Problems and Solutions for Section 3.2 (3.15 through 3.25)

Size: px

Start display at page:

Download "Problems and Solutions for Section 3.2 (3.15 through 3.25)"

Eustacia Boyd
6 years ago
Views:

1 3-7 Problems ad Soluios for Secio 3 35 hrough Calculae he respose of a overdamped sigle-degree-of-freedom sysem o a arbirary o-periodic exciaio Soluio: From Equaio 3: x = # F! h "! d! For a overdamped SDOF sysem see Problem 34 h! " = = F " x x = m#! m#! e!# e!#!" e #!!"!#! e!!" ' d" e!#!" e #!!"!#! e!!" ' d" F " e # " e #!!"!#! e!!" ' m#! d"

2 Calculae he respose of a uderdamped sysem o he exciaio give i Figure P36 Plo of a pulse ipu of he form f = F si Soluio: x = F For, Figure P36 F e #! si! d " ' d e "#! m! d = F si < From Figure P36 x = F sie #! si! d " d m! d e "#! x = F e "#! m! d! d! ' { } e #!! d " si " #! cos' "! " d si! d " #! cos! d e #!! d! {! ' d " si " #! cos'! " d si! d " #! cos! d } ' For! > ", : # f! h "! d! = # f! h "! d! # h "! d!

3 3-9 " # " #! d! x = F m! d e "#! " #! d! sie #! si! d " d = F e "#! ' m! d e! "! d ' si " #! d ' '! cos "! ' # d - #, '! d ' si! d '! cos! d / e! "! d si " #! d '! cos " #! ' d - #,! d ' si! d '! cos! d / Aleraely, oe could ake a Laplace Trasform approach ad assume he uder-damped sysem is a mass-sprig-damper sysem of he form m!!x The forcig fucio give ca be wrie as = F H! H! " F si Normalizig he equaio of moio yields c!x kx = F si!!x!"!x " x = f H # H # where f = F m ad m, c ad k are such ha <! < Assumig iiial codiios, rasformig he equaio of moio io he Laplace domai yields X s = f e!" s s s # s The above expressio ca be covered o parial fracios X s = f e!" s # As B s ' f e!" s where A, B, C, ad D are foud o be # Cs D s s '

4 3-!"# A =! # "# # B =!! # "# "# C =! # "# D =! # "#! # "# Noice ha X s ca be wrie more aracively as = f As B X s # s Cs D s!" s " ' f # As B e s s Cs D s!" s " ' e s G s = f G s Performig he iverse Laplace Trasform yields = f g H! " x g! " where g is give below g = Acos Bsi Ce!"# cos # d D! C"# # d ' e!"# si # d! d is he damped aural frequecy,! d =! " # Le m= kg, c= kg/sec, k=3 N/m, ad F = N The sysem is solved umerically Boh exac ad umerical soluios are ploed below

5 3- Below is he code used o solve his problem Esablish a ime vecor =[::]; Figure Aalyical vs Numerical Soluios Defie he mass, sprig siffess ad dampig coefficie m=; c=; k=3; Defie he ampliude of he forcig fucio F=; Calculae he aural frequecy, dampig raio ad ormalized force ampliude zea=c/sqrkm; w=sqrk/m; f=f/m; Calculae he damped aural frequecy wd=wsqr-zea^; Below is he commo deomiaor of A, B, C ad D parial fracios coefficies dummy=-w^^zeaw^; Hece, A, B, C, ad D are give by A=-zeaw/dummy; B=w^-/dummy; C=zeaw/dummy;

6 3- D=-w^zeaw^/dummy; EXACT SOLUTION for i=:legh Sar by defiig he fucio g gi=acosibsiicexp-zeawicoswdid- Czeaw/wdexp-zeawisiwdi; Before =pi, he respose will be oly g if i<pi xei=fgi; d is he idex of delay ha will correspod o =pi d=i; else Afer =pi, he respose is g plus a delayed g The amou of delay is pi secods, ad i is d icremes xei=fgigi-d; ed; ed; NUMERICAL SOLUTION Sar by defiig he forcig fucio for i=:legh if i<pi fi=fsii; else fi=; ed; ed; Defie he rasfer fucios of he sysem This is give below

