TEST BANK FOR Engineering Vibration 3rd Edition By Daniel J. Inman (Solution Manual)

TEST BANK FOR Engineering Vibration 3rd Edition By Daniel J. Inman (Solution Manual) Problems and Solutions Section 1.1 (1.1 through 1.19) 1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static) displacement is recorded below. Plot the data and calculate the spring's stiffness. Note that the data contain some error. Also calculate the standard deviation. m(kg) 15 16 x(m) 1.14 1.25 1.37 1.48 1.59 1.71 1.82 Solution: Free-body diagram: m k kx mg Plot of mass in kg versus displacement in m Computation of slope from mg/x m(kg) x(m) k(N/m) 10 1.14 86.05 11 1.25 86.33 12 1.37 85.93 13 1.48 86.17 14 1.59 86.38 15 1.71 86.05 16 1.82 86.24 0 1 2 10 15 20 m x From the free-body diagram and static equilibrium: kx = mg (g = 9.81m/ s 2 ) k = mg/ x μ = !ki n = 86.164 The sample standard deviation in computed stiffness is: ! = (ki " μ) 2 i=1 n # n "1 = 0.164 1.2 Derive the solution of m˙x˙ + kx = 0 and plot the result for at least two periods for the case with ωn = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s. Solution: Given: m!x!+ kx = 0 (1) Assume: x(t) = aert . Then: rt x! = are and rt x ar e 2 !! = . Substitute into equation (1) to get: mar2ert + kaert = 0 mr2 + k = 0 r = ± k m i Thus there are two solutions: x1 = c1e k m i ! " # $ % & t , and x2 = c2e ' k m i ! " # $ % & t where (n = k m = 2 rad/s The sum of x1 and x2 is also a solution so that the total solution is: it it x x x c e c e 2 2 2 1 2 1 ! = + = + Substitute initial conditions: x0 = 1 mm, v0 = 5 mm/s x(0) = c1 + c2 = x0 = 1!c2 = 1" c1 , and v(0) = x! (0) = 2ic1 " 2ic2 = v0 = 5 mm/s !"2c1 + 2c2 = 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns): "2c1 + 2 " 2c1 = 5 i!c1 = 1 2 " 5 4 i, and c2 = 1 2 + 5 4 i Therefore the solution is: x = 1 2 ! 5 4 i " # $ % & ' e2it + 1 2 + 5 4 i " # $% & ' e!2it Using the Euler formula to evaluate the exponential terms yields: x = 1 2 ! 5 4 i " # $ % & ' (cos2t + i sin2t ) + 1 2 + 5 4 i " # $ % & ' (cos2t ! i sin 2t ) ( x(t ) = cos2t + 5 2 sin2t = 3 2 sin(2t + 0.7297) Using Mathcad the plot is: x t cos 2. t . 5 2 sin 2. t 0 5 10 2 2 x t t 1.3 Solve m˙x˙ + kx = 0 for k = 4 N/m, m = 1 kg, x0 = 1 mm, and v0 = 0. Plot the solution. Solution: This is identical to problem 2, except v0 = 0. !n = k m = 2 rad/s " # $ % & ' . Calculating the initial conditions: x(0) = c1 + c2 = x0 = 1! c2 =1 " c1 v(0) = x˙ (0) = 2ic1 " 2ic2 = v0 = 0! c2 = c1 c2 = c1 = 0.5 x(t) = 1 2 e2it + 1 2 e"2it = 1 2 (cos2t + isin 2t) + 1 2 (cos2t " i sin2t) x(t)= cos (2t ) The following plot is from Mathcad: Alternately students may use equation (1.10) directly to get x(t ) = 22 (1)2 + 02 2 sin(2t + tan!1[ 2 "1 0 ]) = 1sin(2t + # 2 ) = cos2t x t cos 2. t 0 5 10 1 1 x t t 1.4 The amplitude of vibration of an undamped system is measured to be 1 mm. The phase shift from t = 0 is measured to be 2 rad and the frequency is found to be 5 rad/s. Calculate the initial conditions that caused this vibration to occur. Assume the response is of the form x(t ) = Asin(!nt +"). Solution: Given: A = 1mm, " = 2 rad, ! = 5rad/s . For an undamped system: x(t ) = Asin !n( t +" ) = 1sin(5t + 2) and v(t ) = x! (t ) = A!n cos !n( t +" ) = 5 cos(5t + 2) Setting t = 0 in these expressions yields: x(0) = 1sin(2) = 0.9093 mm v(0) = 5 cos(2) = - 2.081 mm/s 1.5 Find the equation of motion for the hanging spring-mass system of Figure P1.5, and compute the natural frequency. In particular, using static