Prof. G. V. TRAPEZNIKOV

DEFORMATION OF A COLUMN BEYOND THE STABILITY LIMIT

Derivation of the fundamental equation of deformation of a column

The experimental work I carried out in connection with the study of longitudinal bending revealed an interesting feature of the deformation of a column with a large ratio \(l/i\). This feature consists in the fact that the ends of the column, when they are hinged in the supports and loaded by axial compressive forces, described arcs of a circle with centers lying on the initial axis of the column. Of course, the trajectory of the end of the column may be taken as an arc of a circle only with a certain (as we shall see below) approximation. But the most important point is that the central angle of the arc, with which the end of the column almost coincides during deformation, considerably exceeds \(\pi/2\). In Drawing 1 a column is shown in various phases of deformation under longitudinal bending. The trajectories of its ends are indicated by dashed lines.

The need to know the settlement of the ends of a rod when its load exceeds the corresponding critical force arises in the calculation of drill pipes, perforator rods, and those springs that work in longitudinal bending. Knowledge of the magnitude of the settlement of the end of a column and of the analytical relation between the magnitudes of the settlement and the corresponding load acquires particular importance when the latter is applied by impact. Only with such knowledge can we calculate the greatest deformation and stress under impact.

All this compelled me to determine the trajectory of the ends of a column already quite accurately, resorting to the theory of longitudinal bending, and to give, instead of a complicated and irrational dependence between force and deformation, a simple dependence, quite convenient for calculation and, to a sufficient degree, accurate.

Drawing 1.

Since, in an exact calculation, the determination of the deformation of longitudinal bending is connected with tabular computation of the values of elliptic integrals, it became necessary to have tables of their values, determined to the seventh decimal place, with small differences in the modulus \(k\). I reproduce these tables in full in the present article, since they may be needed by some of our colleagues who, in their work, encounter the necessity of using tables of elliptic integrals calculated with accuracy to the eighth significant digit.

Let us briefly derive the equation of the bent axis of the rod.

Consider one half of a rod having free ends. In view of the symmetry of the rod, its half (Fig. 2) will be considered as a rod fixed at one (in the drawing—the lower) end and having at the free end a load by a concentrated force \(P\). The force \(P\) is a longitudinal, central, compressive force.

With the chosen coordinates, with origin at the point of fixing, and with the notation adopted in Fig. 2, the differential equation of the bent axis of the rod, in exact form, is represented as follows:

\[ \frac{1}{\rho} = \frac{y''}{(1+y'^2)^{3/2}} = \frac{P}{EI}(f-y) = \alpha^2(f-y). \tag{1} \]

Multiplying both sides of the equality by \(y'\):

\[ \frac{y'y''}{(1+y'^2)^{3/2}}=\alpha^2(f-y)y' \]

and, integrating, we obtain:

\[ -\frac{1}{(1+y'^2)^{1/2}} = -\frac{\alpha^2}{2}(f-y)^2+C \]

By the notation \(y'=\operatorname{tg}\beta\), we obtain:

\[ -\cos\beta = -\frac{\alpha^2}{2}(f-y)^2+C \qquad\ldots\ldots\ldots\tag{2} \]

The maximum angle of rotation of the axis of the rod at the free end will be denoted by \(\beta_0\). Remembering that \(\alpha^2(f-y)=\dfrac{1}{\rho}\) and using the substitution:

\[ \text{for } y=f;\qquad \beta=\beta_0 \]

Fig. 2.

we bring equation (2) to the form:

\[ \cos \beta - \cos \beta_0 = \frac{1}{2a^2\rho^2} \]

whence:

\[ \rho = \frac{1}{a\sqrt{2(\cos \beta - \cos \beta_0)}} \qquad \ldots\ldots\ldots (3) \]

The radius of curvature can be expressed through the derivatives:

\[ \rho = \frac{ds}{d\beta} = \frac{dx}{\cos \beta\, d\beta} = \frac{dy}{\sin \beta\, d\beta}; \]

Making in (3) substitutions of all three expressions for the radius, we obtain three equations:

\[ dy = \frac{\sin \beta\, d\beta}{a\sqrt{2(\cos \beta - \cos \beta_0)}} \qquad \ldots\ldots\ldots (4) \]

\[ dx = \frac{\cos \beta\, d\beta}{a\sqrt{2(\cos \beta - \cos \beta_0)}} \qquad \ldots\ldots\ldots (5) \]

\[ ds = \frac{d\beta}{a\sqrt{2(\cos \beta - \cos \beta_0)}} \qquad \ldots\ldots\ldots (6) \]

