= M!1/ 2 KM!1/ 2 = 2.2482 0.8246 1.0825 0.8246 0.5329 0.8246 1.0825 0.8246 2.2482 ' # $ $ $% & ' ' ' (10!7 det!K! 9.8255#10!7 ' 2 +1.3645#10!14 '! 4.1382 #10!22 = 0 ' 1 = 4.3142 #10!9 $ 1 = 2.0771#10!5 rad/s ' 2 = 1.1657 #10!7 $ 2 = 3.4143#10!4 rad/s ' 3 = 8.2283#10!7 $ 3 = 9.0710 #10!4 rad/s v 1 = 0.2443!0.9384 0.2443 ' # $ $ $% & ' ' ' v 2 = 0.7071 0!0.7071 ' # $ $ $% & ' ' ' v 3 = 0.6636 0.3455 0.6636 ' # $ $ $% & ' ' ' Use the mode summation method to find the solution. Transform the initial conditions: q 0( ) = M 1/ 2x 0( ) = 0 2.2361 0!'

• Engineering Vibration Inman 3rd
• Engineering Vibration 3rd Edition Daniel J Inman Solution Manual

#$ T!q 0( ) = M 1/ 2!x 0( ) = 0 The solution is given by Eq. (4.103), x t( ) = di sin! It + 'i( )ui i=1 4 # where! I = tan '1 # i v i Tq 0( ) v i T!q 0( ) $% & ' ( ) i = 1,2,3 Eq.

4.97( )( ) d i = v i Tq 0( ) sin! I i = 1,2,3 Eq. 4.98( )( ) u i = M '1/ 2v i Substituting known values yields! 3 = ' 2 rad d 1 = #2.0984 d 2 = 0 d 3 = 0.7726 u 1 = 0.0178!0.02098 0.01728 ' # $ $ $% & ' ' ' u 2 = 0.05 0!0.05 ' # $ $ $% & ' ' ' u 3 = 0.04692 0.007728 0.04692 ' # $ $ $% & ' ' ' The solution is x t( ) =!0.03625 0.04403!0.03625 ' # $ $ $% & ' ' ' cos 9.7044 (10!5t( ) + 0.03625 0.005969 0.0325 ' # $ $ $% & ' ' ' cos 6.1395(10!4 t( ) m The results are very similar to Problem 50. The responses of mass 1 and 3 are the same for both problems, except the amplitudes and frequencies are changed due to the increase in mass 2.

There would have been a greater change if the heavy mass was placed at mass 1 or 3. 4.52 Repeat Problem 4.46 for the case that the airplane body is 10 m instead of 4 m as indicated in the figure. What effect does this have on the response, and which design (4m or 10 m) do you think is better as to vibration? Solution: Given: m 1 0 0 0 10 0 0 0 1! ' # # # $% & & &!!x + EI l 3 3 '3 ' '3 6 '3 0 '3 3! ' # # # $% & & & x = 0 Calculate eigenvalues and eigenvectors: M!1/ 2 = m!1/ 2 1 0 0 0 0.3612 0 0 0 1 ' # $ $ $% & ' ' '!K = M!1/ 2 KM!1/ 2 = EI ml 3 3!0.9487 0!0.9487 0.6!0.9487 0!0.9487 3 ' # $ $ $% & ' ' ' Again choose the parameters so that the coefficient is 1 and compute the eigenvalues: det!K!

6.6'2 +10.8' = 0 ' 1 = 0 ' 2 = 3 ' 3 = 3.6 v 1 =!0.2887!0.9129!0.2887 # $%%% & ' ( ( ( v 2 = 0.7071 0!0.7071 # $%%% & ' ( ( ( v 3 = 0.6455!0.4082 0.6455 # $%%% & ' ( ( ( The natural frequencies are! 2 = 1.7321 rad/s! 3 = 1.8974 rad/s The relationship between eigenvectors and mode shapes is u = M!1/ 2v u 1 = m!1/ 2!0.2887!0.2887!0.2887 ' # $ $ $% & ' ' ' u 2 = m!1/ 2 0.7071 0!0.7071 ' # $ $ $% & ' ' ' u 3 = 0.6455!0.1291 0.6455 ' # $ $ $% & ' ' ' It appears that the mode shapes contain less 'amplitude' for the wing masses. This seems to be a better design from a vibration standpoint. 4.53 Often in the design of a car, certain parts cannot be reduced in mass. For example, consider the drive train model illustrated in Figure P4.44. The mass of the torque converter and transmission are relatively the same from car to car.

However, the mass of the car could change as much as 1000 kg (e.g., a two-seater sports car versus a family sedan). With this in mind, resolve Problem 4.44 for the case that the vehicle inertia is reduced to 2000 kg. Which case has the smallest amplitude of vibration? Solution: Let k1 = hub stiffness and k2 = axle and suspension stiffness. From Problem 4.44, the equation of motion becomes 75 0 0 0 100 0 0 0 2000! ' # # # $% & & &!!x +10,000 1 '1 0 '1 3 '2 0 '2 2! ' # # # $% & & & x = 0 x 0( ) = 0 and!x 0( ) = 0 0 1!'

