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NUS / Mechanical Engineering / ME 3112 / How to determine the original mass or spring stiffness?

How to determine the original mass or spring stiffness?

How to determine the original mass or spring stiffness?

Description

School: National University of Singapore
Department: Mechanical Engineering
Course: Mechanics of Machines
Professor: Ong teo
Term: Fall 2015
Tags:
Cost: 50
Name: ME3112 Part 2 review
Description: A very detailed and structrual summary of the second part of this course. Handwritten. with marking and corrections. If the prof cannot give you the clear thoughts in lecture notes, let this help you to study everything within 2 weeks. Don't miss
Uploaded: 11/14/2015
15 Pages 38 Views 1 Unlocks
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How to determine the original mass or spring stiffness?



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What is damped free vibration?



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4= 8/8422+82 7.2. Damped Force Vibration:

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ndx = kx- c d et

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Xut)= Xhlt) + Xplt) where Xplt)= Xosincut + 0) where

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Damping Factor S = &- anwn = 2 km

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x= è limit (asin wat + Calos wat)

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damped frequency Sl>l: x= cient + C2 e At Ako, a

system returns to its equielbrium in finite time t=0, x= xo and ds = vo define wa = ( 2 ) ( h) = WA? (E)= 1) = will 1921) XXL = e sunt | Votle watwad) xo . e wat vot (Swo-wa) Ko e wat ) x= 1 Gt Czt) e whit (21= 22=- Ce m-Wn)

system return to its equillibrium position in the shortest time without oscillation

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Pa = 22

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viscous damping friction force is directly proporcional and opposite to the velocity of the moving body. k damping force = c dx

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mx + cx + kx=0 x= e At

mai+2+k=0> characterstic equation

2=-2m /)=1k) Critical damping coeffcient radical

Ce=2m/k = amwn (radical=0)

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P= Pm sinust

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md=p-kx

(Let weight balance out the Xo)

mX+kX= (on sinugt

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S= Sm Sin wat

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case 2.

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the origional mass or spring stiffness?

Original: Tu = (22)? (W) R$

Soul Ta'= (22) (when)

Tn. Ta can be measured from experiments

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Am is know, we can sowe two unknown

R.

m.

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Due to the sinousidal property:

x= A sin (Wat+0) A=Xm ant = v= WnA cos Cunt +°) o

Vm= Km. Wn a= te rester - wn? A sin Umt+0)

Am=Xm. we?

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85. Application of Principle of Conservation

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Ve= h k (X+Xo) 2 equillibrium

Vg = -mgx

our T = 4 m (dx 2 x= xm sin (Wat+0)

dx = Wn Km Cos Clint+O) At X=Xm, T,=0 V = 7 kl Xm txo) - mg Xu At x=0, T2 = m ( x Won? V2 = kx 2

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3 k (X m + Xo } - mg xom = K x 2 + m (Xm Won) 2 Ikkunt ko + k XmKong Kn= kx 2 + I'm de che

7 kNm = mxu con

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consider taking moment about point o without conside kinetic force at G and kinetic torque is applicable

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up

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Io = 7 ML2 1. Ia tm 122

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- madsino = In

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1A - mb2 [112) + d2

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solution O= A cos wat + B sin wat

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to, W = We period remains the

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T= m va + 2 Ig 2

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Slider - Crank

Mechanism

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Normal Set up:

ostu x = L cosp + Rcoso o w R L ß Rsino = L siny 3 -one ---eo - derivative of them will give v

V=-BL sing-WR sino ®

WR COJO = Bucose 1 x=/ 42 - (Rsing)2 + Rcoso

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dv

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w a Resin Rw coso = O

get w and substitute into & get Vmax 2. Off-set Slider - Crank Mechanism

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X= 4 cos Q + REDSO H4 Rsine = L sing

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v=

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cos (42-B2) 3/2 - R 22-B?

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a 22-82 - Rew cose where B= HAR sine

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83. Free Vibration of Spring Mass sys No Static displacement due to the weight of mass os

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Static equilibrium - kXoting=0

- kx-kXoting=ma

- kx=ma matkxo

(radis) (Hg) mdx + kx=0 Wn-2afn

dt2 de + anexo, com

m TEPE dax+ winx=0

Cet solution be x = (Acos Wat +13 sinant another expression of solution or x= xm sin(Watto)

j(watto)

x= xm caso sin(Wit) + (xm singlos (wit) = xm cos (Wutto)+j Xm sin (witte) B

either real or the

Xm2 = A + B2 tande imaginary part is

need initial conditions to sowe ODE the solution.

t=0, x=Xo oup to dx = Vo

So A=Xo B V

* = Xocos Wnt + no sinant with xm = x2 + ( )

tand= Xown

x=xml

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• Complex spring systems

extension sane

forces same extension added

kon=k, tkz

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k }

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forces added

I

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Ktot = kitka

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rer

Problem:

a=ra To=lae To Id = I a mg-T=ma mg. Ia ama Tas ng

so if targer the r (with constant m. I)

larger the acceleration, a hence it will drop faster

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from Energy principle of view= way soul initial KE=0

final KE = 4 mv 2 + 3 Iw2

v=rw final KE = 4 mv 2 + 1 - 2 if youyo falls by a distance of

migh = {mv2 + 1 x Luz mgh

3 žm+mt

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