Calculation of the Walls in Tatarinov’s Press

(With 1 table.)

S. K. Konyukhov.

Theoretical investigations of Tatarinov’s press, even in a particular form, present a series of deeply interesting problems for the engineer-technician in many respects. The novelty of the question calls forth the awakening of personal creativity, and this fact is profoundly pedagogical, especially for young engineers, in whom there is very often laid up, figuratively speaking, a reserve of chernozem strength. The beneficial role of the teacher, in our opinion, consists in directing the stream of these forces into the proper channel. He to whom these investigations appeal will not regret the time spent, since, besides being instructive, acquaintance with the problems set forth below is also engaging from a technical point of view. Yet this engaging quality is not obtained gratis: the comparative completeness of the investigation is hampered by such obstacles of a mathematical character that one need not even think of them upon a cursory acquaintance with the construction of the new implement of technology. Hence arises the necessity of confining oneself to more or less probable assumptions, and only with such an approach obtaining answers to the questions that present themselves. In this collateral problem, relating to the choice of probable assumptions, the character of personal creativity may perhaps reveal itself most fully, and a more or less fine technical sense may be displayed. The closer the assumptions chosen are to reality, the more probable is the practical accuracy of the expected solution.

Among the questions not touched upon by me in the preceding article dealing with the press, there remained open the question of determining the thickness of the walls of the folding harmonica. It is to this determination that we now turn, while at the outset it is necessary to stipulate that the proposed method is by no means the only one of its kind, since the solution is of an approximate character.

Let \(AB\) represent the slope of some link of the press (Fig. 1). Per unit surface area, as before, let the pressure be equal to \(p\) kilograms and be directed perpendicular to the slope, inclined to the horizon at an angle \(\varphi\). Resolve this pressure into two components \(p_1\) and \(p_2\), directed parallel to \(AC\) and \(AB\). Then

\[ p_1 = p : \sin \varphi; \qquad p_2 = p : \operatorname{tg}\varphi \]

Denote \(AC\) by \(r_1\), and \(BD\) by \(r_2\), and let us consider a differential annular element of the conical surface, whose height is \(dh\), and whose corresponding radius is \(r\). If the wall thickness is \(\varepsilon\), then the differential bursting force \(dT\), being uniformly distributed over the annular section, will be applied to the area \(2\pi r\varepsilon\).

The magnitude of this force will be found from the equation

\[ dT = p_2\, dh\, 2\pi r = \frac{p\, dh}{\operatorname{tg}\varphi}\cdot 2\pi r = \frac{2\pi r\, p\, \cos\varphi}{\sin\varphi}. \]

Since \(dr = dh \cos\varphi\), then

\[ dT = \frac{2\pi r\, p\, dr}{\sin\varphi} \]

The final force \(T\) will be found as the definite integral, taking the limits for \(r\) as the upper \(r_1\), and the lower \(r_2\). Thus,

\[ T = \frac{2\pi p}{\sin\varphi}\int_{r_2}^{r_1} r\,dr = \frac{\pi p(r_1^2-r_2^2)}{\sin\varphi} \]

Denote the allowable tensile stress, referring it to one square centimeter, by \(K_z\). Then

\[ 2\pi r\varepsilon_1 k_z = \frac{\pi p(r_1^2-r_2^2)}{\sin\varphi}. \]

Hence we have

\[ \varepsilon_1 = \frac{p(r_1^2-r_2^2)}{2r\sin\varphi\, k_z} \tag{1} \]

From consideration of this equation it becomes clear that the thickness of the wall of the press slope \(\varepsilon_1\) is not a constant quantity, but depends on \(r\), and the greatest value for \(\varepsilon\) corresponds to the minimum \(r=r_2\).

