Prof. G. V. Trapeznikov.

ON THE CALCULATION OF THE IMPACT ACTION OF A LOAD

Two formulas: one, for the maximum deformation due to the action of a load falling from a height \(H\):

\[ \Delta l_d=\Delta l_c+\sqrt{\Delta l_c^2+2H\Delta l_c} \]

or:

\[ \frac{\Delta l_d}{\Delta l_c}=1+\sqrt{1+2\frac{H}{\Delta l_c}} \]

and the other, for the maximum stresses under the same conditions:

\[ \sigma_d=\sigma_c+\sqrt{\sigma_c^2+2\frac{H}{l}E\sigma_c} \]

or:

\[ \frac{\sigma_d}{\sigma_c}=1+\sqrt{1+2\frac{HE}{l\sigma_c}} \]

usually exhaust the entire calculation for the impact action of a load.

These formulas are applicable, in calculations for impact, to simple rods subjected to tension or compression. The functional composition of the formulas that may be given for calculating impact under torsion and bending will be the same as in the formulas just written. The entire change will consist in replacing some physical characteristics of the material and geometrical characteristics of the deformable body by others, and stresses of one kind by others.

Under production conditions, the picture of impact is very often, not to say always, much more complicated than what we have in the simple fall of a load. The impact during the recoil of the barrel of an artillery piece, the running of a vehicle (railcar, automobile) against an obstacle, the operation of the valves of a steam engine or an internal-combustion engine—all these examples, and a vast number of others which there is neither need nor possibility to enumerate, indicate the inapplicability of the formulas given at the beginning.

When calculating parts or systems for the impact action of a load, the designer is, in most cases, compelled to create complex analytical expressions in order to exhaust—

by considering the picture of the impact, he had the possibility of determining the effect of the latter.

It is, of course, impossible to encompass (not to mention within the scope of this little article) all cases of the impact action of a load that practice presents to us; it is impossible because individual cases have their own specific character, which precludes the possibility of their exact classification. Thus, for example, the variant of impact in which two, and moreover both deformable, systems with numerically comparable masses collide, under an eccentric application of the impact, can be solved only for a particular case, and such a solution would require an entire book. Therefore, in a general form one can consider only such an impact in which the mass of the deformable system is small in comparison with the striking mass, and only one of the participants in the impact is movable: either the mass or the system.

The purpose of the present article is to give, in a general form, the course of calculation of the deformation of an elastic system under the action upon it of a moving mass. In doing so, we shall not set ourselves particular types of deformations, but shall consider an elastic system in general, in which, within the limits of applicability of Hooke’s proportionality, there will always be the relation:

\[ \alpha \delta = P, \]

where by the letter \(\delta\)*) we shall mean the magnitude of the absolute deformation of the system under the load \(P\). In other words, \(\delta\) is the displacement of the point of application of the load \(P\) to the system deformed by it, in the direction of this load.

The coefficient \(\alpha\) is the factor of proportionality between \(\delta\) and \(P\), and, for simple deformations, has the following functional composition:

Tension and compression

\[ \alpha = \frac{P}{\Delta l} = \frac{EF}{l}. \]

Torsion

\[ \alpha = \frac{M}{s} = \frac{2GJ_p}{al}. \]

Bending

\[ \alpha = \frac{P}{f} = \frac{EJ_z}{\gamma l^3}. \]

In the second formula, \(a\) denotes the magnitude of the linear displacement of the force \(P\) from the couple \(M = Pa\). In the third formula the coefficient \(\gamma\) depends on the manner of loading of the beam and on the conditions of fastening of its supports.

*) Here the standard notation of deformations has deliberately been violated, and the absolute deformation of the system has been denoted by the letter \(\delta\), because we are considering the general case. In particular cases it may be: \(\delta = \Delta l\), \(\delta = f\), etc.

Under the condition of considering the impact action of a mass on a system taken in general form, we have no need to obtain a separate formula for the dynamic stress. Within the elastic range of the material of the system, always:

\[ k=\beta P=\alpha\beta\delta^{**}) \]

where \(k\) is the stress and \(\beta\) is the coefficient of proportionality corresponding to the type of system, the conditions of application of the load, and the position of that element of the system where \(k\) is determined.

Therefore we shall confine ourselves to finding \(\delta_d\)—the absolute dynamic deformation at the point of application of the impact. As for determining the value of the coefficient \(\alpha\) for each particular case, in complex systems we must be able, with the necessary accuracy, to obtain it analytically, graphically, or experimentally.

