9780134361307, Chapter 3, Problem 38P

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Western University *

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344

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Mechanical Engineering

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Oct 30, 2023

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Chapter 3, Problem 38P Problem A turbine rotor is mounted on a stepped shaft that is fixed at both ends as shown in Fig. The torsional stiffness of the two segments of the shaft are given by kt1 = 3,000 N-m/rad and kt2 = 4,000N-m/rad. The turbine generates a harmonic torque given by M(f) = MO cos wt about the shaft axis with MO = 200N-m and w= 500 rad/s. The mass moment of inertia of the rotor about the shaft axis is JO = 0.05 kg-m 2. Assuming the equivalent torsional damping constant of the System as ct = 2.5 N-m-S/rad, determine steady-state response of the rotor, at). Fig: g ) O\ £ Ry / \\ k ‘\’ '\I i 0 { \\ ky A || B F TR S | § Sy 7 ,\ / / MI1) = My cos a 4 Turbine rotor, /, \‘\ /”/‘ e P Step-by-step solution Step 1072 The motion equation of a turbine rotor is expressed as: Jf+cf+(k,+k,)0=M(t) ... (1) Hence the mass moment of inertia of the rotor is .J,,, the torsional damping constantis ¢,, the torsional stiffness of section 1 is I(,I , the torsional stiffness of section 2 is k:z’ the angular acceleration is g, the angular velocity is ¢, the angular displacement is ¢, and the moment function is M (r). The moment function is determined as: M(t)= M, cos ot With the moment constantis ), and the angular frequency of momentis . Substitute M, coser for M(f) into equation (1) Job+¢,0+(k, +k,)0 =M, cos o Step207 5 The steady state response of the above motion equation, 6, () is determined as: 0.(1)= X cos(wt —¢) With the amplitude of the response is X and the phase angle is ¢. Determine the amplitude of the response, X by using the following relation: M, 2 S, 12 [(k,, +k,—J,@ ) + c,'w’] Substitute 200 Nm for M,, 3000 N-m/rad for k,, 4000 N-m/rad for k,, 0.05kg-m* for J,, 2.5N-m-s/rad for ¢, and 500 rad/s for . XY= 200 12 2 2\2 2 [(3ooo+ 4000-0.05(500)’ ) +(2.5)’ (500) ] =0.0355m Therefore, the amplitude of the response, Xis 0.0355m . Step 3072 Determine the phase angle, ¢ by using the following relation: ¢:tan"(—c‘w ] k, +k,—J,0 Substitute 3000 N-m/rad for k,, 4000 N-m/rad for k,, 0.05kg-m’ for J,, 2.5N-m-s/rad for ¢, ,and 500 rad/s for @. (2.5)(500) - 3000+4000—0.05(500)2 =-12.8° Therefore, the phase angle, ¢ is —12.8°. Determine the steady state response of the vibration system, @, (1) by using the following relation: 0.(t)= X cos(wt - @) Substitute 0.0355 mfor ., —12.8° for ¢, and 500 rad/s for @. 0,(1)=0.0355cos(500r - (-12.8°)) =0.0355cos(5007 +12.8°) Therefore, the steady state response of the vibration system ¢, (1) is [0.0355cos(500r +12.8°)
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