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Engineering Vibration Inman 3rd Edition Solution Manual

Engineering Vibration Inman 3rd Edition Solution Manual Average ratng: 9,7/10 4198votes

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CD1 Stamp SOLUTION MANUAL FOR 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.

Engineering Vibration Inman 3rd Edition Solution ManualEngineering Vibration Inman 3rd Edition Solution Manual

Engineering Vibration Solution manual daniel j. Inman Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. Engineering Vibration, 3rd Edition. Serving as both text and reference manual. Engineering Vibrations, 2nd Edition. Related to engineering vibration inman 4th edition solutions slideshare engineering vibration 4th edition solutions engineering vibrations inman 3rd edition pdf engineering vibrations inman 4th edition pdf engineering vibration 4th edition solution pdf engineering.

M(kg) 10 11 12 13 14 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: 0=+ kxxm!! (1) Assume: x(t) = ae rt. Then: rt arex =!

And rt earx 2 =!!. Substitute into equation (1) to get: mar 2 e rt + kae rt = 0 mr 2 + k = 0 r = ± k m i Thus there are two solutions: x 1 = c 1 e k m i! '# $%& t, and x 2 = c 2 e ' 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: itit ececxxx 2 2 2 121! +=+= Substitute initial conditions: x0 = 1 mm, v0 = 5 mm/s x 0( ) = c 1 + c 2 = x 0 = 1!

C 2 = 1' c 1, and v 0( ) =!x 0( ) = 2ic 1 ' 2ic 2 = v 0 = 5 mm/s!' 2c 1 + 2c 2 = 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns): '2c 1 + 2 ' 2c 1 = 5 i! C 1 = 1 2 ' 5 4 i, and c 2 = 1 2 + 5 4 i Therefore the solution is: x = 1 2! 5 4 i ' #$% &' e 2it + 1 2 + 5 4 i ' #$% &' e!2it Using the Euler formula to evaluate the exponential terms yields: x = 1 2!

Engineering Vibration Inman 3rd Edition Solution Manual

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.

Engineering Vibration Inman Solution Manual

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 v 0 = 0.!n = k m = 2 rad/s ' #$% &'. Calculating the initial conditions: x 0( ) = c1 + c2 = x0 = 1! C2 =1 ' c1 v 0( ) = ˙ x 0( ) = 2ic 1 ' 2ic 2 = v 0 = 0! C 2 = c 1 c 2 = c 1 = 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) = 2 2 (1) 2 + 0 2 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: rad/s5,rad2,mm1 ===!' For an undamped system: x t( ) = Asin!nt + '( ) = 1sin 5t + 2( ) and v t( ) =!x t( ) = A!n cos!nt + '( ) = 5cos 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 sincemg = kxs from static equilibrium.

1. 98 Trx450es Service Manual. 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 free- body 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: ( ) ( )!' += tAtx n sin and differentiating yields the velocity: v t( ) = A!n cos!nt + '( ). Realizing that both the sin and cos functions have maximum values of 1 yields: ( ) mm/s 62.8===!'

201 max n Av Likewise for the acceleration: ( ) ( )!' ' +#= tAta nn sin 2 ( ) 2mm/s 3948=== 22 max 201!' N Aa 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( ) = sinacosb + cosasinb Given: ( ) ( ) ( ) ( ) ( )!' Sincoscossinsin tAtAtA nnn +=+ = Bsin!nt + C cos!nt where B = A cos' and C = A sin' 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) =!n Bcos(!nt) '!nCsin(!nt) Now substitute the initial conditions into these expressions for the position and velocity to get: x 0 = x(0) = Bsin(0) + C cos(0) = C v 0 = ˙ x (0) =!nB cos(0) '!nC sin(0) =!nB(1) '!