equilibrium along with Newton’s law, determine what effect gravity has on the equation of motion and the system’s natural frequency. Figure P1.5 Solution: The free-body diagram of problem system in (a) for the static case and in (b) for the dynamic case, where x is now measured from the static equilibrium position. (a) (b) From a force balance in the static case (a): mg = kxs , where xs is the static deflection of the spring. Next let the spring experience a dynamic deflection x(t) governed by summing the forces in (b) to get m!x!(t ) = mg ! k(x(t ) + xs )"m!x!(t ) + kx(t ) = mg ! kxs "m!x!(t ) + kx(t ) = 0"# n = k m since mg = kxs from static equilibrium. 1.6 Find the equation of motion for the system of Figure P1.6, and find the natural frequency. In particular, using static equilibrium along with Newton’s law, determine what effect gravity has on the equation of motion and the system’s natural frequency. Assume the block slides without friction. Figure P1.6 Solution: Choosing a coordinate system along the plane with positive down the plane, the freebody diagram of the system for the static case is given and (a) and for the dynamic case in (b): In the figures, N is the normal force and the components of gravity are determined by the angle θ as indicated. From the static equilibrium: !kxs + mgsin" = 0 . Summing forces in (b) yields: !Fi = m!x!(t )"m!x!(t ) = #k(x + xs ) + mgsin$ "m!x!(t ) + kx = #kxs + mgsin$ = 0 "m!x!(t ) + kx = 0 "% n = k m rad/s 1.7 An undamped system vibrates with a frequency of 10 Hz and amplitude 1 mm. Calculate the maximum amplitude of the system's velocity and acceleration. Solution: Given: First convert Hertz to rad/s: ! n = 2"fn = 2"(10) = 20" rad/s. We also have that A= 1 mm. For an undamped system: x(t)= A (" t +! ) n sin and differentiating yields the velocity: v(t) = A!n cos !n ( t + ") . Realizing that both the sin and cos functions have maximum values of 1 yields: = " = 1(20! )= 62.8mm/s max n v A Likewise for the acceleration: a(t)= #A" (" t +! ) n n sin 2 = = ( ) = 3948mm/s 2 2 2 max " 1 20! n a A 1.8 Show by calculation that A sin (ωnt + φ) can be represented as Bsinωnt + Ccosωnt and calculate C and B in terms of A and φ. Solution: This trig identity is useful: sin(a + b) = sin acosb + cosasinb Given: Asin(" t ! ) Asin(" t)cos(! ) Acos(" t)sin(! ) n n n + = + = Bsin!nt + Ccos!nt where B = Acos" and C = Asin" 1.9 Using the solution of equation (1.2) in the form x(t) = Bsin! nt + Ccos! nt calculate the values of B and C in terms of the initial conditions x0 and v0. Solution: Using the solution of equation (1.2) in the form x(t) = Bsin! nt + Ccos! nt and differentiate to get: x˙ (t) = ! nBcos(! nt) " ! nCsin(! nt) Now substitute the initial conditions into these expressions for the position and velocity to get: x0 = x(0) = Bsin(0) + Ccos(0) = C v0 = x˙ (0) = ! nBcos(0) " ! nCsin(0) = ! nB(1) "! nC(0) =! nB Solving for B and C yields: B = v0 ! n , and C = x0 Thus x(t) = v0 ! n sin! nt + x0 cos! nt 1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition. Solution: Following the lead given in Example 1.1.2, write down the general expression of the velocity by differentiating equation (1.10): x(t) = Asin(!nt + ")# x˙ (t) = A!n cos(!nt + ") # v(0) = A!n cos(!n 0 + ") = A!n cos(") From the figure: Figure 1.6 A = x0 2 + v0 !n " # $ % & ' 2 , cos( = v0 !n x0 2 + v0 !n " # $ % & ' 2 Substitution of these values into the expression for v(0) yields v(0) = A!n cos" = x0 2 + v0 !n # $ % & ' ( 2 (!n ) v0 !n x0 2 + v0 !n # $ % & ' ( 2 = v0 verifying the agreement between the figure and the initial velocity condition. 