Calculate eigenvalues and eigenvectors. M!1/ 2 = 0.1155 0 0 0 0.1 0 0 0 0.0224 ' # $ $ $% & ' ' '!K = M!1/ 2 KM!1/ 2 = 133.33!115.47 0!115.47 300!44.721 0!44.721 10 ' # $ $ $% & ' ' ' det!K!

443.33'2 + 29,000' = 0 ' 1 = 0 # 1 = 0 rad/s ' 2 = 70.765 # 2 = 8.9311 rad/s ' 3 = 363.57 # 3 = 19.067 rad/s v 1 =!0.1857!0.2144!0.9589 $% & & & ' ( ) ) ) v 2 = 0.8758 0.4063!0.2065 $% & & & ' ( ) ) ) v 3 = 0.4455!0.8882 0.1123 $% & & & ' ( ) ) ) Use the mode summation method to find the solution. Transform the initial conditions: q 0( ) = M 1/ 2x 0( ) = 0!q 0( ) = M 1/ 2!x 0( ) = 0 0 44.7214!' #$ T The solution is given by q t( ) = c 1 + c 4 t( )v 1 + c 2 sin! 2 t + ' 2 ( )v 2 + c 3 sin! 3 t + ' 3 ( )v 3 where!

I = tan '1 # iv i Tq(0) v i T!q(0) $% & ' ( ), i = 2,3 c i = v i Tq(0) # i cos! I, i = 2,3 Thus φ2 = φ3 =0, c2 = -1.3042 and c3 = 0.2635.

Next apply the initial conditions: q(0) = c 1 v 1 + c i i=2 3! Sin'iv i and!q(0) = c4v1 + ci i=2 3! Sin'iv i Pre multiply each of these by v1 T to get: c1 = 0 = v1 Tq(0) and c 4 =!42.8845 = v 1 T!q(0) So q(t) =!42.8845tv1!1.3042sin(8.9311t)v2 + 0.2635sin(19.067t)v3 Next convert back to the physical coordinates by x(t) = M! 1 2q(t) = 0.9195t 1 1 1 ' # $ $ $% & ' ' ' +!0.1319!0.05299 0.007596 ' # $ $ $% & ' ' ' sin8.9311t + 0.01355!0.02340 0.0006620 ' # $ $ $% & ' ' ' sin19.067t m Comparing this solution to problem 4.44, the car will vibrate at a slightly higher amplitude when the mass is reduced to 2000 kg. 4.54 Use mode summation method to compute the analytical solution for the response of the 2-degree-of-freedom system of Figure P4.28 with the values where m1 = 1 kg, m2 = 4 kg, k1 = 240 N/m and k2=300 N/m, to the initial conditions of x 0 = 0 0.01! ' # $% &,!x0 = 0 0!

Solution: Following the development of equations (4.97) through (4.103) for the mode summation for the free response and using the values of computed in problem 1, compute the initial conditions for the “q” coordinate system: M 1/2 = 1 0 0 2! ' # $% & ' q 0( ) = 1 0 0 2! ' # $% & 0 0.01! ' # $% & = 0 0.02! ' # $% &,q 0( ) = 1 0 0 2! ' # $% & 0 0! ' # $% & = 0 0!

' # $% & From equation (4.97):! 1 = tan '1 x 0 # $% & '( =! 2 = tan '1 x 0 # $% & '( = ) 2 From equation (4.98): d 1 = v 1 T q 0( ) sin! 2( ) = v 1 T q 0( ),d 2 = v 2 T q 0( ) sin!

2( ) = v 2 T q 0( ) Next compute q t( ) from (4.92) and multiply by M 1/2 to get x t( ) or use (4.103) directly to get q t( ) = d 1 cos! 1 t( )v1 + d2 cos!2t( )v2 = cos!1t( )v1 T q 0( )v 1 + cos! 2 t( )v1 T q 0( )v 1 = cos 5.551t( ) 0.0054 0.0184 ' # $% & ' + cos 24.170t( ) (0.0054 0.0016 ' # $% & ' Note that as a check, substitute t = 0 in this last line to recover the correct initial condition q 0( ).

Engineering Vibration Inman 3rd

Next transform the solution back to the physical coordinates x t( ) = M!1/2 q t( ) = cos 5.551t( ) 0.0054 0.0092 ' # $% & ' + cos 24.170t( )!0.0054 0.0008 ' # $% & ' m 4.55 For a zero value of an eigenvalue and hence frequency, what is the corresponding time response? Or asked another way, the form of the modal solution for a non- zero frequency is Asin(!nt + '), what is the form of the modal solution that corresponds to a zero frequency? Evaluate the constants of integration if the modal initial conditions are: 01.0)0( and,1.0)0( 11 rr!. Solution: A zero eigenvalue corresponds to the modal equation:.

Engineering Vibration 3rd Edition Daniel J Inman Solution Manual

41.