In what follows we shall denote the absolute dynamic deformation by \(\delta_d\), and the corresponding static deformation by \(\delta_c\).

Let us proceed to the calculation. Let the moving mass, having weight \(Q\), at the moment of contact with the system possess velocity \(v_0\) and acceleration \(j_0\). For an impact calculation, we must know the magnitude \(v_0\) and the law of variation of \(j\) over the interval: from the moment of contact of the mass with the system to the moment when the mass has lost all of its kinetic energy, which is expended on the deformation of the system. The proper acceleration of the mass \(j\) must be determined by the conditions of its constraints during motion (drive from a machine, guides with friction, a viscous medium), and among these constraints the “mass–system” constraint is not included. This acceleration, depending on the circumstances and conditions of motion of the mass, may be a function of path, instantaneous velocity, or the time elapsed from the specified initial instant of reckoning of the latter. Therefore, in order to consider the question in a more general form, it is necessary to solve the problem in three variants, according as \(j\) is given as a function of which argument. To distinguish the variants we shall use the indices: path—\(x\), velocity—\(v\), and time—\(t\). The acceleration will be written:

\[ j_x,\quad j_v,\quad j_t. \]

First variant \(j=j_x\).

Taking as the origins of time and path the first instant of contact between the mass and the deforming system, we write the differential equation of motion of the mass:

\[ m\frac{d^2x}{dt^2}=\frac{Q}{g}j_x-P_x. \]

\[ ^{**})\ \text{The same remark also applies to the designation of stress by the letter } k. \text{ Depending on the element of the system, it may turn out that:} \]

\[ k=\sigma_z,\quad k=\sigma_b,\quad k=\tau,\ \text{etc.} \]

Here \(m\) is the magnitude of the mass, equal to: \(m=\dfrac{Q}{g}\), and \(P_x\) is the instantaneous elastic reaction of the system, equal to: \(P_x=\alpha x\).

The equation may be reduced to the following form:

\[ \frac{d^2x}{dt^2}=j_x-\frac{\alpha x g}{Q}. \]

Since, under the static action of the load \(Q\), we have:

\[ Q=\alpha \partial_c, \]

the final and simpler form of the differential equation will be:

\[ \frac{d^2x}{dt^2}=j_x-\frac{g}{\partial_c}\,x. \]

or

\[ \left(\frac{dx}{dt}\right)d\left(\frac{dx}{dt}\right) = j_x\,dx-\frac{g}{\partial_c}\,x\,dx. \]

Integration gives:

\[ \frac{1}{2}\left(\frac{dx}{dt}\right)^2 = \int j_x dx-\frac{g}{2\partial_c}\,x+C_1. \ . . . . . . . . . (A) \]

The terminal conditions are as follows:

\[ \text{when } x=0;\quad \frac{dx}{dt}=v=v_0; \]

\[ \text{when } x=\partial_{\partial};\quad \frac{dx}{dt}=0. \]

Substitution of the quantities corresponding to the terminal conditions gives:

\[ \partial_{\partial}^{\,2} - \frac{2\partial_c}{g} \int_{0}^{\partial_{\partial}} j_x dx - \frac{\partial_c v_0^{\,2}}{g} =0 \ . . . . . . . . . (1) \]

This equation must be solved with respect to \(\partial_{\partial}\). Since \(j_x\) is a function of the path, the integral will be a function of \(\partial_{\partial}\). The greater or lesser difficulty of solving equation (1) depends on the form of the function \(j_x\). In calculations for impact, it is often necessary to know the time during which the deformation has reached its maximum value, since this determines the period of oscillation of the system. This time, denoting its value by the letter \(T\), is determined by further integration.

We have:

\[ dt=\frac{dx}{\sqrt{\,2C_1+2\int_0^x j_x\,dx-\frac{g}{\delta_c}x^2\,}}. \]

But, from \((A')\) we obtain:

\[ 2C_1=v_0^2-2\int_0^0 j_x\,dx. \]

The time required for obtaining the maximum value of the deformation is:

\[ T=\int_0^{\delta_d} \frac{dx}{\sqrt{\,v_0^2+2\int_0^x j_x\,dx-\frac{g}{\delta_c}x^2\,}} \ldots \ldots \ldots (2) \]

And in this case, the difficulty of solving the integral depends entirely on the form of the function \(j_x\).

Particular cases.