For the numerical calculation of the integral values obtained below, it is more convenient to express the functions \(y\), \(x\), and \(s\) in the angles

\[ \frac{\beta}{2} \quad \text{and} \quad \frac{\beta_0}{2}. \]

Since:

\[ \sin \beta = 2\sin \frac{\beta}{2}\sqrt{1 - \sin^2 \frac{\beta}{2}} \]

and

\[ \cos \beta = 1 - 2\sin^2 \frac{\beta}{2}, \]

then, setting:

\[ \sin \frac{\beta_0}{2} = k \quad \text{and} \quad \sin \frac{\beta}{2} = k\sin \varphi, \]

after substitution and transformations, we obtain:

\[ dy = \frac{2k}{a}\sin \varphi\, d\varphi \]

\[ dx = \frac{1}{a}\left(2\sqrt{1 - k^2\sin^2 \varphi} - \frac{1}{\sqrt{1 - k^2\sin^2 \varphi}}\right)d\varphi \]

\[ ds = \frac{1}{a\sqrt{1 - k^2\sin^2 \varphi}}\, d\varphi \]

Neglecting the deformation due to compression, which, for small deflection arrows, is not difficult to take into account separately, we shall have the following for the setting of the limits:

\[ \text{for } \quad y=0;\ x=0;\ s=0;\ \beta=0;\ \varphi=0; \]

\[ \text{for } \quad y=f;\ x=\frac{l-\Delta l}{2};\ s=\frac{l}{2};\ \beta=\beta_0;\ \varphi=\frac{\pi}{2}. \]

Therefore:

\[ f=\frac{2k}{\alpha}\int_{0}^{\frac{\pi}{2}}\sin\varphi\,d\varphi \qquad \ldots\ldots\ldots (7) \]

\[ \frac{l-\Delta l}{2} = \frac{1}{\alpha} \left[ 2\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^{2}\sin^{2}\varphi}\,d\varphi - \int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1-k^{2}\sin^{2}\varphi}} \right] \qquad \ldots (8) \]

\[ \frac{l}{2} = \frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1-k^{2}\sin^{2}\varphi}} \qquad \ldots\ldots (9) \]

Integral (7) is evaluated simply and gives:

\[ f=\frac{2k}{\alpha} \qquad \ldots\ldots\ldots (10) \]

Integrals (8) and (9) are complete elliptic integrals of the first and second kinds in normal trigonometric form. In the table given below, their numerical values are given for various \(k\), from \(\beta_0=0\) to \(\beta_0=180^\circ\), in steps of \(1^\circ\).

Thus:

\[ \frac{l-\Delta l}{2} = \frac{1}{\alpha}\,[2E(k)-F(k)] \qquad \ldots\ldots (11) \]

and

\[ \frac{l}{2} = \frac{1}{\alpha}F(k) \qquad \ldots\ldots (12) \]

Solving the last three equations simultaneously and having the expression for the critical force for the given column in the form

\[ P_k=\frac{\pi^2 EI}{l^2}, \]

we obtain the following equations:

For the relative load:

\[ \frac{P}{P_k} = \frac{[2F(k)]^2}{\pi^2} \qquad \ldots\ldots (13) \]

For the relative deflection arrow:

\[ \frac{f}{l} = \frac{k}{F(k)} \qquad \ldots\ldots (14) \]

relative settlement of the ends:

\[ \frac{\Delta l}{l}=2\left[1-\frac{E(k)}{F(k)}\right]\ldots\ldots\ldots (15) \]

In addition to these quantities, let us determine also the value of the maximum bending moment. The latter is also obtained in the form of a ratio:

\[ \frac{M}{P_k l}=\frac{P f}{P_k l}=\frac{4k}{\pi^2}F(k)\ldots\ldots\ldots (16) \]

The tables given below contain:

I—Table—numerical values of the integrals:

\[ F(k)=\int_0^{\frac{\pi}{2}}\frac{d\varphi}{\sqrt{1-k^2\sin^2\varphi}} \quad\text{and}\quad E(k)=\int_0^{\frac{\pi}{2}}\sqrt{1-\sin^2\varphi}\,d\varphi, \]

given for various \(k\) as \(\frac{\beta_0}{2}\) varies from \(0^\circ\) to \(90^\circ\) in steps of \(0^\circ 5\).

