1. TWO DEGREE OF FREEDOM SYSTEM

2. INTRODUCTION Systems that require two independent coordinates to describe their motion; Two masses in the system X two possible types of motion of each mass. Example: motor pump system. There are two equations of motion for a 2DOF system, one for each mass (more precisely, for each DOF). They are generally in the form of couple differential equation that is, each equation involves all the coordinates.

3. E QUATION OF MOTION FOR FORCED VIBRATION Consider a viscously damped two degree of freedom spring-mass system, shown in Fig.5.3. Figure 5.3: A two degree of freedom spring-mass-damper system

4. 4 Both equations can be written in matrix form as The application of Newton’s second law of motion to each of the masses gives the equations of motion: E QUATIONS OF M OTION FOR F ORCED V IBRATION where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by

5. 5 And the displacement and force vectors are given respectively: E QUATIONS OF M OTION FOR F ORCED V IBRATION It can be seen that the matrices [m], [c], and [k] are all 2 x 2 matrices whose elements are known masses, damping coefficient and stiffnesses of the system, respectively.

6. 6 where the superscript T denotes the transpose of the matrix. E QUATIONS OF M OTION FOR F ORCED V IBRATION o The solution of Eqs.(5.1) and (5.2) involves four constants of integration (two for each equation). Usually the initial displacements and velocities of the two masses are specified as o Further, these matrices can be seen to be symmetric, so that, x 1 ( t = 0) = x 1 (0) and 1 ( t = 0) = 1 (0), x 2 ( t = 0) = x 2 (0) and 2 ( t = 0) = 2 (0).

7. 7 Assuming that it is possible to have harmonic motion of m 1 and m 2 at the same frequency ω and the same phase angle Φ, we take the solutions as F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM By setting F 1 (t) = F 2 (t) = 0, and damping disregarded, i.e., c 1 = c 2 = c 3 = 0, and the equation of motion is reduced to:

8. 8 Since Eq.(5.7)must be satisfied for all values of the time t, the terms between brackets must be zero. Thus, F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM Substituting into Eqs.(5.4) and (5.5),

9. 9 or which represent two simultaneous homogenous algebraic equations in the unknown X 1 and X 2. For trivial solution, i.e., X 1 = X 2 = 0, there is no solution. For a nontrivial solution, the determinant of the coefficients of X 1 and X 2 must be zero: F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM

10. 10 The roots are called natural frequencies of the system. which is called the frequency or characteristic equation. Hence the roots are: F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM

11. 11 The normal modes of vibration corresponding to ω 1 2 and ω 2 2 can be expressed, respectively, as To determine the values of X 1 and X 2, given ratio which are known as the modal vectors of the system. F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM

12. 12 Where the constants,, and are determined by the initial conditions. The initial conditions are The free vibration solution or the motion in time can be expressed itself as F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM

13. 13 Thus the components of the vector can be expressed as The resulting motion can be obtained by a linear superposition of the two normal modes, Eq.(5.13) where the unknown constants can be determined from the initial conditions: F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM

14. 14 Substituting into Eq.(5.15) leads to The solution can be expressed as F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM

15. 15 from which we obtain the desired solution F REE V IBRATION A NALYSIS OF AN U NDAMPED S YSTEM

16. 16 Solution : For the given data, the eigenvalue problem, Eq.(5.8), becomes E XAMPLE 5.3:F REE V IBRATION R ESPONSE OF A T WO D EGREE OF F REEDOM S YSTEM Find the free vibration response of the system shown in Fig.5.3(a) with k 1 = 30, k 2 = 5, k 3 = 0, m 1 = 10, m 2 = 1 and c 1 = c 2 = c 3 = 0 for the initial conditions or

17. 17 from which the natural frequencies can be found as By setting the determinant of the coefficient matrix in Eq.(E.1) to zero, we obtain the frequency equation, E XAMPLE 5.3S OLUTION The normal modes (or eigenvectors) are given by

18. 18 By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain The free vibration responses of the masses m 1 and m 2 are given by (see Eq.5.15): E XAMPLE 5.3S OLUTION

19. 19 while the solution of Eqs.(E.10) and (E.11) leads to The solution of Eqs.(E.8) and (E.9) yields Equations (E.12) and (E.13) give E XAMPLE 5.3S OLUTION

20. 20 E XAMPLE 5.3S OLUTION Thus the free vibration responses of m 1 and m 2 are given by

21. 21 Figure 5.6: Torsional system with discs mounted on a shaft T ORSIONAL S YSTEM Consider a torsional system as shown in Fig.5.6. The differential equations of rotational motion for the discs can be derived as

