Solution Manual for Vibrations 3rd Edition

Solution Manual for Vibrations 3rd Edition Solutions to Exercises Chapter 2 Section 2.1 2.1 Examine Eqs. (2.1) and (2.5) and verify that the units (dimensions) of the different terms in the respective equations are consistent. Solution 2.1 In Eqs. (2.1), the units of JG are kgm2 and the units of md2 are kgm2 . In Eq. (2.5), the units of energy are Nm, and the units of mv2 are kg(m/s)2 = kg(m/s2 )m = Nm. Section 2.2 2.2 Consider the slider mechanism of Example 2.2 and show that the rotary inertia JO about the pivot point O is also a function of the angular displacement . Solution 2.2 The rotary inertia is JO  Jm ( )  Jm  Jm  Jm where l s b e J  1 m e 2 J  m b 2 J  1 m b 2 me 3 e ms s mb 3 b l 2  l 2  l  2  J ( )  m  m d 2  m    a 2  al cos( ) ml l 12 l l  12  2    To determine how  depends on , we use geometry. First r 2 ()  a 2  b 2  2abcos a 2  r 2 ()  b 2  2r()bcos(   ) Combining the above equations, we obtain r 2 ()  r 2 ()  b 2  2r()bcos(    )  b 2  2abcos which leads to and, hence, b  a cos  r() cos(    ) cos(    )  b  a cos r() 2 l  m r p G thus, resulting in      cos1  b  a cos   r()    Thus, we have expressed the angle  in terms of the angle . Thus, since JO is also a function of . Jm is a function of , 2.3 Consider the crank-mechanism system shown in Figure E2.3. Determine the rotary inertia of this system about the point O and express it as a function of the angular displacement . The disc has a rotary inertia Jd about the point O. The crank has a mass mG and rotary inertia JG about the point G at the center of mass of the crank. The mass of the slider is mp. Solution 2.3 The total inertia of the system is JO  Jd  J p  J g where Jp is the rotary inertia of the slider about the point O, Jg is the rotary inertia of the crank about the point O and 2 p p p J  J  m r 2 g G G G and rG is the distance from O to the point G and rp is the distance from O to the point p. From geometry, we see that r 2  r cos  l cos  2  d 2 In addition, therefore, r  r cos  a cos  2  rsin  a sin  2 rsin  d  lsin   sin1  rsin  d    l   Thus, the total inertia of the system JO can be expressed as a function of . Section 2.3.2 2.4 Find the equivalent length Le of a spring of constant cross section of diameter d2 that has the same spring constant as the tapered spring shown in Case 2 of Table 2.3. Both springs have the same Young’s modulus E. Solution 2.4 From Cases 1 and 2 of Table 2.3, we have that J 3 L 2  (h  x) 2  a  2   2    x 2  2hx 1  a / 2    (a / 2)2  x  2 2hx e A E d 2E  Ed d e  2  1 2 Therefore, the equivalent length is L 4L e m Case 1 "– 4#L–% Case 2 L  d2 L e d1 2.5 Extend the spring combinations shown in Figures 2.6b and 2.6c to cases with three springs. Verify that the equivalent stiffness of these spring combinations is consistent with Eqs. (2.14) and (2.16), respectively. Solution 2.5 In Figure 2.5b, we extend Eq. (2.13) to obtain F(x)  F1 (x)  F2 (x)  F3 (x)  k1x  k2 x  k3x  (k1  k2  k3 )x  kex which is consistent with Eq. (2.14). In Figure 2.6c, we extend Eq. (2.15) to obtain x  x1  x2  x3  F  F  F   1  1  1  F  F k k k  k k k  k 1 2 3 which is consistent with Eq. (2.16).