II—Table—numerical values of the relative quantities given by equations: (13), (14), (15), (16).

I—Table of numerical values of the integrals:

\[ F(k)=\int_0^{\frac{\pi}{2}}\frac{d\varphi}{\sqrt{1-k^2\sin\varphi}} \quad\text{and}\quad E(k)=\int_0^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2\varphi}\,d\varphi \]

II—Table of values of relative quantities: load, deflection, settlement of the ends, and maximum bending moment

Table II gives values for the relative quantities: load, deflection arrow, settlement of the ends, and the maximum bending moment only up to $\dfrac{\beta_0}{2}=30^\circ$ (i.e., up to $\beta_0=60^\circ$). Further calculation of the quantities mentioned above is of no practical interest, since already at the value of $\beta_0$ limiting for this table the column has a very large deflection. Drawing 3 shows how the column is deformed when the angle $\beta_0$ reaches $60^\circ$.

Fig. 3.

On the basis of the numerical material of the above tables, the graph shown in drawing 4 has been compiled. For the plotted functions, among others, the function $\dfrac{f}{l}$ is given at double scale.

Graph of the functions: $\dfrac{\Delta l}{l}$, $\dfrac{f}{l}$, $\dfrac{P}{P_k}$ and $\dfrac{M}{P_k l}$.

Using Table II, let us construct the trajectory of the ends of the column. For this it is sufficient to take

(Graph)

Visible labels:

• \(\dfrac{T}{P_k}\)
• \(\dfrac{M}{P_k l}\)
• \(\dfrac{\Delta l}{l}\)
• \(\dfrac{f}{l}\)
• \(0\)
• \(1\)
• \(2\)
• \(0^\circ\)
• \(90^\circ\)
• \(180^\circ\)

one half of the column, as is shown in Fig. 5. From the figure it is clear that the trajectory of the end of the column differs very little from an arc of a circle whose center is placed at a certain point on the initial axis of the column. The agreement with an arc of a circle is remarkable not only because it is achieved (at least to the eye) satisfactorily, but also because it is maintained over a large deformation path. Such agreement of the circle with the trajectory of the end of the column could have been achieved with still greater success if the center of the circle were placed outside the \(X\) axis. However, finding by the method of least squares the “best” center led to a negligible value of the displacement of the center from the axis and, at the same time, complicated, so to speak quite disproportionately to the refinement obtained, the expression of the relation between the deformations.

Fig. 5.

Figure 6. Placing the origin of coordinates at the end of the column, we shall there have the point of tangency of the circle. Its equation will be:

\[ y^2+(\rho-x)^2=\rho^2 \]

Fig. 6.

or:

\[ \rho=\frac{y^2+x^2}{2x} \]

Replacing the coordinates by the corresponding deformation quantities, we obtain:

\[ \frac{\rho}{l} = \frac{\left(\frac{f}{l}\right)^2+\left(\frac{\Delta l}{2l}\right)^2}{2\frac{\Delta l}{2c}} = \frac{4\left(\frac{f}{l}\right)^2+\left(\frac{\Delta l}{l}\right)^2}{4\frac{\Delta l}{c}} \qquad\ldots\ldots (17) \]

Let us denote the quantity: \(\frac{\rho}{l}=\rho_0\) and call it the relative radius of the trajectory of the end of the column under longitudinal bending.

To determine the relative radius of the trajectory, which, as is now clear, is a characteristic of the deformation of longitudinal bending, we shall not resort to the method of least squares, since then we would have to specify in advance the intervals of the arcs within whose limits one or another, most accurate, radius is obtained. Instead, using equation (17), we shall compile a table of relative radii (see p. 14). This will be better, since, on the one hand, it will give an idea of the changes in this quantity depending on the course of the deformation and, on the other, it will give an approximate but reliable numerical value for it within the selected interval.

