CALCULATION OF PRESSURE-GAUGE SPRINGS

M. P. Shumskii

(Presented by the scientific seminar of the Department of Gyroscopic Instruments and Devices)

In instrument-making practice, pressure-gauge springs with various forms of cross section are used; the method of calculation for them is often unknown. The question of the optimal shape of the section of pressure-gauge springs has not been solved. In this connection, the problem of stresses and displacements under the action of pressure \(p\) in a spring whose sectional shape is given by the equation \(y = y(x)\), where \(y(x)\) is an arbitrary function, is of interest.

In order to simplify the investigation, we shall adopt the assumptions usual in problems of this kind:

1. All portions of the spring separated by sections normal to the axis of the spring are in identical conditions.
2. The hypotheses of non-pressure of the layers and invariability of the normal are valid.
3. The axial line of the tube is not stretched.
4. The minor semiaxis of the section is small in comparison with the radius of curvature of the spring axis \((a \ll R)\).
5. The spring is symmetric with respect to the plane in which the spring axis lies.

In what follows it is assumed that the section has two axes of symmetry. The absence of symmetry with respect to the second axis does not create any fundamental difficulties in using the proposed method.

The following notation is used in the work:

\[ \begin{aligned} p&\text{ — excess pressure in the spring;}\\ a,\ b&\text{ — the major and minor semiaxes of the section;}\\ h&\text{ — wall thickness of the spring;}\\ R&\text{ — radius of curvature of the spring axis;}\\ x,\ y&\text{ — coordinates of a point of the midline of the section;}\\ z,\ z_y&\text{ — coordinate of a point on the normal to the wall of the spring,}\\ &\quad\text{measured from the midline, and its projection onto the }y\text{-axis;}\\ E,\ \mu&\text{ — modulus of elasticity and Poisson’s ratio;}\\ \delta\gamma&\text{ — relative angle of unbending of the spring;}\\ w,\ w_0&\text{ — displacement of a point of the midline in the direction}\\ &\quad\text{of the }y\text{-axis and the value of this displacement at the point }x=0;\\ T&\text{ — normal force in the section, referred to a unit}\\ &\quad\text{length of the midline;}\\ N&\text{ — resultant of the stresses in the section }x=a\\ &\quad\text{of a unit ring.} \end{aligned} \]

\(\sigma_1\) — normal stress in the spring section;

\(\varepsilon_1\) — relative elongation of the middle surface of the wall in the direction of the stress \(\sigma_1\);

\(F\) — area of the spring cross section;

\(s\) — arc of the center line of the section;

\(4l\) — perimeter of the center line of the section;

\(J\) — moment of inertia of the cross-sectional area with respect to the \(x\)-axis;

\(g\) — intensity of the load on a unit ring, caused by the presence of the normal force \(T\) in the spring section;

\(v,\ \varphi,\ \psi,\ \Phi,\ \Psi,\ C\) — functions depending on the shape of the center line of the section;

\(A\) — coefficient depending on the shape and dimensions of the section;

\[ x=\frac{Rh}{a^2} \]

— spring parameter.

Under the action of the pressure \(p\), the fiber \(AB\) of the part of the spring bounded by two neighboring transverse sections assumes the position \(A_1B_1\) (Fig. 1). The relative elongation is

\[ \varepsilon_1=\frac{CB_1}{AB}=\frac{CD-Db_1}{AB} =\frac{w\Theta-(y-z_y+w)\vartheta}{(R+y+z_y)\Theta}. \tag{1} \]

Fig. 1.
a — deformed cross section of the spring; b — deformation of a longitudinal fiber

By the conditions of the problem \(y \ll R_1,\ z_y \ll R\). In the region of small displacements \(w \ll y\). Neglecting small terms in formula (1),

\[ \varepsilon_1=\frac{w-y\delta\gamma}{R}-\frac{z_y\delta\gamma}{R}. \tag{2} \]

Here \(\delta\gamma=\frac{\vartheta}{\Theta}\) is the relative angle of unbending of the spring, and \(w\) is the projection onto the \(y\)-axis of the displacement of point \(A'\) of the center line of the section.

Let us find the normal force in the section, referred to a unit length of the center line of the section.

\[ T=\frac{Eh}{R}(\delta\gamma\cdot y-w)\cdot\Theta . \tag{3} \]

Here \(E\) is the modulus of elasticity, \(h\) is the wall thickness of the spring.

By two normal sections we cut out from the spring a unit ring (Fig. 2). Owing to the nonparallelism of the sections, the normal

Fig. 2

forces on the element \(ds\) of the unit ring give a resultant

\[ \frac{Tds}{R\Theta}, \]

parallel to the \(y\)-axis.