7 3-3 s^zeaww^ Defie he umeraor ad deomiaor um=[]; de=[ zeaw w^]; Esablish he rasfer fucio sys=fum,de; Obai he soluio usig lsim x=lsimsys,f,; Plo he resuls figure; segcf,'color','whie'; plo,xe,,x,'--'; xlabel'timesec'; ylabel'respose'; leged'forcig Fucio','Exac Soluio','Numerical Soluio'; ex6,5,'\uparrow','fosize',8; axes'posiio',[55 3/8 5 5] plo6:63,xe6:63,6:63,x6:63,'--'; 37 Speed bumps are used o force drivers o slow dow Figure P37 is a model of a car goig over a speed bump Usig he daa from Example 4 ad a udamped model of he suspesio sysem k = 4 x 5 N/m, m = 7 kg, fid a expressio for he maximum relaive deflecio of he car s mass versus he velociy of he car Model he bump as a half sie of legh 4 cm ad heigh cm Noe ha his is a movig base problem

8 3-4 Figure P37 Model of a car drivig over a speed bump Soluio: This is a base moio problem, so he firs sep is o raslae he equaio of moio io a useable form Summig forces yields i he verical direcio yields m!!x k x! y = were he displaceme y is prescribed Nex defied he relaive displaceme o be z = x-y, he relaive moio bewee he car s wheel ad body The equaio of moio becomes: m!!z m!!y kz =! m!!z kz = "m!!y Subsiuio of he form of y io his las expressio yields: m!!z kz = my! b si! b " # " # where Φ is he Heavyside sep fucio ad ω b is he frequecy associaed wih he bump The relaioship bewee he bump frequecy ad he car s cosa velociy is! b = "! v = "! v where v is he speed of he car i m/s For cosa velociy, he ime = v!, whe he car fiishes goig over he bump Here, z is From equaio 33 wih zero dampig he soluio is: Subsiuio of f =y yields: z = m! f " # si! #d# < z = Y! b! si! b "! b # si! #d# = = Y! b! "!! b si! b "!! b # " si!! "! # b b ' '! "! b = Y! b!! "!! si! "! si! b b < b where he iegral has bee evaluaed symbolically Clearly a resoace siuaio prevails Cosider wo cases, high speed! b >>! ad low speed! b <<! as whe he wo frequecies are ear each oher ad obvious maximum occurs For high speed, he ampliude ca be approximaed as Y! b # Y! b!! b! "! b! /! b si! b " si! For he values give, his has magiude:! b!! "! si! b

9 3-5 Zv! # Y "! ' 3 v 3 b This icreases wih he cube of he velociy Thus he faser he car is goig he more sever he bump is larger relaive ampliude of vibraio, hece servig o slow mooris dow A plo of magiude versus speed shows bump size is amplified by he suspesio sysem For slow speed, magiude becomes Zv! Y " #! ' v b A plo of he approximae magiude versus speed is give below

10 3-6 Clearly a speeds above he desiged velociy here is srog amplificaio of he bump s magiude, causig discomfor o he driver ad passegers, ecouragig a slow speed whe passig over he bumb 38 Calculae ad plo he respose of a udamped sysem o a sep fucio wih a fiie rise ime of for he case m = kg, k = N/m, = 4 s ad F = N This fucio is described by F = " F!! # F > Soluio: Workig i Mahcad o perform he iegrals he soluio is:

11 3-7

12 A wave cosisig of he wake from a passig boa impacs a seawall I is desired o calculae he resulig vibraio Figure P39 illusraes he siuaio ad suggess a model This force i Figure P39 ca be expressed as = F! # F, " ' > Calculae he respose of he seal wall-dike sysem o such a load Soluio: From Equaio 3: x =! F " h # " d" From Problem 38, h! " = si# m#! " for a udamped sysem For < : x = m! " F # ' si! # - d,- / x = F m! " si! # d # " si! # - d,- / Afer iegraig ad rearragig, x = F # si! k! " F ' k # " cos! ' < For > : # f! h "! d! = # f! h "! d! # h "! d! x = - -, -,- m! x = F m! " F # ' " si! # si! # d / d # " si! # d /

13 3-9 Afer iegraig ad rearragig, x = F k! # si! " si! " " F k # cos! > 3 Deermie he respose of a udamped sysem o a ramp ipu of he form F = F, where F is a cosa Plo he respose for hree periods for he case m = kg, k = N/m ad F = 5 N Soluio: From Eq 3: x =! F " h # " d" From Problem 38, h! " = si# m#! " for a udamped sysem Therefore, x = m! ' Afer iegraig ad rearragig, '" F # si! # d# = F x = F m! " # si! # d# " # m!!! si! " ' = F k # F si! k! Usig he values m = kg, k = kg, ad F = 5 N yields x = 5! 5si m

