale markings: \(20,\ 30,\ 40,\ 50,\ 60,\ 70\)

Lower diagram labels:

• \(b = 3.5\ \mathrm{cm/kg}\)
• \(3.00\)
• \(2.50\)
• \(2.25\)
• \(2.00\)
• \(1.75\)
• \(1.50\)
• \(1.25\)
• \(1.00\)
• \(0.75\)
• \(0.50\)
• \(\varphi = 16\)
• \(14\)
• \(12\)
• \(10\)

Bottom-left scale: \(n\)

Bottom scale markings: \(15,\ 10,\ 9,\ 8,\ 7,\ 6,\ 5,\ 4,\ 3,\ 2,\ 1\)

Engineer V. Nechaev. “Calculation of Springs”

Nomogram

for calculating leaf springs.

89 × 9.5

P
t, n

k₍B₎ = 2000 kg/cm²
3000
4000
5000
6000
7000

Δ = 0.5 cm/kg
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.50

l
cm

Δ = 3.5 cm/t
3.00
2.50
2.25
2.00
1.75
1.50
1.25
1.00
0.75
0.50

ψ = 1.8
1.5
1.0

n

Eng. V. Nechaev, “Calculation of Springs”

Nomogram for Calculating Leaf Springs

100 × 13

\(P\)
t

Top scale: \(15,\ 10,\ 9,\ 8,\ 7,\ 6,\ 5,\ 4,\ 3,\ 2,\ 1\)

Diagonal scale labels:

\(R_i = 2000\ \mathrm{kg/cm^2}\)

\(1000\)

\(4000\)

\(5000\)

\(6000\)

\(7000\)

\(\Delta = 0.5\ \mathrm{cm/t}\)

\(0.75\)

\(1.00\)

\(1.25\)

\(1.50\)

\(1.75\)

\(2.00\)

\(2.25\)

\(2.50\)

\(2.75\)

\(3.00\)

\(3.50\)

\(\ell\)
cm

Scale: \(20,\ 30,\ 40,\ 50,\ 60,\ 70\)

\(\Delta = 0.5\ \mathrm{cm/t}\)

\(3.00\)

\(2.75\)

\(2.50\)

\(2.25\)

\(2.00\)

\(1.75\)

\(1.50\)

\(1.25\)

\(1.00\)

\(0.75\)

\(0.50\)

\(\varphi = 16\)

\(20\)

\(30\)

\(40\)

Bottom scale: \(10,\ 9,\ 8,\ 7,\ 6,\ 5,\ 4,\ 3,\ 2,\ 1\)

Eng. V. Nechaev. “Calculation of Springs”

Nomogram

for calculating leaf springs

110 × 13

\(P\)
t

Top scale: 15, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

\(k = 2000\ \mathrm{kg/cm^2}\)

3000
4000
5000
6000
7000

\(\Delta = 0.5\ \mathrm{cm/kg}\)

0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.50

\(l\)
cm

Scale: 20, 30, 40, 50, 60, 70

\(\Delta = 3.5\ \mathrm{cm/kg}\)

3.0
2.75
2.50
2.25
2.00
1.75
1.50
1.25
1.00
0.75
0.50

\(\varphi = 16\)
14
12
10

Bottom scale: 15, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Eng. V. Nechaev, “Calculation of Springs”

1. The angle at the apex of the spring:

\[ 2\varphi_0 = 57.4\,\frac{S}{R}\ \text{degrees}. \]

1. The length of the remaining leaves of the spring is determined in the usual way:

\[ \text{Length of the main leaf} = S. \]

\[ \text{Length of the second leaf} = S\,\frac{n-1}{n}. \]

\[ \text{Length of the third leaf} = S\,\frac{n-2}{n},\ \text{etc.} \]

II. Determination of the response time of the spring.

(For the parachute mechanism according to Fig. 11).

Designations (except those indicated above):

\(P_0\)—preload of the spring, i.e. the spring force during the braking period (obviously \(P_0 < P\)),

\(z\)—the amount by which the spring straightens during the response period, in cm,

\(f - z\)—the remaining elastic deflection corresponding to the load \(P_0\).

\(\tau\)—response time of the spring in seconds, i.e. the interval of time elapsing from the moment the cable breaks until the knives come into contact with the guides. During this time the deflection \(f\) of the spring will decrease to the value \(f - z\).

\(G_A\)—weight of the spring itself in kilograms,

\(G_B\)—weight of cross-piece \(B\) in kg,

\(G_C\)—weight of shackle \(C\) in kg (each shackle separately),

\(J_D\)—mass moment of inertia of each lever \(D\) with respect to the axis \(E—E\), in \(kg\ cm\ sec^2\);

\(J_E\)—mass moment of inertia of shaft \(E\) in \(kg\ cm\ sec^2\),

\(J_F\)—mass moment of inertia of each knife, with respect to the axis \(E—E\), in \(kg\ cm\ sec^2\).

(The mass moments of inertia are calculated according to the formulas of Hütte, vol. I, p. 248, 13th ed. In the calculation all dimensions are taken in cm, acceleration \(g = 981\ cm/sec^2\)),

\(r_D\)—radius of lever \(D\) in centimeters, selected from design considerations.

1. Reduced mass of the spring:

\[ m_{A\,red}=\frac{11}{15}\frac{G_A}{981}\ \frac{kg\ sec^2}{cm}; \]

where the weight of the spring can be calculated by the formula:

\[ G_A \cong S \frac{n+1}{2} b h \gamma \; kg, \]

if \(S\), \(b\), and \(h\) are in centimeters, and \(\gamma\) is in \(kg/cm^3\).

1. Reduced mass of the crossmember \(B\):

\[ m_B = \frac{G_B}{981} \frac{kg\;sec^2}{cm}. \]

1. Reduced mass of each shackle \(C\):

\[ m_C = \frac{G_C}{981} \frac{kg\;sec^2}{cm}. \]

1. Reduced mass of each lever \(D\):

\[ m_{D\,red} = \frac{J_D}{r_D^2} \frac{kg\;sec^2}{cm}. \]

1. Reduced mass of each shaft \(E\):

\[ m_{E\,red} = \frac{J_E}{r_D^2} \frac{kg\;sec^2}{cm}. \]

1. Reduced mass of each knife \(F\):

\[ m_{F\,red} = \frac{J_F}{r_D^2} \frac{kg\;sec^2}{cm}. \]

1. Reduced mass of the entire system:

\[ m = m_{A\,red} + m_B + 2m_C + 2m_{D\,red} + 2m_{E\,red} + 4m_{F\,red}. \]

1. The quantities \(P_0\), \(P\), \(f\), and \(z\) are related to one another by the relation:

\[ \frac{P_0}{P} = \frac{f-z}{f}. \]

Specifying the necessary tightening of the spring \(P_0\) (of course less than \(P\)), from this one can determine \(z\), or, conversely, specifying \(z\) from the structural transmission ratios to the knives, one can determine the corresponding \(P_0\).

1. Response time of the spring:

\[ \tau = \sqrt{\frac{m\Delta}{1000}} \arccos \left(1 - \frac{z}{f}\right) \text{ seconds}, \]

or

\[ \tau = \sqrt{\frac{m\Delta}{1000}} \arccos \left(\frac{P_0}{P}\right) \text{ seconds}, \]

where \(\Delta\) is the flexibility of the spring in \(cm/tn\).

1. Timoshenko S. P. Theory of Vibrations in Engineering. 1931. ↩↩↩↩↩↩