If the deformation of the system has a small magnitude, and the acceleration \(j_x\) changes along the path not too rapidly, then one may assume that the acceleration \(j_x=j_0=\mathrm{const}\). It must be stipulated that the process of deformation of the system itself must not change the conditions of the external connection of the mass. Considering, in this case, the acceleration constant and equal to that which existed at the initial moment of contact of the mass with the system, we obtain, for equations (1) and (2), the expressions:

\[ \delta_d^2-2\frac{j_0}{g}\delta_c\delta_d-\frac{\delta_c v_0^2}{g}=0 \ldots \ldots \ldots (1)' \]

\[ T=\int_0^{\delta_d} \frac{dx}{\sqrt{\,v_0^2+2j_0x-\frac{g}{\delta_c}x^2\,}} = \sqrt{\frac{\delta_c}{g}} \left| \arcsin \left( \frac{\frac{gx}{j_0\delta_c}-1} {\sqrt{\,1+\frac{gv_0^2}{j_0^2\delta_c}\,}} \right) \right|_0^{\delta_d} \ldots \ldots \ldots (2)' \]

from which the calculation equations are obtained:

\[ \frac{\delta_d}{\delta_c} = \frac{j_0}{g} \left( 1+\sqrt{\frac{gv_0^2}{j_0^2\delta_c}} \right) \ldots \ldots \ldots (1)'' \]

\[ T= \sqrt{\frac{\delta_c}{g}} \left[ \frac{\pi}{2} + \arcsin \left( -\frac{1}{\sqrt{\,1+\frac{gv_0^2}{j_0^2\delta_c}\,}} \right) \right] \ldots \ldots \ldots (2)'' \]

In the case of impact during free fall of a mass, equations (1) and (2) take an even simpler form:

We have: \(j_0=g;\ v_0^2=2gH\).

We obtain:

\[ \frac{\delta_0}{\delta_c}=1+\sqrt{1+\frac{2H}{\delta_c}}\ldots\ldots\ldots (1)''' \]

and

\[ T=\sqrt{\frac{\delta_c}{g}} \left[ \frac{\pi}{2}+\arcsin\left( \frac{1}{\sqrt{1+\frac{2H}{\delta_c}}} \right) \right]\ldots\ldots (2)'' \]

Examples of Calculation

In solving these and the following examples of calculating impact, I must stipulate that all intermediate data not directly related to operations with the impact formulas will, in most cases, not be calculated, owing to the small amount of space allotted to the article. Therefore, almost all examples will contain ready numerical data.

Drawing 1.

Example 1. A wagon has a total weight \(G=38\ t\). Two buffers, operating simultaneously, have springs with the characteristic:

\[ f=0.000214\,P\ (cm,\ kg). \]

The limiting working deformation of the buffer springs (before contact of the coils) is \(f_{max}=6\ cm\). The coefficient of wheel friction is:

\[ \mu=0.05. \]

Determine the limiting impact velocity of the wagon against a dead-end stop, neglecting the deformation of the latter.

Acceleration:

\[ j=\mathrm{const.}=-\frac{\mu G}{g} =-\frac{0.05\cdot 38000}{981} =-1.99\simeq -2\ \frac{cm}{sec^2}, \]

Static deformation:

\[ \delta_c=0.000214\cdot 38000=8.13\ cm. \]

The limiting velocity must correspond to the limiting deformation of the spring:

\[ \delta_0=f_{max}=6\ cm. \]

Apply equation (1):

\[ 6^2+\frac{2\cdot2\cdot8.13\cdot6}{981} -\frac{8.13}{981}v_0^2=0. \]

whence:

\[ v_0=\sqrt{4280}=65.4\,\frac{cm}{sec}=2.35\,\frac{km}{h}. \]

Example 2. The flywheel of a cam press has: \(n=120\ \dfrac{rev}{min}\). The outline of the cam is expressed by the formula: \(\rho=a e^\varphi+r\), where: \(a=2\ cm\) and \(r=5\ cm\).

Determine the deformation (within the elastic range) of the stamped part, if, under the action of the static load applied to the punch, it is deformed by:

\[ \delta_c=0.5\ cm. \]

Let, reckoning from the instant of contact of the punch and the deformable object, the stroke of the punch be determined by:

\[ x=\rho-2r=a e^\varphi-r \qquad (\varphi=\omega t). \]

Fig. 2.