The shape of the center line of the spring section after deformation can be found by solving the problem of bending of a unit ring by a normal load \(p\) and by a load

\[ g=\frac{T}{R\Theta}=\frac{Eh}{R^2}(\delta\gamma\cdot y-w), \tag{4} \]

parallel to the \(y\)-axis (Fig. 2). In Fig. 2,

\[ N=pa+\int_0^a g\,ds(x). \]

To find the relation between the unknowns \(w\) and \(\delta\gamma\) in the right-hand side of formula (4), we use the condition of absence of a moment in the spring section

\[ \int_F \sigma_1 (y+z_y)\,dF=0, \tag{5} \]

where \(F\) is the area of the section, and \(\sigma_1\) is the normal stress in the section.

For a section symmetric with respect to the \(x\)-axis, conditions (5) are equivalent to the following:

\[ \int_F \varepsilon_1 (y+z_y)\,dF=0. \tag{6} \]

Taking into account that \(dF=ds\cdot dz\), and that \(\varepsilon_1\) is determined by formula (2), and carrying out the integration, we find

\[ h\int_0^{4l} yw\,ds-J\cdot\delta\gamma=0. \tag{7} \]

Here

\[ J=\int_{\breve F}^{c} (y+z_y)^2\,dF,\quad 4l \text{ is the perimeter.} \tag{8} \]

From equation (7)

\[ \delta_\gamma=\frac{4w_0h}{J}\int_0^e y\,\frac{w}{w_0}\,ds . \tag{9} \]

Let us denote

\[ v=\frac{w}{w_0};\quad A=\frac{4ah}{J}\int_0^e yv\,ds . \tag{10} \]

Transforming formula (4), taking (9) and (10) into account,

\[ g=\frac{w_0Eh}{R^2}\left(\frac{A}{a}y-v\right). \tag{11} \]

The bending moment in the section of a unit ring (Fig. 2) is

\[ M=-M_a+N\cdot(a-x)-p\,\frac{(a-x)^2}{2}-p\,\frac{y^2}{2} -\int_x^a g(t)(t-x)\,ds(t). \tag{12} \]

The sections \(x=0\) and \(x=a\), by virtue of symmetry, do not rotate; therefore,

\[ \int_0^e M\,ds=0. \tag{13} \]

From (13), taking (11) and (12) into account and introducing the notations

\[ \frac12\int_0^x\left(1-\frac{x^2+y^2}{a^2}\right)ds(x)=a\psi, \tag{14} \]

\[ \int_0^x\left[(a-x)\int_0^e y\,ds-\int_x^a xy\,ds(x)+x\int_x^a y\,ds(x)\right]ds(x)=ba^3\varphi_1, \tag{15} \]

\[ \int_0^x\left[(a-x)\int_0^e v\,ds-\int_x^a xv\,ds(x)+x\int_x^a v\,ds(x)\right]ds(x)=a^3\varphi_2, \tag{16} \]

\[ \varphi_2-A\,\frac{b}{a}\,\varphi_1=\varphi, \tag{17} \]

we find

\[ w_0=\frac{R^2}{Eh}\left(p\,\frac{\psi(a)}{\varphi(a)}-\frac{M_a l}{a^3\varphi(a)}\right). \tag{18} \]

The moment \(M_a\) is found from the condition that the vertical displacement of the section \(x=a\) is absent,

\[ w(a)=w_0-\frac{12(1-\mu^2)}{Eh^3}\int_0^a\left[\int_0^s M\,ds\right]dx=0. \tag{19} \]

Carrying out the termwise integration in (19) and replacing \(w_0\) by its value from (18), we find

\[ M_a=pa^2\, \frac{ \dfrac{x^2}{12(1-\mu^2)}\,\psi(a)+\Phi\cdot\psi'(a)-\Psi\cdot\varphi(a) }{ \dfrac{l}{a}\left[\dfrac{x^2}{12(1-\mu^2)}+\Phi\right]-C\cdot\varphi(a) }. \tag{20} \]

Here

\[ \Phi=\frac{1}{a}\int_{0}^{a}\varphi\,dx,\qquad \Psi=\frac{1}{a}\int_{0}^{a}\psi\,dx,\qquad C=\frac{1}{a^{2}}\int_{0}^{a}s\,dx,\qquad \varkappa=\frac{Rh}{a^{2}}. \tag{21} \]

Knowing \(M_a\), from (18) we find \(w_0\)

\[ w_0=\frac{pR^2}{Eh}\, \frac{\dfrac{a}{e}C\cdot\psi(a)-\Psi} {\dfrac{\varkappa^2}{12(1-\mu^2)}+\Phi-\dfrac{a}{e}\cdot C\cdot\varphi(a)}. \tag{22} \]

The maximum value of the stress in the section \(x=a\), without taking account of stress concentration, is

\[ \sigma_a=6\frac{M_a}{h^2}. \tag{23} \]

The relative angle of bending of the spring is

\[ \delta\gamma=w_0\frac{A}{a}. \tag{24} \]

We prescribe the unknown function \(v\), entering into the solution, approximately in the form

\[ v=1-\frac{7}{4}\frac{x^2}{a^2}+\frac{3}{4}\frac{x^4}{a^4} +\varepsilon\frac{x^2}{a^2}\left(1-\frac{x^2}{a^2}\right), \tag{25} \]

where

\[ \varepsilon=\frac{16}{3}\frac{w'\left(\dfrac{a}{2}\right)}{w'_0}-\frac{13}{4}. \tag{26} \]

Here \((\,{}'\,)\) means that the quantities are calculated for \(\varepsilon=0\).