22. 22 which upon rearrangement become T ORSIONAL S YSTEM For the free vibration analysis of the system, Eq.(5.19) reduces to

23. 23 Find the natural frequencies and mode shapes for the torsional system shown in Fig.5.7 for J 1 = J 0, J 2 = 2J 0 and k t1 = k t2 = k t. Solution : The differential equations of motion, Eq.(5.20), reduce to (with k t3 = 0, k t1 = k t2 = k t, J 1 = J 0 and J 2 = 2J 0 ): E XAMPLE 5.4:N ATURAL F REQUENCIES OF T ORSIONAL S YSTEM Fig.5.7: Torsional system

24. 24 The solution of Eq.(E.3) gives the natural frequencies gives the frequency equation: E XAMPLE 5.4S OLUTION Rearranging and substituting the harmonic solution:

25. 25 Equations (E.4) and (E.5) can also be obtained by substituting the following in Eqs.(5.10) and (5.11). The amplitude ratios are given by E XAMPLE 5.4S OLUTION

26. 26 Generalized coordinates are sets of n coordinates used to describe the configuration of the system. Equations of motion Using x(t) and θ(t). C OORDINATE C OUPLING AND P RINCIPAL C OORDINATES

27. 27 and the moment equation about C.G. can be expressed as From the free-body diagram shown in Fig.5.10a, with the positive values of the motion variables as indicated, the force equilibrium equation in the vertical direction can be written as Eqs.(5.21) and (5.22) can be rearranged and written in matrix form as C OORDINATE C OUPLING AND P RINCIPAL C OORDINATES

28. 28 The lathe rotates in the vertical plane and has vertical motion as well, unless k 1 l 1 = k 2 l 2. This is known as elastic or static coupling. From Fig.5.10b, the equations of motion for translation and rotation can be written as Equations of motion Using y(t) and θ(t). C OORDINATE C OUPLING AND P RINCIPAL C OORDINATES

29. 29 These equations can be rearranged and written in matrix form as If, the system will have dynamic or inertia coupling only. Note the following characteristics of these systems: C OORDINATE C OUPLING AND P RINCIPAL C OORDINATES

30. 30 1.In the most general case, a viscously damped two degree of freedom system has the equations of motions in the form: 2.The system vibrates in its own natural way regardless of the coordinates used. The choice of the coordinates is a mere convenience. 3.Principal or natural coordinates are defined as system of coordinates which give equations of motion that are uncoupled both statically and dynamically. C OORDINATE C OUPLING AND P RINCIPAL C OORDINATES

31. 31 E XAMPLE 5.6:P RINCIPAL C OORDINATES OF S PRING -M ASS S YSTEM Determine the principal coordinates for the spring-mass system shown in Fig.5.4.

32. 32 We define a new set of coordinates such that Approach : Define two independent solutions as principal coordinates and express them in terms of the solutions x 1 (t) and x 2 (t). The general motion of the system shown is E XAMPLE 5.6 S OLUTION

33. 33 Since the coordinates are harmonic functions, their corresponding equations of motion can be written as E XAMPLE 5.6 S OLUTION

34. 34 The solution of Eqs.(E.4) gives the principal coordinates: From Eqs.(E.1) and (E.2), we can write E XAMPLE 5.6 S OLUTION

35. 35 The equations of motion of a general two degree of freedom system under external forces can be written as Consider the external forces to be harmonic: F ORCED V IBRATION A NALYSIS where ω is the forcing frequency. We can write the steady-state solutions as

36. 36 Substitution of Eqs.(5.28) and (5.29) into Eq.(5.27) leads to We defined as in section 3.5 the mechanical impedance Z re (iω) as F ORCED V IBRATION A NALYSIS

37. 37 And write Eq.(5.30) as: Where, F ORCED V IBRATION A NALYSIS

38. 38 where the inverse of the impedance matrix is given Eqs.(5.33) and (5.34) lead to the solution F ORCED V IBRATION A NALYSIS Eq.(5.32) can be solved to obtain:

39. 39 Find the steady-state response of system shown in Fig.5.13 when the mass m 1 is excited by the force F 1 (t) = F 10 cos ωt. Also, plot its frequency response curve. E XAMPLE 5.8:S TEADY -S TATE R ESPONSE OF S PRING -M ASS S YSTEM

40. 40 The equations of motion of the system can be expressed as E XAMPLE 5.8 S OLUTION We assume the solution to be as follows. Eq.(5.31) gives

41. 41 Eqs.(E.4) and (E.5) can be expressed as Hence, E XAMPLE 5.8 S OLUTION

42. 42 Fig.5.14: Frequency response curves E XAMPLE 5.8 S OLUTION