As for the force \(p_1\), it likewise cannot be neglected, and therefore we shall determine what influence it exerts on the wall of the funnel. This force also tends to rupture our differential annular

element by the section \(2r\,dh\). Let us denote the corresponding stress in the material of the harmonic by \(K^1_z\); generally speaking, \(k^1_z \lesseqgtr k_z\). The mathematical dependence between the elements of the ring, the load \(p_1\), and the stress \(K^1_z\) is found without any difficulty:

\[ k^1_z=\frac{p_1\cdot 2r\,dh}{2\varepsilon_2\cdot dh} =\frac{p_1 r}{\varepsilon_2} =\frac{pr}{\varepsilon_2\sin\varphi}; \quad \varepsilon_2=\frac{pr}{k^1_z\sin\varphi} \]

We shall now clarify the question of which of these formulas, and in what cases, should be used to determine the thickness of the walls of the press. From the scientific point of view this question would not have sufficient weighty raison d’être, but here practical significance is involved, and the engineer is not permitted to ignore it. It is self-evident that here one can only be speaking of the greatest values of the wall thickness. In view of the fact that in the formula

\[ \varepsilon_1=\frac{p(r^2_1-r^2_2)}{2r\sin\varphi k_z} \]

the only variable quantity is \(r\), which enters in the denominator, it may be said that the greatest value of \(\varepsilon_1\) will correspond to the smallest \(r=r_2\):

\[ \text{Max. }\varepsilon_1=\frac{p(r^2_1-r^2_2)}{2r_2\sin\varphi k_z} \]

For \(\varepsilon_2\) the situation is otherwise. Here the greatest value of \(\varepsilon_2\) corresponds to the greatest \(r=r_1\). Therefore one may write

\[ \text{Max. }\varepsilon_2=\frac{pr_1}{K^1_z\sin\varphi} \tag{2} \]

Let us assume, for simplicity, that \(K_z=K^1_z\). Then

\[ \varepsilon_1:\varepsilon_2=\frac{r^2_1-r^2_2}{2r_1}:r_2 \]

This equality serves as a criterion in clarifying the question of which formula, in the sense of reliability, should be given preference in practical cases when determining the thickness of the walls of the harmonic of the press. Indeed, if

\[ \frac{r^2_1-r^2_2}{2r_1r_2}>1,\quad \text{then } \varepsilon_1>\varepsilon_2 \]

This means that the wall thickness determined by formula (1) fully guarantees against rupture by the forces \(p_1\). Solving the written inequality, we obtain the following result:

\[ r^2_1-r^2_2>2r_1r_2 \]

or

\[ r^2_1-2r_1r_2+r^2_2>2r^2_2;\quad r_1-r_2>r_2\sqrt{2};\quad r_1>r_2(1+\sqrt{2});\quad r_1>2.41r_2 \]

Thus, in this case the check of the wall strength must be carried out only by formula (1)

\[ \varepsilon_1=\frac{p\left(r_1^2-r_2^2\right)}{2r_2\sin\varphi\cdot k_z} \]

But if

\[ \frac{r_1^2-r_2^2}{2r_1r_2}<1, \]

then \(\varepsilon_1<\varepsilon_2\).

Solving this inequality, we find:

\[ r_1<r_2(1+\sqrt{2});\qquad r_1<2.41\,r_2 \]

For this case, in determining the thickness of the walls of the press, preference must be given to the second formula (2)

\[ \varepsilon_2=\frac{pr_1}{k'_z\sin\varphi} \]

And only under the condition \(r_1=2.41\,r_2\) do we have one and the same wall thickness that will well guarantee strength against rupture. If we now introduce the condition that \(k_z>k'_z\), or \(\alpha k_z=k'_z\), where \(\alpha=0.8\), then the relation derived earlier for \(\varepsilon_1:\varepsilon_2\) is transformed into the following:

\[ \varepsilon_1:\varepsilon_2=(r_1^2-r_2^2)k'_z:2r_1r_2k_z \]

Solving this equality leads to the following result: if \(\varepsilon_1:\varepsilon_2>1\), then \((r_1^2-r_2^2)0.8k_z>2r_1r_2k_z\), or \(r_1^2-r_2^2>2.5\cdot r_1r_2\);

\[ r_1^2-2.5\,r_1r_2-r_2^2>0 \]

Whence

\[ r_1>2.85\,r_2 \]

All these considerations, minor at first glance, are of interest in that the greatest supplementary force from the press harmonica comes very close to the conditions just considered, since, as we saw earlier, the ratio \(a_1:a_2=4\) \((r_1=4r_2)\), when the supplementary force from the press harmonica reaches its maximum. The practical meaning of attention to the press harmonica consists in the fact that, by a proper choice of the form of the harmonica, we not only achieve reliable strength of the walls, but also gain in the volume of the press, so that, with a smaller expenditure of working fluid per stroke, we attain the very same force effect. Here, obviously, we have a rational economy of energy, and it is unnecessary to dwell on how important a factor this is in engineering.