The solution obtained has been applied to the calculation of springs of flat-oval section.

For a flat-oval section, putting \(\varepsilon=0\) in (25), we find

\[ J=4hb^2(a-b)\left(1-\frac{1}{12}\frac{h^2}{b^2}\right) +\frac{\pi}{4}\left[\left(b+\frac{h}{2}\right)^4-\left(b-\frac{h}{2}\right)^4\right], \tag{27} \]

\[ A=\frac{34}{15}\frac{a^2bh}{J},\qquad C=\frac{1}{2}+\frac{\pi-3}{2}\frac{b^2}{a^2}, \tag{28} \]

\[ \varphi(a)=\frac{5}{21}+0.04\frac{b^2}{a^2} -\frac{ab^2h}{J}\left[\frac{34}{45}+0.16\frac{b^2}{a^2}-0.023\frac{b^3}{a^3}\right], \tag{29} \]

\[ \psi(a)=\frac{1}{3}-\left(2-\frac{\pi}{2}\right)\frac{b^2}{a^2} +\left(\frac{5}{3}-\frac{\pi}{2}\right)\frac{b^3}{a^3}, \tag{30} \]

\[ \Psi(a)=\frac{5}{24}-\frac{1}{4}\frac{b^2}{a^2} +\left(\frac{3\pi}{4}-\frac{7}{3}\right)\frac{b^3}{a^3} +\left(\frac{19}{8}-\frac{3\pi}{4}\right)\frac{b^4}{a^4}, \tag{31} \]

\[ \Phi=0.1523+0.0113\frac{b^3}{a^3} -\frac{ab^2h}{J}\left[0.472+0.04\frac{b^3}{a^3}\right]. \tag{32} \]

The maximum stress and the relative angle of bending are found from formulas (20)–(24), after first computing \(\psi(a)\), \(\varphi(a)\), \(\Phi\), \(\Psi\), \(C\) from formulas (27)–(32). Taking into account that the problem of finding the relati-

relative bending angle \(\delta_\gamma\) of a flat-oval spring has been solved earlier with sufficient accuracy [1], and this solution is widely used; let us consider another variant for determining the stresses.

From formulas (18), (23), (24), (28) we find

\[ \sigma_a = p \frac{6a^3}{l h^2} \left[ \psi(a) - \frac{E \cdot \delta_\gamma}{p} \frac{15 J \varphi(a)}{34abR^2} \right]. \tag{33} \]

Having computed \(\delta_\gamma\) from the known formulas [1], and \(\varphi(a)\), \(\psi(a)\), \(J\) from formulas (27), (29), (30), we find the maximum bending stress \(\sigma_a\).

Example. Determine the maximum stress in a spring of flat-oval section with dimensions \(R = 40\) mm, \(a = 4\) mm, \(b = 1\) mm, \(h = 0.16\) mm at pressure \(p = 1\ \mathrm{kg/cm^2}\). Poisson’s ratio \(\mu = 0.3\).

From formulas (27)–(32) we find:

\[ J = 0.243 \cdot 10^{-3}\ \mathrm{cm^4},\quad \varphi(a) = 0.308,\quad \psi(a) = 0.039, \tag{34} \]

\[ C = 0.504,\quad \Psi = 0.193,\quad \Phi = 0.0282. \]

The maximum stress according to formulas (20), (22) is

\[ \sigma_a = 728\ \frac{\mathrm{kg}}{\mathrm{cm^2}}. \]

The relative bending angle according to formulas (22), (24) is

\[ \delta_\gamma = \frac{1.33 \cdot 10^4}{E}. \]

According to the known formulas of L. E. Andreeva for a flat-oval section [2],

\[ \sigma_a = 575\ \frac{\mathrm{kg}}{\mathrm{cm^2}},\quad \delta_\gamma = \frac{1.31 \cdot 10^4}{E}. \]

According to Wuest’s formulas for thin-walled springs [2],

\[ \sigma_a = 727\ \frac{\mathrm{kg}}{\mathrm{cm^2}},\quad \delta_\gamma = \frac{1.39 \cdot 10^4}{E}, \tag{35} \]

Conclusions

The paper has considered the problem of calculating pressure-gauge springs with an arbitrary cross-sectional shape. The problem is reduced to the integration of known functions. The solution found has been applied to the calculation of flat-oval springs. Simple working formulas have been obtained for determining stresses and displacements. A comparison, on a specific example, of different calculation methods shows that the proposed method gives results close to Wuest’s sufficiently rigorously substantiated results [4].

References

1. L. E. Andreeva. Elastic Elements of Instruments. Mashgiz, 1962.
2. W. Wuest. Die Berechnung von Bourdonfedern, “VDJ—Forschungsheft 489,” supplement to “Forschung auf dem Gebiete des Ingenieurwesens,” edition B., vol. 28, 1962.