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Unformatted text preview: 6Problems and Solutions Section 6.4 (6.30 through 6.39)6.30Calculate the first three natural frequencies of torsional vibration of a shaft of Figure 6.7 clamped at x= 0, if a disk of inertia J= 10 kg m2/rad is attached to the end of the shaft at x= l. Assume that l= 0.5 m, J= 5 m4, G= 2.5 109Pa, = 2700 kg/m3.Solution:The equation of motion is&&=. Assume separation of variables: =(29(29to get &&= &&== 2so that&&q+2= 0 +2= 0where 2=2.The clampedinertia boundary condition is (0,t) = 0, and (,29=&&(,29 .This yields that (0) = 0 andGJ(29(29=(29 &&(29=(292(29orJ(29=2(29The solution of the spatial equation is of the form(29= + but the clamped boundary condition yields B= 0. The inertia boundary condition yieldsJA=2==15 410 2(2700/329 (0.5 29So the frequency equation istan=675Using the MATLAB function fsolve; this has the solutions1= 1.56852= 4.70543= 7.84241= 3.13692= 9.41083= 15.6847Thus1= 3018.5 rad/s f1= 480.4 Hz2=9055.6 rad/s f2= 1441.2 Hz3= 15092.6 rad/s f3= 2402.1 Hz2766.31Compare the frequencies calculated in the previous problem to the frequencies of the lumpedmass singledegreeoffreedom approximation of the same system.Solution:First calculate the equivalent torsional stiffness of the rod.k==(2.5 10929 (5290.5= 2.5 1010&&= &&+= 010&&+2.5 1010= 0&&+2.5 109= 0so that 2= 2.5 109, = 5 105rad/s or about 80,000 Hz, far from the 482 Hz of problem 6.30.2866.32Calculate the natural frequencies and mode shapes of a shaft in torsion of shear modulus G, length l, polar inertia J, and density that is free at x= 0 and connected to a disk of inertia Jat x= l.Solution:Assume zero initial conditions, i.e. (x,0) = &(,029= 0. From equation 6.662(,292=2(,292(1)The boundary condition at x= land at x= 0 isGJ(,29= 2(,292(0,29= 0Using separation of variable in (1) of form (x,t) = (x)T(t) yields:...
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This note was uploaded on 11/28/2010 for the course ME 4440 taught by Professor Hill during the Winter '09 term at Detroit Mercy.
 Winter '09
 Hill
 Torsion

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