In the case when it is desirable to determine the wall thickness of the press harmonica more precisely, taking into account the weight of the walls and atmospheric pressure, one may proceed as follows: let the pressure of the fluid inside the conical branch pipe be \(p\), and the external atmospheric

pressure be \(q\); let the specific weight of the material of the tube walls be \(g_1\), and the specific weight of the liquid inside the press \(g_2\); let the thickness of the walls remain, as before, \(\varepsilon\). Cut out from our pipe a conical ring of height \(dx\), at a distance \(x\) from the upper edge, Fig. 2. The upper base of the truncated cone has radius \(r_2\), the lower \(r_1\), and \(y\) denotes the variable radius corresponding to the distance \(x\). The bevel is, as before, inclined to the horizontal at the angle \(\varphi\). Without particular difficulty we shall write the following equality.

\[ G'_x + G''_x + G'''_x = 2\pi y \varepsilon k_x + \pi y^2 p_x \tag{3} \]

Here \(G'_x\) denotes the weight of the tube of height \(x\), so that

\[ G'_x = \int_0^x 2\pi y\,dx\,\varepsilon g_1 \]

The second term of the left-hand side, \(G''_x\), expresses the pressure of the atmosphere on the projection onto the horizontal plane of the lateral surface of the truncated cone of height \(x\), so that

\[ G''_x = \pi (y^2 - r_2^2) q \]

Finally, the third term \(G'''_x\) represents the weight of the liquid in the volume of the truncated cone of height \(x\), so that

\[ G'''_x = \int_0^x \pi y^2\,dx\,g_2 \]

These three forces, acting vertically downward, are balanced by the elastic forces in the annular section \(2\pi y \varepsilon\), i.e. by the force \(2\pi y \varepsilon k_x\), and by the pressure of the liquid due to elasticity \(p_x\), i.e. by the force \(\pi y^2 p_x\).

For the successful integration of the expressions for \(G'_x\) and \(G'''_x\), we shall use the following consideration:

\[ y = r_2 + x\cot\varphi;\qquad dy = dx\cot\varphi \]

Then

\[ G'_x = \int_0^x 2\pi (r_2 + x\ctg\varphi)\varepsilon g_1\,dx = 2\pi \varepsilon g_1\left(r_2 x + \frac{x^2\ctg\varphi}{2}\right) \]

In exactly the same way we find without difficulty that

\[ G'''_x = \int_0^x \pi (r_2 + x\ctg\varphi)^2 g_2\,dx = \pi g_2\left(r_2^2 x + r_2\ctg\varphi\,x^2 + \frac{x^3\cotg^2\varphi}{3}\right) \]

Substituting the values found for the weights into equation (3), we shall have

\[ 2\pi\varepsilon g_1\left(r_2 x + \frac{x^2\ctg\varphi}{2}\right) + \pi q\left[(r_2 + x\ctg\varphi)^2 - r_2^2\right] + \pi g_2\left(r_2^2 x + r_2\ctg\varphi\,x^2 + \frac{x^3\ctg^2\varphi}{3}\right) = 2\pi\varepsilon k_x(r_2 + x\ctg\varphi) + \pi p_x(r_2 + x\ctg\varphi)^2 \]

Performing the reduction by \(\pi\) and assuming that \(p_x\) is a linear function of \(x\), so that \(p_x=xg_2\), we obtain

\[ \varepsilon= \frac{ xg_2(r_2+x\operatorname{ctg}\varphi)^2 - q(2r_2x\operatorname{ctg}\varphi+x^2\operatorname{ctg}^2\varphi) - g_2\left(r_2^2x+r_2\operatorname{ctg}\varphi\,x^2+\dfrac{x^3\operatorname{ctg}^2\varphi}{3}\right) }{ 2\left[ g_1\left(r_2x+\dfrac{x^2\operatorname{ctg}\varphi}{2}\right) - k_x(r_2+x\operatorname{ctg}\varphi) \right] } \]