Then:

\[ v_x=\frac{dx}{dt}=a\omega e^\varphi=\omega(x+r), \]

\[ j_x=\frac{d^2x}{dt^2}=a\omega^2 e^\varphi=\omega^2(x+r). \]

At \(x=0\)

\[ v_0=\omega r=\frac{\pi n r}{30}=\frac{\pi 120\cdot 5}{30}=62.8\,\frac{cm}{sec}. \]

\[ \omega=12.56\ sec^{-1}. \]

We compute, for use in equation (1), the value of the integral:

\[ \int_0^{\delta_{\partial}} j_x\,dx = \omega^2\int_0^{\delta_{\partial}}(x+r)\,dx = \omega^2\left[\frac{\delta_{\partial}^2}{2}+r\delta_{\partial}\right]. \]

Upon substituting the obtained value of the integral into (1) and transforming, we obtain:

\[ (g-\omega^2\delta_c)\delta_{\partial}^{\,2} -2\omega^2\delta_c r\delta_{\partial} -v_0^2\delta_c=0. \]

Substitution of the numerical values of the quantities gives:

\[ \delta_{\partial}^{\,2}-0.873\delta_{\partial}-2.183=0. \]

whence:

\[ \delta_{\partial}=1.98\ cm. \]

Second variant \(j = j_v\).

The differential equation of motion of the mass will be written as:

\[ \frac{d^2 x}{dt^2} = j_v - \frac{g}{\partial_c}x. \]

Let us rewrite it in the following form:

\[ \frac{dx}{dx}\frac{dv}{dt} + \frac{g}{\partial_c}x - j_v = 0, \]

or, finally:

\[ \left(\frac{g}{\partial_c}x - j_v\right)dx + vdv = 0. \]

In view of the fact that the function \(j_v\) is in general unknown, for the integrating factor there may be taken a function of one variable. Such a function can only be:

\[ \frac{1}{N}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial v}\right) = \frac{\partial j_v}{v\partial v}. \]

Since the integrating factor must have the form:

\[ \mu = e^{-\int \frac{\partial j_v}{v\partial v}\, dx}, \]

the quantity \(\frac{\partial j_v}{v\partial v}\) must be constant. This will occur only in the single case, namely, when: \(j_v = av^2 + b\). Then:

\[ \mu = e^{-\int 2a\,dx} = e^{-2ax}. \]

This case, indeed, has a very wide occurrence. The resistance force of a liquid medium to a body moving in it is proportional to the square of the velocity of the body. Thus, the acceleration (deceleration) received by the body from an external constraint (the liquid), being proportional to the force of this constraint, will be proportional to the square of the velocity. For generality of the solution, we retain in the expression for \(j_v\) the constant \(b\).

Acceleration \(j_v = av^2 + b\).

The integrating factor will have the form: \(\mu = e^{-2ax}\).

On multiplying by \(e^{-2ax}\), the equation must become a complete differential.

\[ e^{-2ax}\left(\frac{g}{\partial_c}x - av^2 - b\right)dx + e^{-2ax}v\,dv = 0. \]

Solving the equation

\[ e^{-2ax}\int v\,dv=\frac{1}{2}e^{-2ax}v^{2}+\varphi(x) \]

\[ e^{-2ax}\left(\frac{g}{\delta_c}x-av^{2}-b\right) =-ae^{-2ax}v^{2}+\varphi'(x). \]

Whence:

\[ \varphi'(x)=e^{-2ax}\left(\frac{g}{\delta_c}x-b\right) \]

and

\[ \varphi(x)=-\frac{1}{4a^{2}}\left[2ab-\frac{g}{\delta_c}(2ax+1)\right]e^{-2ax}. \]

In final form, the equation is represented as:

\[ \left[2a^{2}v^{2}+2ab-\frac{g}{\delta_c}(2ax+1)\right]e^{-2ax}=C. \]

The boundary conditions are as follows:

\[ \begin{aligned} &\text{for } x=0; \quad v=v_0;\\ &\text{for } x=\delta_d; \quad v=0. \end{aligned} \]

Their fulfillment gives:

\[ \left[2ab-\frac{g}{\delta_c}(2a\delta_d+1)\right]e^{-2a\delta_d} =2aj_0-\frac{g}{\delta_c}\ldots\ldots(3) \]

A transcendental equation has been obtained with respect to the sought quantity \(\delta_d\).

In view of the inconveniences associated with using the equation in transcendental form, let us give it a form that would make it possible to use a nomogram for finding \(\delta_d\).

\[ e^{(2a\delta_d)} = \frac{(2a\delta_d)\left(\dfrac{b}{g}\right)-(2a\delta_d)-1} {(2a\delta_d)\left(\dfrac{j_0}{g}\right)-1} \ldots\ldots(3)' \]

It is useful to note that, in the present case, there is no proportionality between the dynamic and static deformations.