As is evident from consideration of Table III, the value of the relative radius of the trajectory, within the angle \(\beta_0\) from \(0^\circ\) to \(130^\circ\), is fairly stable, and the relative radius, as the arithmetic mean of all the figures in the given interval, is given by:

\[ \rho_0=0.40423 \]

We take:

\[ \rho_0=0.4=\frac{2}{5} \]

Thus, for broad ranges of variation of the elastic deformation of longitudinal bending of the column, we establish the following very simple formula relating the relative deformations:

\[ 4f_0^2+\varepsilon_b^2=4\rho_0\varepsilon_b \qquad\ldots\ldots (18) \]

where \(f_0\) denotes the relative deflection.

For small deformations we may write simply: \((\varepsilon_b^2\simeq 0)\)

\[ f_0^2=\rho_0\varepsilon_b\simeq 0.4\,\varepsilon_b \qquad\ldots\ldots (19) \]

III—Table of Relative Radii

To simplify the computational operations, we must also have a simple dependence between the deformations and the load. For this purpose we shall compute the values of the following quantity:

\[ \frac{P - P_k}{P_k} = P_0 - 1. \]

And we shall plot the graph (drawing 6) of the variation of $\xi_b$ with the variation of $P_0 - 1$.

From the graph just presented (drawing 7) we see that the dependence between $\xi_b$ and $P_0 - 1$ is close to linear. Therefore, within the limits from $\beta_0 = 0^\circ$ to $\beta_0 = 30^\circ$, i.e., over a range of bending deformations sufficient for practical purposes and with a sufficient degree of accuracy, we may regard this dependence as expressed analytically as follows:

\[ \xi_b = 2(P_0 - 1) \ldots\ldots\ldots\ldots (20) \]

The maximum value of the error will occur at \(\beta_0 = 30^\circ\). Its magnitude is:

\[ \eta = \frac{\varepsilon_b - 2(P_0 - 1)}{\Sigma_b} = \frac{2 \cdot 0.03512087 - 0.06756812}{0.06756812} = 0.0397 \]

To determine the growth of the error with increasing \(\beta_0\), the graph below is given (Drawing 8).

Fig. 8.

Calculation of the column under a static load.

The deformation of the column, within the range from $\beta_0 = 0^\circ$ to $\beta_0 = 30^\circ$, can be expressed by the diagram shown in Fig. 9. When the load is increased from $0$ to the value $P_k$, the shortening of the ends of the column will occur only from the compressive deformation, which at the critical load will reach the value $\Delta l_k$. With a further increase of the load, the increase in the shortening of the ends will proceed considerably faster, since the shortening due to longitudinal bending will be added to the compressive deformation. Therefore, since we do not neglect the compressive deformation, we must consider separately both phases of the compressive deformation; we must consider

Fig. 9.

to treat separately both phases of deformation: one within the magnitude of the critical force and the other beyond it.

When a column is loaded by a force greater than the critical one, the stress in the cross section will be composed of the compressive stress and the bending stress. The greatest stress, which we must equate to the design stress, will occur in the cross section at the middle of the column, at the place of the greatest deflection.

Denoting by \(R\) the design stress, by \(F\) the area of the cross section, and by \(W\) the minimum section modulus of the latter, we shall write:

\[ R = \frac{P}{F} + \frac{Pf}{W} \]

Transform this equation as follows:

\[ R = \frac{P}{F}\left(1 + \frac{Ff}{W}\right) \]

Calling \(e\) the distance to the most remote fiber and having:

\[ W = \frac{I}{e} = \frac{Fi^2}{e}, \]

where \(i\) is the corresponding radius of inertia, we obtain:

\[ R = \frac{P}{F}\left[1 + \frac{ef}{i^2}\right] \ . . . . . . . . . . (21) \]

Let us now express the allowable stress through one of the variables characterizing the phenomenon of longitudinal bending, i.e., through \(\varepsilon_b\), \(f_o\), and \(P_o\). In doing so we obtain:

or

\[ R = \frac{P_k}{F}\left(1 + \frac{\varepsilon_b}{2}\right)\left[1 + \frac{el}{i^2}\sqrt{\rho_0 \varepsilon_b}\right] \ . . . . . (22) \]

or

\[ R = \frac{P_k}{F}\left(1 + \frac{f_0^2}{\rho_0}\right)\left[1 + \frac{el}{i^2} f_0\right] \ . . . . . (23) \]

or

\[ R = \frac{P_k}{F} P_0 \left[1 + \frac{el}{i^2}\sqrt{\rho_0(P_0 - 1)}\right] \ . . . . . (24) \]

Example of calculation.

