Answers and Solutions 1 Vibrations of Systems with a Single Degree of Freedom 1.1 Free Vibrations 1 (a) p = ~3EJx(a+b)1 ma2b2; (b) p = ~/l2EJx Ima2(4a+3b); (c) p = ~(cl +3EJx I mI3 ). Ins t r u c t ion. In accordance with the force method, the canonical form of the differ­ ential equation of motion for a system with the single degree of freedom is y = 811 (-my). Here, 811 is the displacement of the load under the action of a unit force in the direction coinciding with this displacement. Therefore, the frequency of free vibrations is p =11 Jm811 • 2 (a) p =~EJx Im1 3 ; (b) p = ~48EJx I 7ml3 ; (c) p = ~6EJx I mh2(h+b); (d) p = ~12EJx Ima2 (4a+3b). As an example of the determination of the compliance (coefficients for stati­ cally indeterminate systems), we consider Problem 2d. Breaking the constraint (under the assumption that the constraint between the load of mass m and the base surface is bilateral), we apply the force N (the constraint reaction 104 1 Vibrations of Systems with a Single Degree of Freedom , tl N Fig. 218. force) to the load. As a result, we arrive at the frame loaded by both the force N and the inertial force J (Fig. 218). Using the force method, we obtain two equa­ tions in terms of the displacement of the load along the axes y and z: y = d}J + d22N; (1) 0= d2}J +d22 N; (2) where J = -my. Excluding the force N, we arrive at the expression y = -o}\my, where £5;\ = (£5;}b;2 - £5;2b;\) / b;2· Thus, the vibration frequency is p = 1/ ~mo;l. The approach used makes it possible the determination of the system compli­ ance 0 1 for statically indeterminate problems. This approach does not require 11 traditional calculations (as a rule, rather cumbersome) and preliminarily uncover­ ing the static indeterminacy with subsequently determining the displacement of the unit force in the direction of the load shift under vibrations. If several con­ straints are imposed on the system, (e.g., the system is trice statically indetermi­ nate), then the same number of algebraic equations can be obtained for finding constraint reaction forces. Furthermore, these forces are excluded from the differ­ ential equation. 3 (a) (b) 1.1 Free Vibrations 105 4 Similarly to the solution to Problem 2, the differential equation for small tor­ sional vibrations in all the systems has the form .. 1 rp+-rp=O. J81l Therefore, the eigenfrequencies of the systems 4a-4d (see Fig. 4) are corre­ spondingly equal to (a) (b) (c) (d) Ins t r u c t ion: For variants (c) and (d), the systems are statically indeterminate. Thus, in order to find q, it is necessary initially to exclude the static indeterminacy. 5 ~ ml~"'2· 6 7 P =~iC9 m· 8 The rotation of the gear wheel through a certain small angle rp causes a dis­ placement of the gear-wheel center by the distance rpD / 2 . In this case, the point A (the point of fixation of the springs to the carrier) shifts by the distance 8 = rpl - rpD / 2. With allowance for the moments of inertia of the gear wheel J1 = mlD 2 /8 and of the carrier J2 = m2D 2 112, we can find the kinetic en­ ergy of the system T _ (rfIJ+"'2)D2¢ + (rflJD2 + mlP) ¢2 _ miD¢2 - 8 8324· We express the variation of the system potential energy in terms of the spring deformation and of the carrier helix angle: II = 2 cf + m2g -t (1-cos rp) =[ c (I - ~ )2 + ~gl ] rp2 . Furthermore, using the Lagrange equation of the second kind, we arrive at the expression 106 1 Vibrations of Systems with a Single Degree of Freedom .. + 12 "'2g1+4c(l-DI2)2 = 0 rp (9mJ+6"'2)d2+8m2z2-12"'2lD rp . Hence, it follows that the eigenfrequency of the system is p= 12m.gl-48c(l-DI2)2... " (9mJ+ 6"'2)D2+8m212-12"'2lD . 