SolSec 2.7 - 2-44Problems and Solutions Section 2.7 (2.55...

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Unformatted text preview: 2-44Problems and Solutions Section 2.7 (2.55 through 2.71)2.55Consider a spring-mass sliding along a surface providing Coulomb friction, with stiffness1.2 ×104N/m and mass 10 kg, driven harmonically by a force of 50 N at 10 Hz.Calculate the approximate amplitude of steady-state motion assuming that both the massand the surface that it slides on, are made of lubricated steel.Solution: Given:m= 10 kg, k= 1.2x410 N/m, oF= 50 N, drω=10(2π) = 20πrad/sω= mk= 34.64 rad/sfor lubricated steel, µ= 0.07From Equation 2.100 XFkmgFroo=−−14122µπ(())Xx=−−501 2 1014 07 10 9 815012034 64422.(.)()( .)()(.ππX=1.79 × 310−m2.56A spring-mass system with Coulomb damping of 10 kg, stiffness of 2000 N/m, andcoefficient of friction of 0.1 is driven harmonically at 10 Hz. The amplitude at steadystate is 5 cm. Calculate the magnitude of the driving force.Solution:Given:m= 10 kg, k= 2000 N/m, µ= 0.1, ω=10(2π) = 10(2π) = 20πrad/s,ωn= mk= 14.14 rad/s, X= 5 cmEquation 2.106XFkrmgkXo=−+()1422µπFXkrmgkXo=−+()1422µπFo=−+( .)().( . )()( .)()(.)0 05 200012014 144 0 1 10 9 81200005222ππoF= 1874 N2-452.57A system of mass 10 kg and stiffness 1.5 ×104N/m is subject to Coulomb damping. Ifthe mass is driven harmonically by a 90-N force at 25 Hz, determine the equivalentviscous damping coefficient if the coefficient of friction is 0.1.Solution:Given:m= 10 kg, k= 1.5x410 N/m, oF= 90 N, ω =25(2π) = 50πrad/s,ωn= mk= 38.73 rad/s, µ= .1Steady-state Amplitude – Equation 2.107XFkmgFroo=−−=×−−=×−141901 51014 0 1 10 9 819015038 733 8510224224µπππ()().( . )()( .)()..mEquivalent Viscous Damping Coefficient:CmgXeq==×−44 0 1 10 9 81503 85104µπωππ( . )()( .)()(.)eqC= 206.7 Ns/m2-462.58a. Plot the free response of the system of Problem 2.57 to initial conditions of x(0) = 0and x&(0) = |F/m| = 9 m/s using thesolution in Section 1.10.b.Use the equivalent viscous damping coefficient calculated in Problem 2.57 and plotthe free response of the “equivalent” viscously damped system to the same initialconditions.Solution: See Problem 2.57(a)x(0) = 0 and x&(0) = mFo= 9 m/s10105.14xmk==ω=38.73 rad/sFrom Equations 2.88:mgkxxmµ=+&&for <x&mgkxxmµ−=+&&for >x&Let mgFdµ== (.1)(10)(9.81) = 9.81 NTo start, ˙( )xBn91==ωTherefore, kFAd=1and Bn19=ωSo, x(t) = FkttFkdnndcossinωωω+−9This will continue until x&= 0, which occurs at time 1t:x(t) = AtBtFknnd22cossinωω++x&(t) = −+ωωωωnnnnAtBt22sincosx tAtBtFknnd( )cossin12121=++ωω˙( )sincosx tAtBtnnnn12121== −+ωωωωTherefore, AxtFktdn211=−( )( )/cosωand BxtFktdn211=−( )( )/sinωSo, x tx tFkttx tFkttFkdnndnnd( )( )/coscos( )/sinsin=−( )[ ]+−( )[ ]+1111ωωωωAgain, when x&= 0 at time 2t, the motion will reverse:x(t) = AtBtFknnd33cossinωω+−x&(t) = −+ωωωωnnnnAtBt33sincosx tAtBtFknn...
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This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 2.7 - 2-44Problems and Solutions Section 2.7 (2.55...

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