An iron round column of solid cross section has, for a length \(l = 4\ m\), a critical force \(P_k = 15\ kg\). Determine the stress when a load equal to \(1.001\) times the critical load is applied.

We determine from the formula for the critical force the moment of inertia of the cross section of the column.

\[ I=\frac{P_k l^2}{\pi^2 E}=\frac{15\cdot(400)^2}{\pi^2\cdot 2\cdot 10^6}=0.1217\ \mathrm{cm}^4 \]

The diameter of the section will be:

\[ d=\sqrt[4]{\frac{64I}{\pi}}=\sqrt[4]{\frac{64\cdot 0.1217}{\pi}}=\sqrt[4]{2.48}=1.255\ \mathrm{cm} \]

The area of the section:

\[ F=\frac{\pi d^2}{4}=\frac{\pi(1.255)^2}{4}=1.237\ \mathrm{cm}^2 \]

and

\[ i^2=\frac{I}{F}=\frac{0.1217}{1.237}=0.0983\ \mathrm{cm}^2;\quad e=\frac{d}{2}=0.627\ \mathrm{cm} \]

The critical stress will be:

\[ \sigma_k=\frac{P_k}{F}=\frac{15}{1.237}=12.15\ \mathrm{kg}/\mathrm{cm}^2 \]

The stress due to the action of a force equal to \(1.001P_k\) will be: (24)

\[ R=\frac{15}{1.237}\,1.001\left[1+\frac{0.627\cdot 400}{0.0983}\sqrt{0.8\cdot 0.001}\right]=874\ \mathrm{kg}/\mathrm{cm}^2. \]

Let us determine the magnitude of the deflection, which will occur under the given loading conditions. For this we use formula (23), in which we may make a certain simplification. In view of the slight excess of the applied load over the critical force, we are entitled to expect that the quantity:

\[ \frac{f_0^2}{\rho_0}=2.5\left(\frac{f}{l}\right)^2 \]

will be very small in comparison with unity, and in the present case it may freely be neglected.

From the formula thus simplified we obtain:

\[ f=\frac{i^2}{e}\left[\frac{RF}{P_k}-1\right]=\frac{0.0983}{0.627}\left[\frac{874\cdot 1.237}{15}-1\right]=\frac{0.983\cdot 71.4}{0.627}=11.2\ \mathrm{cm} \]

The relative deflection will be

\[ f_0=\frac{f}{l}=\frac{11.2}{400}\simeq \frac{1}{36}. \]

Calculation of a Column under a Dynamic Load

The potential energy will be expressed by the area of the diagram, i.e., by the sum of the areas of the triangle and the trapezoid. According to the notation adopted in drawing 7, we write:

\[ U=\frac{P_k\Delta l_k}{2}+\frac{P_m+P_k}{2}(\Delta l_m-\Delta l_k) \]

Since:

\[ \varepsilon_m=\frac{\Delta l_m}{l}=\varepsilon_d+\varepsilon_b=2(P_0-1)+\frac{P_0P_k}{EF}=P_0(2+\varepsilon_k)-2 \]

then:

\[ U=\frac{P_kl}{2}\left[(P_0^2-1)(2+\varepsilon_k)+\varepsilon_k\right] \]

In the case where the bending deformation is very small, we have the right to assume that the bending occurred under the action of the constant load \(P_k\), since the change of load, for small deflections, is negligible. Then the expression for the potential energy takes the form:

\[ U=\frac{P_k\Delta l_k}{2}+P_k(\Delta l_m-\Delta l_k) =\frac{P_kl}{2}\left[4(P_0-1)+(2P_0-1)\varepsilon_k\right]. \tag{25} \]

Let us consider the impact action of a load on a column. We shall analyze the simplest case—a free column. By a free column we shall mean one whose bending is not constrained by any lateral obstacles. It will deform as a column with hinged ends.

In order that, upon impact, the column begin to bend, it is necessary that the magnitude of the impact be no less than:

\[ U_{\min}=\frac{P_k\Delta l_k}{2}=\frac{\sigma_k^2 V}{2E} \]

Drawing 10 represents a superposition of the deformation diagram of the column and the scheme of this deformation under the influence of a certain moving mass \(Q\).