The nomogram is given in a binary field. Its use is best clarified in solving the example presented below.

Calculation example.

Example 3. A seaplane lands on water while parachuting, with a vertical velocity

\[ v_0=3.25\,\frac{m}{sec}. \]

Under conditions of calm remain-

... on water, the specific load on the bottom part is \(p = 3\ \mathrm{kg}/\mathrm{dm}^2\).

The coefficient of resistance for water is

\[ k = 40\ \frac{\mathrm{kg}\ \mathrm{sec}^2}{\mathrm{m}^4}. \]

Determine the dynamic deformation of the bottom plating, for some element of it, if the latter, under conditions of free floating, has:

\[ d_c = 0.2\ \mathrm{cm} \]

\[ j_v = -\frac{ksv}{ps}\, g = -\frac{gk}{p}\, v^2 \]

Consequently:

\[ a = -\frac{gk}{p} = \frac{981 \cdot 40 \cdot 10^2}{3 \cdot 10^8} \simeq 0.0131\ \mathrm{cm}^{-1} \]

\[ 2ad_c = -2 \cdot 0.0131 \cdot 0.2 = 0.00524 \]

\[ \frac{j_0}{g} = \frac{av_0^2}{g} = -\frac{0.0131(325)^2}{981} = -1.410 \qquad \frac{b}{g} = 0. \]

Nomogram

For finding the dynamic deformation we use the nomogram in the following way. Having marked on the axis \(\dfrac{j_0}{g}\) the point

\[ \frac{j_0}{g}=-1.310, \]

we draw a straight line through it and through the point \(\dfrac{b}{g}=0\) on the second axis. On the continuation of this straight line we find the point which would correspond, in the field of negative \(2a\delta_0\) (since \(a<0\)), to a position between the negative curves of the line \(2a\delta_c\) at the mark: \(2\delta_c=-0.00524\). Determining the position of the found point by its mark between the straight lines \(2a\delta_0\), we obtain:

\[ 2a\delta_0=-0.13. \]

The dynamic deformation will be:

\[ \delta_0=-\frac{0.13}{2a}=\frac{-0.13}{-2.0\cdot 0131}\simeq 5\ \text{cm}. \]

Acceleration \(j_v=av+b\).

In the case of the motion of a body with comparatively small velocities in a liquid possessing a large coefficient of viscosity, we may freely assume that the resistance to its motion will be proportional to the velocity. It is therefore necessary to consider one more case of motion, when:

\[ j_v=av+b. \]

The equation of motion of the mass will be written:

\[ \frac{d^2x}{dt^2}-(av+b)+\frac{g}{\delta_c}x=0. \]

We differentiate it so that it includes only two variables.

\[ \frac{d^2v}{dt^2}-a\frac{dv}{dt}+\frac{g}{\delta_c}v=0. \]

The obtained linear equation with constant coefficients has the characteristic equation:

\[ \lambda^2-a\lambda+\frac{g}{\delta_c}=0 \]

with roots:

\[ \lambda_1=\frac{1}{2}\left(a+\sqrt{a^2-\frac{g}{\delta_c}}\right) \quad \text{and} \quad \lambda_2=\frac{1}{2}\left(a-\sqrt{a^2-\frac{g}{\delta_c}}\right). \]

The roots are real, since only \(a\) can be negative, but it is under a radical of the second degree and \(a^2 > \dfrac{g}{\partial_e}\). Therefore the solution of the equation is written:

\[ v=Ae^{\lambda_1 t}+Be^{\lambda_2 t}. \]

Hence, by integration and differentiation, we obtain:

\[ x=\frac{A}{\lambda_1}e^{\lambda_1 t}+\frac{B}{\lambda_2}=e^{\lambda_2 t} \qquad \frac{dv}{dt}=\lambda_1 A e^{\lambda_1 t}+\lambda_2 B e^{\lambda_2 t}. \]

The terminal conditions to be satisfied are as follows:

\[ \text{at } t=0;\quad x=0;\quad v=v_0;\quad \frac{dv}{dt}=av_0+b \]

\[ \text{and at } t=T;\quad x=\partial_0;\quad v=0. \]

After substitutions and the corresponding operations, we obtain the calculation formulas:

\[ T=\frac{1}{\lambda_1-\lambda_2}\ln \frac{\lambda_1 v_0-j_0}{\lambda_2 v_0-j_0} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (5) \]

\[ \partial_0= \frac{(\lambda_1 v_0-j_0)^{\frac{\lambda_1}{\lambda_1-\lambda_2}}} {\lambda_1\lambda_2(\lambda_2 v_0-j_0)^{\frac{\lambda_2}{\lambda_1-\lambda_2}}} \ .\ .\ .\ .\ .\ .\ .\ .\ (6) \]

Example of calculation.