1.11 (a)A 0.5 kg mass is attached to a linear spring of stiffness 0.1 N/m. Determine the natural frequency of the system in hertz. b) Repeat this calculation for a mass of 50 kg and a stiffness of 10 N/m. Compare your result to that of part a. Solution: From the definition of frequency and equation (1.12) (a) ! n = k m = .5 .1 = 0.447 rad/s fn = ! n 2" = 2.236 2" = 0.071 Hz (b) ! n = 50 10 = 0.447rad/s, fn = ! n 2" = 0.071 Hz Part (b) is the same as part (a) thus very different systems can have same natural frequencies. 1.12 Derive the solution of the single degree of freedom system of Figure 1.4 by writing Newton’s law, ma = -kx, in differential form using adx = vdv and integrating twice. Solution: Substitute a = vdv/dx into the equation of motion ma = -kx, to get mvdv = - kxdx. Integrating yields: v2 2 = !"n 2 x2 2 + c2 , where c is a constant or v2 = !"n 2x2 + c2 # v = dx dt = !"n 2x2 + c2 # dt = dx !"n 2x2 + c2 , write u ="nx to get: t ! 0 = 1 "n du c2 ! u2 $ = 1 "n sin!1 u c % & ' ( ) * + c2 Here c2 is a second constant of integration that is convenient to write as c2 = -φ/ωn. Rearranging yields !nt +" = sin#1 !nx c $ % & ' ( ) * !nx c = sin(!nt +" )* x(t ) = Asin(!nt +" ), A = c !n in agreement with equation (1.19). 1.13 Determine the natural frequency of the two systems illustrated. (a) (b) Figure P1.13 Solution: (a) Summing forces from the free-body diagram in the x direction yields: -k1x +x -k2x Free-body diagram for part a m˙x˙ = !k1 x ! k2 x" m˙x˙ + k1x + k2 x = 0 m˙x˙ + x k1 + k2 ( ) = 0, dividing by m yields : ˙x˙ + k1 + k2 m #$ %& x = 0 Examining the coefficient of x yields: ! n = k1 + k2 m (b) Summing forces from the free-body diagram in the x direction yields: -k1x -k2x +x -k3x Free-body diagram for part b m!x! = !k1x ! k2x ! k3x," m!x!+ k1x + k2x + k3x = 0" m!x!+ (k1 + k2 + k3 )x = 0" !x!+ (k1 + k2 + k3 ) m x = 0 "# n = k1 + k2 + k3 m 1.14* Plot the solution given by equation (1.10) for the case k = 1000 N/m and m = 10 kg for two complete periods for each of the following sets of initial conditions: a) x0 = 0 m, v0 = 1 m/s, b) x0 = 0.01 m, v0 = 0 m/s, and c) x0 = 0.01 m, v0 = 1 m/s. Solution: Here we use Mathcad: a) all units in m, kg, s parts b and c are plotted in the above by simply changing the initial conditions as appropriate m 10 x0 0.0 T 2. ! fn "n !n 2. " x t A. sin . !n t " A . 1 !n x0 . 2 !n 2 v0 2 ! atan . "n x0 v0 0 0.5 1 1.5 0.2 0.1 0.1 0.2 x t xb t xc t t k 1000 ! n := k m v0 1 1.15* Make a three dimensional surface plot of the amplitude A of an undamped oscillator given by equation (1.9) versus x0 and v0 for the range of initial conditions given by –0.1 x0 0.1 m and -1 v0 1 m/s, for a system with natural frequency of 10 rad/s. Solution: Working in Mathcad the solution is generated as follows: !n 10 N 25 i 0 .. N j 0 .. N v0 j 1 . 2 N j A x0 , v0 . 1 !n !n . 2 x0 2 v0 2 M i , j A x0 , i v0 j Amplitude vs initial conditions 0 10 20 0 10 20 0 0.05 0.1 M x0 i 0.1 . 0.2 N i 1.16 A machine part is modeled as a pendulum connected to a spring as illustrated in Figure P1.16. Ignore the mass of pendulum’s rod and derive the equation of motion. Then following the procedure used in Example 1.1.1, linearize the equation of motion and compute the formula for the natural frequency. Assume that the rotation is small enough so that the spring only deflects horizontally. Figure P1.16 Solution