9 In the process of small free vibrations of the cylinder, its kinetic energy is the sum of the kinetic energies of the rotational and translational motions T = (1 12)J¢/ + (11 2)m(rrp)2. With allowance for the moment of inertia J = Qr2 12g of the cylinder, the equation for the total kinetic energy can be rewritten as T=(3/4)(QI g)r2rp2. The variation of the system potential energy is caused by stretching (com­ pressing) the springs and lifting the cylinder when it moves over the concave sur­ face. In the case of rotation of the cylinder through angle rp, its upper point A shifts by the distance 8 = 2rrp . The potential energy of the deformed springs is IIJ = 2c82 12 = 4cr2 rp2 . While deviating the cylinder from its equilibrium position through angle rp, the angular coordinate If/ of the cylinder center of mass is Iff = rpr I(R - r). In this case, the variation of the potential energy of the cylinder depends on its position Q rrp2 II2 = Q(R - r)(1- cos '1') = 2 (R-r) . Substituting expression for kinetic and potential energies into the Lagrange equation, we arrive at the equation for small free vibrations rp+( t (R~r) + 136 d)rp = 0, whence it follows that the system eigenfrequency is _ 2 q + 16 cq P - "3 (R-r) 3Q· As follows from the formula obtained, p ~ 00 as R ~ r , i.e., the eigenfre­ quency increases with a decrease in the radius of curvature of the concave surface. 10 p = ~c I (2m) - g I (2/). 11 p = .J c I Jm = 224 S·I. 1.1 Free Vibrations 107 12 To solve the problem, it is convenient to employ the Lagrange equation of the second kind. In the case of deviation of rod AB in the course of the vibration process through a certain angle rp from the equilibrium position, the kinetic energy of the disk is T = mv 2 /2+ Jmqi; (1) where rp is the torsion angle of the rod. The linear velocity v of the disk displacement and the angular velocity iff are related to the rod deviation velocity ip by the dependences v = lip; iff = (l / R)ip. (2) The potential energy of the system is II = cc/ / 2 + mgy, where y is the variation of the vertical coordinate of the disk center of mass. It is evident that y = 1(1- cos rp) sin a ~ (I rp2 sin a) / 2. (3) Substituting these relationships into the Lagrange equation of the second kind, we arrive at the differential equation .. + c(lIR)2+mglsina - 0 rp (3/2)mI2 rp - . Hence, the system eigenfrequency is r---=----- c(ll R)2+mglsina p= (3/2)mI2 It follows from the latter relationship that the frequency p vanishes (i.e., the system becomes unstable) when sin a = -cl(mgR2). 13 In the case of shifting the float by the distance L1x in the vertical direction, e.g., downwards, a complementary buoyancy force (mJ2 / 4)rL\x arises. Thus, the differential equation of motion can be represented in the form [~L\X+(mJ2 /4)rL\x]/1 +m2L\x(l; /11)+cL\x(l;/I) = 0, Therefore, the system eigenfrequency is p = ~r-(mJ-2-r-/-4-+-c-I;-:-/-/-:-12-)(-~-+-m-2-1;-I-/ -/12-). 14 We set up the differential equation of motion using the Lagrange equation of the second kind. The kinetic energy of the disk is T = mv2 /2 + J (P12 / 2, where v is the velocity of the disk motion and (PI is its angular velocity. 108 1 Vibrations of Systems with a Single Degree of Freedom The potential energy of the disk is II = mg( R - r )(1- cos rp). The angles rp and rpl (see Fig. 219a) and also the velocity v are related to each other by the relationships rpl = R;:r rp; V= (R - r)ifJ. (1) With allowance for the latter expression, the disk kinetic energy is T = [(R - r) / r2](mr2 + J)(ifJ2 /2). We substitute the expressions for the kinetic and potential energies into the Lagrange equation (mr2 + J)rp + mg[r2 / (R - r)]sin rp = O. (2) In the case of small vibrations ( sin rp ~ rp), the vibration frequency is p = ~mgr2 / [(R - r)(mr2 + J)]. We now derive the differential equation of motion using the d' Alembert method. According to Fig. 219b, we may write out mX = - N cos rp + F cos rp; (3) my = N cos rp + F sin rp - mg; (4) JijJ = -Fr, (5) Here, N is the normal pressing force and F is the friction force between the hub and the guide, which hampers slipping the disk. r a b Fig. 219. Multiplying Eqs. (3) and (4) by costp and sintp, respectively, we obtain after summation 1.1 Free Vibrations 109 m(x cos cP + ji sin cp) = F - mg sin cpo (6) Furthermore, using Eqs. (1) and (5), we write out the expression for the fric­ tion force (7) We now express the coordinates of the disk center of mass in terms of the an­ gle rp: x = (R - r)sin cP; y = (R - r)(1- cos cp). (8) Substituting relationships (7) and (8) into Eq. (6), we arrive at Eq. (2) derived above by the Lagrange method. 15 The solution of this problem is similar to that of Problem 14. The differential equation for small vibrations has the form (mR2 + 2J)ijJ + CR2 cp = 0, where ({J is the wheel rotation angle. 16 .. 3(R-a)g - 0' cP + [2+4a2 cP - , p = 31\:~~2g; CPOmin = f.l(l2 + 4a 2 ) / (12 + a2 + 3Ra).