In Prof. J. Perry’s book Applied Mechanics a somewhat different solution of this question is given. The starting equation is an equation identical with the third (3), namely

\[ 2\pi r\varepsilon\Delta x\, g_1+2\pi r^2\Delta x\, g_2+2\pi r\Delta r\,q = \Delta x\,\frac{d}{dx}\left(\pi r^2p+2\pi rk_d\right) \]

By means of not entirely clear substitutions and reductions (in the American adaptation all of these have been omitted), he brings this fundamental equation to the following differential equation

\[ \frac{dr}{dx} + \frac{ \dfrac{r}{2}\left[\dfrac{dp}{dx}(1+a)+\dfrac{2g_1}{k_d}\right] }{ q+pa } =0 \]

The indefinite integral has the following form

\[ \frac{2g_1}{ak}x+\frac{(1+a)^{\varepsilon}}{a}\log(1-bx)+2\log r=C. \]

The few indications given by J. Perry reduce to the following: \(\varepsilon\), \(g_1\), \(g_2\), \(p\), \(q\), and \(k_d\) have the same meanings as those adopted by us. The quantity \(q\) is considered by him as a definite function of \(x\), \(g_1=\mathrm{const}\), \(g_2=\dfrac{cp_1}{\gamma}\).

Then

\[ a=1+\frac{2k_d}{k}, \]

where \(k\) is the greatest stress in the material of the walls. Apparently, here the discussion concerns \(k_d=k_z\) and \(k=k_z^1\), the values of which were given earlier. In addition,

\[ \varepsilon=\frac{\gamma}{\gamma-1};\qquad p=p_0(1-bx)^{\varepsilon};\qquad q=q_0(1-cx)^{\varepsilon} \]

Despite every desire to proceed, by transformations, from the fundamental equation to the differential one, I did not succeed in doing this. Perhaps one of the engineers will be more fortunate than I in this respect and will indicate the solution of the puzzle, for which I shall be grateful in advance. The differential equation took, for me, the following form:

\[ 2r\varepsilon g_1+r^2g_2+2rq\frac{dr}{dx} = 2\varepsilon\left(\frac{dr}{dx}k_x+\frac{r\,dk_x}{dx}\right) + 2r\frac{dr}{dx}p_x + r^2\frac{dp_x}{dx} \]

or

\[ \frac{dr}{dx}\left[2rq-2\varepsilon k_x-2rp_x\right]+2r\varepsilon g_1+r^2g_2-\frac{2\varepsilon r\,dk_x}{dx}-\frac{r^2\,dp_x}{dx}=0. \]

whence

\[ \frac{dr}{dx} + \frac{ \frac{r}{2}\left[2\varepsilon g_1+rg_2-\frac{2\varepsilon\,dk_x}{dx}-\frac{r\,dp_x}{dx}\right] }{ rq-\varepsilon k_x-rp_x } =0 \]

This equation reduces to the following

\[ \frac{dr}{dx} + \frac{ \frac{r}{2}\left[\frac{dp_x}{dx}+\frac{2\varepsilon\,dk_x}{r\,dx}-g_2-\frac{2\varepsilon g_1}{r}\right] }{ p_x+\frac{\varepsilon k_x}{r}-q } =0 \]

or, finally,

\[ \frac{dr}{dx} + \frac{ \frac{r}{2}\left[\frac{dp_x}{dx}+\frac{2\varepsilon\,dk_x}{r\,dx}-g_2-\frac{2\varepsilon g_1}{r}\right] }{ p_x\left(1+\frac{\varepsilon k_x}{p_x r}\right)-q } =0 \]

The integral of this expression is reduced to the form

\[ 2\log_n r+\frac{1}{a}\log_n(ap_x-q) + \int \frac{2\varepsilon\,dk_x-rg_2\,dx-2\varepsilon g_1\,dx}{r(pa-q)} =C \]

Assuming that \(k_x\) is a constant quantity, and \(r=r_2+x\cotg\varphi\), we obtain

\[ 2\log_n r+\frac{1}{a}\log_n(ap_x-q) -\frac{g_2x}{pa-q} -\frac{2\varepsilon g_1}{pa-q}\cdot\frac{1}{\cotg\varphi} \log_n(r_2+x\cotg\varphi) =C. \]