Example 4. When mooring a steamship to a wharf, while bringing its side up to the piles, it presses with its upper rigid wale against the end of one of the piles. At the moment of contact we have:

\[ v_0=0.05\ \frac{m}{sec}. \]

Drawing 3.

The displacement of the steamship is \(120\ t\).
The length of the free end of the pile is \(l=2\,m\).
Its diameter is \(d=30\ cm\). Its material is pine.

\[ a=-50\ sec^{-1} \]

\[ j=av \]

\[ j_0=-50\cdot 5=-250\ \frac{cm}{sec^2}. \]

Static deformation:

\[ \delta_c=\frac{Ql^3}{3EJ}=\frac{120000\cdot(200)^3 64}{3\cdot 10^5\cdot \pi\cdot(30)^4}=80.5\ \text{cm}. \]

The fact that, in the present case, the static deformation assumes enormous dimensions should not confuse the person making the calculation, since here we are dealing not with a falling load, for which the static deformation must always be less than the dynamic one. Here, by static deformation we mean that which would occur if the moving mass (but not moving vertically and not under the influence of gravity) loaded the system statically by its weight.

The quantities entering into equation (6) are:

\[ \frac{g}{\delta_c}=\frac{981}{80.5}=13.2\ \text{sec}^{-2}. \]

\[ \lambda_1=\frac{1}{2}\left[-50-\sqrt{2500-12.2}\right]=-0.06\ \text{sec}^{-1} \]

\[ \lambda_2=\frac{1}{2}\left[-50+\sqrt{2500-12.2}\right]=-49.94\ \text{sec}^{-1}. \]

The time spent on the impact is determined from (5):

\[ T= \frac{1}{-0.06+49.94} \ln\frac{-0.05\cdot 5+250}{-49.94\cdot 5+250} = \frac{1}{49.88}\ln 832.3 = \frac{6.724}{49.88} = 0.135\ \text{sec}. \]

The dynamic deformation will be:

\[ \delta_d= \frac{\left(-0.06\cdot 5+250\right)^{\frac{-0.06}{49.88}}} {0.06\cdot 49.94\left(-49.94\cdot 5+250\right)^{\frac{-49.94}{49.88}}} = 0.165\ \text{cm}. \]

Stress in the pile:

\[ k_d=\frac{M_0}{W}=\frac{P_\delta l}{W} =\delta_d\frac{3Ed}{2l^2} = 0.165\frac{3\cdot 10^5\cdot 30}{2\cdot(200)^2} = 18.6\ \text{kg}/\text{cm}^2. \]

Third variant \(j=j_t\).

The acceleration depending on time will take place chiefly when the body moves, wholly or partly para-

magnetic, in a variable magnetic field caused by the likewise varying electrical conditions of its occurrence and variation.

The differential equation will be written:

\[ \frac{d^{2}x}{dt^{2}}=j_t-\frac{g}{\partial_c}\,x \]

or:

\[ \frac{d^{2}x}{dt^{2}}+\frac{g}{\partial_c}x=j_t. \]

The equation without the right-hand side gives the solution:

\[ x=A\sin\varphi+B\cos\varphi \]

where, for brevity, we denote:

\[ \varphi=\sqrt{\frac{g}{\partial_c}}\,t, \]

remembering that:

\[ \frac{d\varphi}{dt}=\sqrt{\frac{g}{\partial_c}}. \]

To find the complete integral it is necessary to solve a system of two equations with respect to \(A'\) and \(B'\)—derivatives taken with respect to time:

\[ A'\sin\varphi+B'\cos\varphi=0 \]

\[ A'\cos\varphi-B'\sin\varphi=j_t. \]

The solution of the system gives:

\[ A'=\sqrt{\frac{\partial_c}{g}}\,j_t\cos\varphi \]

\[ B'=-\sqrt{\frac{\partial_c}{g}}\,j_t\sin\varphi. \]

Hence \(A\) and \(B\) are easily determined.