Upon contact of the moving mass with the end of the column (1), the latter will first begin to be compressed. The compressive deformation will increase according to the law of the straight line (2):

\[ x_1=\frac{P_{x1}l}{EF}\ldots\ldots\ldots (26) \]

where \(P_{x1}\) is the variable dynamic load.

The increase of the compressive deformation will continue until the dynamic load reaches the magnitude of the critical force of the column. The value of the absolute compressive deformation at this moment will be (3):

\[ x_{1\max}=\frac{P_kl}{EF}=\Delta l_k\ldots\ldots\ldots (27) \]

The further motion of the mass will cause longitudinal bending, and the magnitude of the displacement of the end of the column is found from (4):

\[ \frac{P_m - P_{x_2}}{P_m - P_k} = \frac{\Delta l_m - x_2}{\Delta l_m - \Delta l_k}; \]

\[ x_2 = \frac{\Delta l_m - \Delta l_k}{P_m - P_k} \]

\[ P_{x_2} = \frac{P_k \Delta l_m - P_m \Delta l_k}{P_m - P_k} \tag{28} \]

where \(P_{x_2}\) is the variable dynamic load in the second phase of deformation.

Fig. 10.

Together with the increasing deformation, the velocity of motion of the mass decreases and, finally, becomes equal to zero, which will correspond to the greatest deformation and the greatest dynamic force (5).

Let us write the equation of motion of the mass for the first phase of deformation. Let us agree that we know: the velocity of the mass at the moment of its contact with the end of the column, the acceleration at that moment, and the law of variation of this acceleration as the result of the action on the mass of external constraints (of course, the column is not included here as a constraint).

Let the moving mass have weight \(Q\) and have velocity, at the instant of contact, \(V_0\), and acceleration \(j_0\). The acceleration, dependent on external constraints and, generally speaking, variable, shall be denoted by \(j\).

The force of resistance of the strut to the motion of the mass is determined from (26). We write the equation of motion of the mass:

\[ \frac{d^2 x_1}{dt} = j-\frac{P_{x1}g}{Q} = j-\frac{gEF}{Ql}x_1 . \]

Since the quantity

\[ \frac{Ql}{EF} \]

is the compressive deformation from the static loading of the strut by the load \(Q\), denoting this quantity by \(\Delta l_c\), we obtain:

\[ \frac{d^2 x_1}{dt^2} = j-\frac{g}{\Delta l_c}x_1 = j-\frac{g}{l\varepsilon_c}x_1 . \]

We integrate the equation of motion of the mass within the limits of the first phase, i.e., up to the moment when the resistance force of the strut reaches the value of its critical force \(P_k\).

In this case \(x_1\) will vary from \(0\) to \(\Delta l_k\); \(\dfrac{dx_1}{dt}\) will vary from \(V_0\) to \(V_k\), where \(V_k\) is the velocity at the end of the first phase.

\[ \frac{1}{2} \left[ \left(\frac{dx_1}{dt}\right)^2 \right]_{V_0}^{V_k} = \int_0^{\Delta l_k} j\,dx_1 - \frac{g}{l\varepsilon_c} \left[ \frac{x_1^2}{2} \right]_0^{\Delta l_k}, \]

or

\[ V_k^2 - V_0^2 = 2\int_0^{\Delta l_k} j\,dx_1 - \frac{g l \varepsilon_k^2}{\varepsilon_c} \qquad \ldots\ldots (29) \]

The resistance force of the strut during motion in the second phase is determined from (22):

\[ P_{x2} = \frac{P_m-P_k}{\Delta l_m-\Delta l_k}x^2 + \frac{P_k\Delta l_m-P_m\Delta l_k}{\Delta l_m-\Delta l_k}. \]

We write the equation of motion of the mass:

\[ \frac{d^2x_2}{dt^2} = j-\frac{gP_{x2}}{Q} = j-\frac{g}{Q} \left[ \frac{P_m-P_k}{\Delta l_m-\Delta l_k}x_2 - \frac{P_m\Delta l_k-P_k\Delta l_m}{\Delta l_m-\Delta l_m} \right]. \]