Or

\[ 2\log_n r+\frac{1}{a}\log_n(ap_x-q) -\frac{g_2x}{pa-q} -\frac{2\varepsilon g_1}{pa-q}\cdot\frac{1}{\cotg\varphi} \log_n r =C \]

Taking the limits for \(x=0\) and \(x_2=h\); \(p_x=p_0\) and \(p_x=p_h\), we determine \(C\), and then also \(\varepsilon\). Thus, in the first case,

\[ 2\log_n r_2+\frac{1}{a}\log_n(ap_0-q_0) -\frac{2\varepsilon g_1}{ap_0-q_0}\cdot\frac{1}{\cotg\varphi} \log_n r_2 =C \]

And the thickness of the walls \(\varepsilon\) is found from the equation

\[ 2\log_n r_1+\frac{1}{a}\log_n(ap_h-q_h) -\frac{g_2h}{p_ha-q_h} -C = \frac{2\varepsilon g_1}{p_ha-q_d}\cdot\frac{1}{\cotg\varphi} \log_n r_1 \]

A very essential problem in determining the thickness of the harmonic press walls is the question of the resistance of the walls to bending. In fact, the smallest deformation of the walls, in the sense

bending, determines to a considerable degree the presence of one or another magnitude of additional force from the harmonica itself. Hence the necessity of checking the walls for bending is understandable. Here the question is complicated by very many circumstances, and in particular by the fact that each link changes its form, sometimes increasing and sometimes decreasing its height during the operation of the harmonica; sometimes increasing its volume and sometimes reducing it, owing to the variable tension of the walls under the influence of both external and internal pressure, while for each fold or link these pressures are functions not only of the height of the individual links, but also of the height of the entire press, the weight of the pressure plate (matrix), etc. It is therefore necessary to make a whole series of assumptions and to consider the link at the greatest permissible height of the link, \(h\max\), at the moment of equilibrium, assuming a priori that the lower link will be in the most unfavorable conditions; all further derivations refer to it.

Let \(AB\) (Fig. 4) represent the fold of the press harmonica; \(r\), \(r_2\), and \(x\) are the radii corresponding to the various levels of the working liquid \(H\), \(H_x\), and \(H_1\). Let us cut from the bottom a strip \(QMNP\), corresponding to the central angle \(d\alpha\). Denote \(NP\) by \(S_1\), \(MQ\) by \(S_2\); then cut out an area \(FEDG\) at the distance \(BC = y\), with \(IC = dy\).

The pressure on this area will be expressed as

\[ dp_x = g_2 \cdot S_y \cdot dy \, H_x = g_2 H_x \cdot \omega \]

In order to give this expression a more visual form, we shall use the following substitutions:

\[ 1)\ \omega = S_y \cdot dy;\quad 2)\ S_y - s_1 : S_2 - s_1 = y : h;\quad S_y = s_1 + \frac{y(S_2 - s_1)}{h} \]

\[ 3)\ H_x - H_1 : H - H_1 = y : h;\quad H_x = H_1 + \frac{y(H - H_1)}{h} \]

Here \(h = AB\). After the corresponding substitutions and integration we obtain

\[ \int dp_x = \int g_2 \cdot S_y \cdot H_x\, dy = g_2 \left[ H_1 s_1 y + \frac{y^2}{2h}(H s_1 - 2H_1 s_1 + H_1 s_2) + \frac{y^3}{3h^2} \right. \]

\[ \left. (H - H_1)(s_2 - s_1) \right] + C \]

Putting here \(y = 0\), we obtain the equation for determining \(C\).

Thus,

\[ C = p_B \]

i.e. the arbitrary constant \(C\) represents the pressure at point \(B\).

After this one may write

\[ p_x=p_{\mathrm{B}}+g_2\left[H_1S_1y+\frac{y^2}{2h}(HS_1-2H_1S_1+H_1S_2)+\frac{y^3}{3h^2}(H-H_1)(s_2-s_1)\right] \]

Or, in general form,

\[ p_x=Ay^3+By^2+Cy+D \]

This equation shows that, in passing from point \(B\) to point \(A\), the pressure changes according to the law of a cubic parabola. The right-hand side of the last equation, as a function of \(y\), has neither a max. nor a min., because their existence is conditioned by conditions inadmissible in practice.