\[ A=\sqrt{\frac{\partial_c}{g}}\int j_t\cos\varphi\,dt+C_1 \]

\[ B=-\sqrt{\frac{\partial_c}{g}}\int j_t\sin\varphi\,dt+C_2. \]

The complete integral of the path of the mass is expressed as:

\[ x=\sqrt{\frac{\partial_c}{g}}\left[\sin\varphi\int j_t\cos\varphi\,dt-\cos\varphi\int j_t\sin\varphi\,dt\right]+C_1\sin\varphi+C_2\cos\varphi. \]

Velocity:

\[ \frac{dx}{dt}=\cos\varphi \int j_t \cos\varphi\,dt+\sin\varphi \int j_t \sin\varphi\,dt+ \sqrt{\frac{g}{\delta_c}}\,(C_1\cos\varphi-C_2\sin\varphi). \]

Acceleration:

\[ \frac{d^2x}{dt^2} = j_t+\sqrt{\frac{g}{\delta_c}} \left[ \cos\varphi \int j_t \sin\varphi\,dt - \sin\varphi \int j_t \cos\varphi\,dt \right] - \frac{g}{\delta_c}(C_1\sin\varphi+C_2\cos\varphi). \]

The final conditions are as follows:

\[ \text{for } i=0;\quad x=0;\quad \frac{dx}{dt}=0;\quad \frac{dx}{dt}=v_0; \]

\[ \text{for } t=T;\quad x=\delta_{\partial};\quad \frac{dx}{dt}=0. \]

Substitution of the particular values of the factors of motion of the mass gives:

\[ 0=\cos\left(\sqrt{\frac{g}{\delta_c}}\,T\right) \int_0^T j_t\cos\varphi\,dt + \sin\left(\sqrt{\frac{g}{\delta_c}}\,T\right) \int_0^T j_t\sin\varphi\,dt + v_0\cos\sqrt{\frac{g}{\delta_c}}\,T \ldots\ldots\ldots\ldots\ldots (7) \]

\[ \delta_{\partial} = \sqrt{\frac{\delta_c}{g}} \left[ \sin\left(\sqrt{\frac{g}{\delta_c}}\,T\right) \int_0^T j_t\cos\varphi\,dt - \cos\left(\sqrt{\frac{g}{\delta_c}}\,T\right) \int_0^T j_t\sin\varphi\,dt + v_0\sin\left(\sqrt{\frac{g}{\delta_c}}\,T\right) \right] \ldots\ldots\ldots\ldots\ldots (8) \]

where

\[ \varphi=\sqrt{\frac{g}{\delta_c}}\,t. \]

The first equation will serve to determine the time required for the deformation to occur. Having determined the time, we must substitute its value into the second equation, from which the magnitude of the dynamic deformation is obtained.

Special cases.

If the magnitude of the acceleration \(j_t\) does not vary with time, then we must obtain simpler equations.

Let

\[ j_t=j_0=\mathrm{const}. \]

Then:

\[ \int_0^T j_t \sin\varphi\,dt = j_0\sqrt{\frac{\partial_c}{g}} \left[ 1-\cos\left(\sqrt{\frac{g}{\partial_c}}\,T\right) \right] \]

\[ \int_0^T j_t \cos\varphi\,dt_0 = j_0\sqrt{\frac{\partial_c}{g}} \sin\left(\sqrt{\frac{g}{\partial_c}}\,T\right) \]

and the equations take the form:

The first, which serves to determine the time:

\[ 0 = j_0\sqrt{\frac{\partial_c}{g}} \sin\left(\sqrt{\frac{g}{\partial_c}}\,T\right) + v_0\cos\left(\sqrt{\frac{g}{\partial_c}}\,T\right) \ldots \tag{7'} \]

It coincides completely with equation \((2)''\).

Substitution of \(T\), determined from this equation, into the second gives:

\[ \frac{\partial_0}{\partial_c} = \frac{j_0}{g} \left( 1+ \sqrt{1+\frac{g v_0^2}{j_0^2\partial_c}} \right) \ldots\ldots\ldots \tag{8} \]

i.e. the equation which we already had—\((1)''\).

Example of calculation.

Example 5. The hammer of a solenoid perforator works at 20 periods per second. Its weight is \(G=2\ kg\). The acceleration varies according to the law:

\[ j_t=3g\cos\frac{\pi t}{10}. \]

What will be the stress in the rod of the perforator, having length \(l=2\ m\) and diameter \(d=2\ cm\), if the spear has struck hard rock, into which, practically, it does not penetrate (the most unfavorable case). The rod is of steel. \(v_0=20\,\frac{m}{sec}\).