Since we have:

\[ \varepsilon_m = 2(P_0-1)+P_0\varepsilon_k, \]

then:

\[ \frac{\Delta l_d}{\Delta l_k} = \frac{\varepsilon_d}{\varepsilon_k} = \frac{P_m}{P_k}. \]

The equation of motion of the mass, in the second phase, will be:

\[ \frac{d^2 x_2}{d t^2} = j-\frac{g\varepsilon_k}{\varepsilon_c(2+\varepsilon_k)} \left(2+\frac{x_2}{l}\right). \]

We integrate the equation over the limits of variation of \(x_2\), from \(\Delta l_k\) to \(\Delta l_m\), and over the limits of variation of \(V\), from \(V_k\) to \(0\), when all the kinetic energy of the moving mass is exhausted and the latter comes to rest.

\[ \frac{1}{2} \left| \left(\frac{d x_2}{d t}\right)^2 \right|_{V_k}^{0} = \int_{\Delta l_k}^{\Delta l_m} j\,d x^2 - \left| \frac{g\varepsilon_k}{\varepsilon_c(2+\varepsilon_k)} \left(2x_2+\frac{x_2^2}{2l}\right) \right|_{\Delta l_k}^{\Delta l_m} \]

or

\[ - V_k^2 = 2\int_{\Delta l_k}^{\Delta l_m} j\,d x_2 - \frac{2g\varepsilon_k(\Delta l_m-\Delta l_k)} {\varepsilon_c(2+\varepsilon_k)} \left(2+\frac{\Delta l_m+\Delta l_k}{2l}\right). \tag{30} \]

Now summing, respectively (pairwise), the left- and right-hand sides of the equalities obtained in (23) and (24), we obtain:

\[ - V_0^2 = 2\int_{0}^{\Delta l_m} j\,d x - \frac{g l\varepsilon_k}{\varepsilon_c} \left[(2+\varepsilon_k)P_0^2-2\right]. \tag{31} \]

Since in the present case we are considering a strut having a considerable bending deformation, in equation (25), even when taken in its general form, one may make a simplification by neglecting, on the right-hand side of the equality, the quantity \(\varepsilon_k\), as small in comparison with two, and write:

\[ V_0^2 = \frac{2g\varepsilon_k l}{\varepsilon_c}(P_0^2-1) - 2\int_{0}^{\Delta l_m} j\,d x \tag{32} \]

From this equation one can determine the magnitude of the greatest dynamic load developed by the mass at the end of the impact. The quantity \(P_0\) cannot be expressed in general form, owing to the fact that the upper limit of the integral

\[ \int_{0}^{\Delta l_m} j\,d x \]

is itself a function of \(P_0\). Knowing, however, the law of variation of the function \(j\) along the path of motion of the mass, we can always determine the value or compute the numerical magnitude of the integral.

Let us consider the simplest, but at the same time the most important, special cases.

Case I: \(j=j_0=\mathrm{const}\).

Then:

\[ \int_0^{\Delta l_m} j\,dx = j_0 \int_0^{\Delta l_m} dx = j_0 \Delta l_m = j_0 l \varepsilon_m = j_0 l \{2(P_0-1)+P_0\varepsilon_k\} \]

Equation (26) becomes:

\[ V_0^2= \frac{2gl\varepsilon_k}{\varepsilon_c}(P_0^2-1) - 2j_0l\{2(P_0-1)+P_0\varepsilon_k\}\, . \tag{33} \]

whence we obtain the quadratic equation with respect to \(P_0\):

\[ P_0^2-\frac{2Qj_0}{P_k g}P_0 - \left( \frac{V_0Q}{2gP_kl} + 1 - \frac{2Qj_0}{P_kg} \right) =0. \]

Its solution gives the magnitude of the greatest dynamic load, in relative units:

\[ P_0= \frac{Qj_0}{P_kg} \pm \sqrt{ \left(1-\frac{Qj_0}{P_kg}\right)^2 + \frac{QV_0^2}{2P_kgl} } \ . . . \tag{34} \]

It is necessary to remember that \(P_m=P_kP_0\).

Case II. \(j_0=g\)

This case corresponds to the free fall of the striking mass.

\[ P_0= \frac{Q}{P_k} + \sqrt{ \left(1-\frac{Q}{P_k}\right)^2 + \frac{QH}{P_kl} } \ . . . . \tag{35} \]

Here \(H\) is the height of fall of the mass.