The cubic parabola itself can easily be constructed by points, assigning to \(y\) the values \(0,\ 0.1h,\ 0.2h\), and so on. Its approximate form will be \(FHg\) (Fig. 5). We replace this curve by the broken line \(FEg\), so that point \(E\) falls above \(D\), the midpoint of \(AB\).

The load on our strip will under such conditions be expressed by the area

\[ AFEGB=AFB+B\Sigma G=\omega_1+\omega_2 \]

But

\[ \omega_1=\frac{AF\cdot AB}{2}=\frac{P_A\cdot h}{2};\quad \omega_2=\frac{P_{\mathrm{B}}\cdot h}{4}; \]

If our plate is regarded as a beam with fixed ends, then the support reactions at points \(A\) and \(B\) are found by the rule of moments of forces, taking points \(B\) and \(A\) as the centers of moments. Then the support moment is

\[ R_Ah=\omega_1\cdot\frac{2}{3}h+\omega_2\cdot\frac{h}{6};\quad R_A=\frac{4\omega_1+\omega_2}{6}\quad(\text{point }A) \]

In exactly the same way the support moment is

\[ R_B\cdot h=\frac{\omega_1h}{3}+\frac{5}{6}\omega_2h;\quad R_B=\frac{2\omega_1+5\omega_2}{6}\quad(\text{point }B) \]

If point \(C\) (Fig. 5) corresponds to the section where the bending moment is greatest, then the ordinate \(CK\) is found from the proportions

\[ CK:AF=y:h;\quad CK=p_x=\frac{P_A\cdot y}{h} \]

If we confine ourselves to the usual formulas of strength of materials (see Khudyakov, “Strength of Materials”), then for the moments of forces we have the equation \(EIy''=M_x\).

The maximum value of \(M_x\) is reached when

\[ \frac{d M_x}{EI\,dx}=\Sigma p=0 \]

The latter condition indicates that in section \(C\) the shearing forces are equal to zero. Expressing this position analytically, we write

\[ R_B-\omega_2-\Delta CKB=0 \]

After substituting the value, we obtain

\[ \frac{2\omega_1+5\omega_2}{6}-\omega_2-\frac{\omega_1 y^2}{h^2}=0 \]

Whence

\[ y=h\sqrt{\frac{2\omega_1-\omega_2}{6\omega_1}} \]

This condition is sufficient to write the condition for the bending moment

\[ \begin{aligned} M_{\max}={}&R_B\cdot y-\omega_2\left(y-\frac{h}{6}\right) -\frac{\omega_1 y^2}{h^2}\frac{4}{3} \\ ={}&\frac{2\omega_1+5\omega_2}{6}h \sqrt{\frac{2\omega_1-\omega_2}{6\omega_1}} \\ &-\omega_2 h\left(\sqrt{\frac{2\omega_1-\omega_2}{6\omega_1}}-\frac{1}{6}\right) -\frac{\omega_1 h^3}{3h^2} \left(\sqrt{\frac{2\omega_1-\omega_2}{6\omega_1}}\right)^3 \\ ={}&\frac{h}{3}\left( \frac{\omega_2}{2} +\frac{2\omega_1-\omega_2}{3} \sqrt{\frac{2\omega_1-\omega_2}{6\omega_1}} \right) \end{aligned} \]

The final solution of the problem, i.e., the selection of a section according to a given bending moment, presents no difficulty if one passes to a finite central angle \(\alpha\) equal to one degree, one minute, or sets \(\alpha=2\pi\). Then the arc \(Sy\) is determined. In order to pass from \(\omega=Sydy\) to the finite \(\Omega\), one may assume that we are operating with an area whose height is not \(dy\), but some unit: a millimeter, a centimeter. After these preparations, one may write and apply the equation

\[ M_{\max}=W\cdot K_b \]

Such is the solution of the problem.

Fig. 4.

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 5.

Table I

Graph for determining \(I =\) [[unclear: formula]] from prototypes.