Fig. 4.

Let us compute separately the integrals entering into equations (7) and (8).

\[ \int_0^T j_t\sin\varphi\,dt = 3g\int_0^T \cos\frac{\pi t}{10} \sin\sqrt{\frac{g}{\partial_c}}\,t\,dt = \]

\[ = \frac{3g}{2} \left| -\frac{ \cos\left(\sqrt{\frac{g}{\partial_c}}+\frac{\pi}{10}\right)t }{ \sqrt{\frac{g}{\partial_c}}+\frac{\pi}{10} } - \frac{ \cos\left(\sqrt{\frac{g}{\partial_c}}-\frac{\pi}{10}\right)t }{ \sqrt{\frac{g}{\partial_c}}-\frac{\pi}{10} } \right|_0^T \simeq \]

\[ \simeq 3\partial_c\left[\sqrt{\frac{g}{\partial_c}}\left(1-\cos \sqrt{\frac{g}{\partial_c}}\,T\right)\cos\frac{\pi T}{10} -\frac{\pi}{10}\sin \sqrt{\frac{g}{\partial_c}}\,T\cdot \sin\frac{\pi T}{10}\right]. \]

\[ \int_0^T j_t\cos\varphi\,dt = 3g\int_0^T \cos\frac{\pi t}{10}\,\sin \sqrt{\frac{g}{\partial_c}}\,t\,dt = \]

\[ =\frac{3g}{2} \left| \frac{\sin\left(\sqrt{\frac{g}{\partial_c}}+\frac{\pi}{10}\right)t} {\sqrt{\frac{g}{\partial_c}}+\frac{\pi}{10}} + \frac{\sin\left(\sqrt{\frac{g}{\partial_c}}-\frac{\pi}{10}\right)t} {\sqrt{\frac{g}{\partial_c}}-\frac{\pi}{10}} \right|_0^T \simeq \]

\[ \simeq 3\partial_c\left[ \sqrt{\frac{g}{\partial_c}} \left(1+\sin \sqrt{\frac{g}{\partial_c}}\,T\cos\frac{\pi T}{10}\right) -\frac{\pi}{10}\cos \sqrt{\frac{g}{\partial_c}}\,T\cdot \sin\frac{\pi T}{10} \right]. \]

In view of the great cumbersomeness of the equations, we shall give only the final results of the substitutions.

Substitution of the found values of the integrals into equation (7) gives:

\[ 3\partial\left[ \sqrt{\frac{g}{\partial_c}} \left(\sin \sqrt{\frac{g}{\partial_c}}\,T + \cos \sqrt{\frac{g}{\partial_c}}\,T\right) -\frac{\pi}{10}\sin\frac{\pi T}{10} \right] + \]

\[ +v_0\cos \sqrt{\frac{g}{\partial_c}}\,T=0. \]

We determine the numerical part of the arguments:

\[ \partial_c=\frac{Gl}{EF} = \frac{2\cdot 200\cdot 4}{2.2\cdot 10^6\pi\cdot 2^2} = 0.000058\ \mathrm{cm} \]

\[ \sqrt{\frac{g}{\partial_c}} = \sqrt{\frac{981}{0.000058}} = 4110\ \mathrm{sec}^{-1}. \]

Substitution of the numbers into (7) and neglect of very small (relatively) quantities gives:

\[ \operatorname{tg}\sqrt{\frac{g}{\partial_c}}\,T=-2800. \]

Whence:

\[ T=0.00038\ \mathrm{sec}. \]

We compute the values of the sines and cosines.

\[ \sin \sqrt{\frac{g}{\delta_c}}\,\tau \cong 1, \]

\[ \cos \sqrt{\frac{g}{\delta_c}}\,\tau = 0.012, \]

\[ \sin \frac{\pi T}{10} = 0.00012, \]

\[ \cos \frac{\pi T}{10} \cong 1. \]

Substitution of all the obtained and computed quantities into equation (8) gives:

\[ \delta_c \cong 0.487\ \mathrm{cm}. \]

We determine the magnitude of the dynamic stress:

\[ k_d=\frac{P_d}{F}=\delta_c\frac{E}{l} =\frac{0.487\cdot 2\cdot 2\cdot 10^6}{200} =5360\ \mathrm{kg}/\mathrm{cm}^2. \]