14 3-3 3 A machie resig o a elasic suppor ca be modeled as a sigle-degree-offreedom, sprig-mass sysem arraged i he verical direcio The groud is subjec o a moio y of he form illusraed i Figure P3 The machie has a mass of 5 kg ad he suppor has siffess 5x 3 N/m Calculae he resulig vibraio of he machie Soluio: Give m = 5 kg, k = 5x 3 N/m,! = k = 548 rad/s ad ha m he groud moio is give by: 5!! y = 75 " 5!! 6 ' # 6 The equaio of moio is m!!x kx! y = or m!!x kx = ky = F The impulse respose fucio compued from equaio 3 for a udamped sysem is h! " = si# m#! " This gives he soluio by iegraig a yh across each ime sep: x = ky" si! m! # " d" =! y" si! # " d" For he ierval < < : x =! For he ierval < < 6: 5" si! # " d" x = 5 # 456si548 mm x =! 5" si! # " d"! 75 # 5" si! # " d" = 75 # 5cos548 # # 5 8si 548 # Combiig his wih he soluio from he firs ierval yields: x = 75 5! 5cos548! 648si548!! 456si548! mm " " 6 Fially for he ierval >6:

15 3-3 6 x =! 5 si! " # d#! 75 " 5si! " # d#! si! " # d# = "5cos548 " " 8si 548 " 6 8si 548 " Combiig his wih he oal soluio from he previous ime ierval yields: x =!5cos548! 684si 548!! 8si 548! 6! 456si 548 mm " 6

16 p = "! d Cosider he sep respose described i Figure 37 Calculae p by oig ha i occurs a he firs peak, or criical poi, of he curve Soluio: Assume = The respose is give by Eq 37: x = F k! F e!"# cos # k! " d! To fid p, compue he derivaive ad le!x =!x =!F!"# k! " e!"# cos # d!!"# cos # d! e!"#!# d! # d si # d! = a # d! =!"# # d si# d! ' = "!! d " # " = a " '! π ca be added or subraced wihou chagig he d age of a agle =,! " # a /! d '! - d # ' Bu,! = a " " # So, = " a #! d ' # # - a# - ' #, /

17 Calculae he value of he overshoo os, for he sysem of Figure P37 Soluio: The overshoo occurs a p =! " d Subsiue io Eq 37: x p = F k! F e!"# /# d cos, # k! " d ' # d! / - The overshoo is os = x p! x ss os = F k! F k! " e!"# /# d # ' Sice! = a ", he cos! = -# " # F os =! k! "!cos! F k e!"# /# d! " os = F k e!"# /# d 34 I is desired o desig a sysem so ha is sep respose has a selig ime of 3 s ad a ime o peak of s Calculae he appropriae aural frequecy ad dampig raio o use i he desig Soluio: Give s = 3s, p = s Selig ime: Peak ime: s = 35 = 3 s #!"!" = 35 = 667 rad/s 3 p =! " d = s # " d = " =! rad/s # " 667 ' " =! # " 667, / ' " =! -, # " 36 / - " =! # " 3 =! # " = 335 rad/s

18 3-34 Nex use he selig ime relaioship o ge he dampig raio:! = 667 = 667 " 335 #! = 3483

19 Plo he respose of a sprig-mass-damper sysem for his ipu of Figure 38 for he case ha he pulse widh is he aural period of he sysem ie, = π ω Soluio: The values from Figure 37 will be used o plo he respose F = 3 N k = N/m! = " = 36 rad/s From example 3 ad Figure 37, wih =! " we have for = o, x = F k! F e!"# k! " cos # d! ' " where = a! ' x = 3-35e -36 cos344 - < For >, x = F e!"# k! " e "#! ", cos # d! ' #! / 64! cos # d! 7-84 x = 35e -36 {369cos cos344 - } > The plo i Mahcad follows:

Vibration 2-1 MENG331

Vibration 2-1 MENG331 Vibraio MENG33 Roos of Char. Eq. of DOF m,c,k sysem for λ o he splae λ, ζ ± ζ FIG..5 Dampig raios of commo maerials 3 4 T d T d / si cos B B e d d ζ ˆ ˆ d T N e B e B ζ ζ d T T w w e e e B e B ˆ ˆ ζ ζ