Case III. \(j=0\).

\[ P_0= \sqrt{ 1+\frac{QV_0^2}{2P_kgl} } \ . . . . . . . . \tag{36} \]

Example of calculation.

A thin round pine pile is driven by a freely falling ram. Determine the necessary impact power and the height of fall of the ram, calculating for the case in which the pile bears against an obstacle in the ground.

The pile has:

\[ l=12\ m,\quad d=25\ cm,\quad E=10^5\ kg/cm^2. \]

Weight of the ram:

\[ Q=200\ kg \]

Allowable stress during driving:

\[ R=\ kg/cm^2. \]

Let us first determine all the auxiliary quantities needed in the calculation.

\[ F=\frac{\pi d^2}{4}=\frac{\pi\cdot 25^2}{4}=490\ \mathrm{cm}^2, \]

\[ I=\frac{\pi d^4}{64}=\frac{\pi\cdot 25^4}{64}=19150\ \mathrm{cm}^4, \]

\[ e=\frac{d}{2}=12.5\ \mathrm{cm}, \]

\[ P_k=\frac{\pi^2 E I}{l^2} =\frac{\pi^2\cdot 10^5\cdot 19150}{1200^2} =13100\ \mathrm{kg}. \]

Taking the strength equation (24), we find from it the value \(P_0\):

\[ 200=\frac{13100}{490}P_0\left(1+ \frac{12.5\cdot 1200\cdot 490}{19150}\sqrt{0.8(P_0-1)}\right) \]

Hence we obtain:

\[ P_0=1.00035 \]

To determine the height from which the ram must fall, we apply equation (35), from which we obtain:

\[ H=\frac{(P_0-1)l}{Q}\{P_k(P_0+1)-2Q\}= \]

\[ =\frac{0.00035\cdot 1200}{200}\{13100\cdot 2.00035-2.200\}=54.1\ \mathrm{cm}. \]

Before obtaining those comparatively simple relations which have been derived in the present work, in the example just given we would have been forced to carry out the calculation for impact compression of piles by the formula:

\[ R_{\partial}=\sigma_c+\sqrt{\sigma_c^2+\frac{2\sigma_c EH}{l}}\ldots\ldots\ldots (37) \]

Here the subscripts \(c\) and \(\partial\) denote, respectively, the static and dynamic stresses due to the action of the load \(Q\).

Since \(\sigma_c\) is very small, namely equal to

\[ \sigma_c=\frac{Q}{F}=\frac{200}{490}=0.408\ \mathrm{kg/cm^2}, \]

one may simplify equation (37) and write:

\[ R_{\partial}=\sqrt{\frac{2\sigma_c EH}{l}} \]

Hence we determine \(H\):

\[ H=\frac{R_0^2 l}{2E\sigma_c} =\frac{200^2\cdot1200}{2\cdot10^5\cdot0.408} =589\ \mathrm{cm}. \]

Thus, a calculation without taking longitudinal bending into account gives a drop height of the ram approximately 11 times greater than is necessary.

Let us determine what stress is obtained in this case. We find the value \(P_0\) from equation (35):

\[ P_0=\frac{200}{13100} +\sqrt{\left(1-\frac{200}{13100}\right)^2 +\frac{200\cdot589}{13100\cdot1200}} +1.0063. \]

The stress in the pile, when the ram falls from a height \(H=5.89\,m\), is determined from (24):

\[ R=\frac{13100}{490}\,1.0063 \left(1+\frac{12.5\cdot1200\cdot490}{19150} \sqrt{0.8\cdot0.0063}\right) =1020\ kg/cm^2. \]

The answer indicates that the pile would be destroyed by bending.

SUMMARY.

1. An exact solution of the problem of a support deformed by stresses exceeding the critical yield by their values leads to an extremely simple relation between loads and deformations.

2. The ends of the support circumscribe circular arcs, having a radius:

\[ \rho=0.404\,l. \]

1. The load and the deformation (set of the ends of the support) are related by the following equation:

\[ 2\,\frac{P-P_k}{P_k}=\frac{\Delta l}{l}. \]

1. The obtained relations allow one to make a calculation of strongly deformed rods-supports both under the static and the dynamic action of the load.