ME260W Mid-Term Exam Instructor: Xinyu Huang Date: Mar

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1 ME60W Mid-Term Exam Istrutor: Xiyu Huag Date: Mar Name: Grade: /00 Problem. A atilever beam is to be used as a sale. The bedig momet M at the gage loatio is P*L ad the strais o the top ad the bottom surfae of the beam are ε =+ My EI ad ε = My EI, i whih E is the Youg s modulus of the material, y ad I are ostats related to the shape of the beam ross-setio. Give L = 0. m, E = 36 GPa, ad I/y = 0.5E-8 m 3. L R R gage U o gage R 3 R 4 P Assume gage ad gage are the same 0 Ohm gage with gage fator GF =. Usig a Wheatstoe bridge (assume 0 Ohm dummy resistors available o ay u-used arms) with exitatio U i = 0V. a. Desig a quarter-bridge arragemet usig oly gage ad determie the rage of the output voltage U o whe the load P hages from 0 to 5 Newto. (0 pts) b. Desig a half-bridge arragemet usig both gage ad gage ad determie the rage of the output voltage U o whe the load P hages from 0 to 5 Newto. (0 pts). Calulate ad ompare the sesitivity (U o /P) for the quarter-bridge ad the halfbridge desig. (0 pts) d. Temperature is kow to ause uiform expasio or otratio of the atilever beam ad it also affets gage fator. Disuss the temperature effet o the auray of the sale with quarter-bridge ad half-bridge desig (0 pts) U i R R3 Hit: Defletio or output of a Wheatstoe bridge is Uo = U i R + R R + R R R the gage fator is defied as GF = δ. ε 3 4 ad Solutio: First determie the rage of strai for gage ad gage whe load P hage from 0 to 5 Newto. Without loadig, ε = ε = 0. Whe 5N load is applied,

2 [ N] [ m] My PL 5 * ε = = = = = = 5556µε EI I 9 N 8 3 E *0 * 0.5*0 m y m ε = ε = 5556µε The, determie the hage of resistae for gage ad gage, δ R = GF * ε δ R = R * GF * ε R 6 δrgage = δrgage = 0 Ω * * 5556 *0 =.3334 Ω [ ] [ ] a.) To form a quarter-bridge, gage a be plaed i R positio. As suh, after loadig, R will hage to.3334 Ohm ad R =R 3 =R 4 =0 Ohm. The output voltage is R R Uo = Ui = 0 = [ Volt] R + R R3 + R The rage of output will be from 0 to Volt. b.) To from a half-bridge, gage ad gage a be plae i the R ad R positio ad R 3 =R 4 =0 Ohm. The output voltage is R R Uo = Ui = 0 = Volt R + R R3 + R The rage of output will be from 0 to Volt..) The sesitivity for the quarter bridge desig will be U o [ Volt] [ N ] [ Volt N ] / P = = The sesitivity for the half bridge desig will be U o [ Volt] [ N ] [ Volt N ] / P = = The sesitivity of the half-bridge desig is roughly twie that of the quarter-bridge desig. d.) Assume the apparet strai ( ε a ) i the gage is the sum of mehaial strai (ε due to bed loadig) ad expasio strai due to temperature hage ( ε T ), ε = ε + ε I a quarter-bridge arragemet, ε T is a soure of error. I a half-bridge arragemet, the output is approximately proportioal to the arithmeti differee of the two gages plaed i R ad R positio, assumig the gages are the same type. I this ase, Total gage gage gage gage gage gage εa = εa + εa = ε + εt ε εt = ε ε So the error due to temperature idued expasio aels out i half bridge desig. However, the error due to the slight hage of gage fator (GF) a ot be elimiated i either quarter-bridge or half-bridge desig. a T [ ]

3 Problem. Suppose U(t) is the uit step futio. Determie the 90% rise time (time required for y to reah 90% of y ) of the systems give below, assume y(t) = 0 whe t = 0: a) 0.4 y& + y = 4U t (5pts) () y& + y = U() t + & + = () () ( ω ) b) 5 5 (5pts) ) && y y y U t (hit: step respose of a seod order system is Solutio: y t = y y + t e ζω whe ζ = ) (0 pts) t t = + 0 exp. Read τ from the first-order equatio, the time ostat for this system, τ =0.4, ad the steady state respose is y = 4 a.) The step respose of a first order system is: y() t y ( y y ) t90 t ( ) ( ) y t = 0.9 * y = y + (0 y ) e 0. = e t = 0.4 * l 0. = b.) Use the same approah as of a.), ote that τ =. t90 90 ( ) 0. = e t = * l 0. =.303.) First determie atural frequey ω, ad dampig ratioζ reall that So the step respose a be used. k ω = = = m v ζ = = = = mω ** ζω ( ) = 0.9 = ( + ω ) t 0. = ( + ) y t y y y t e t e 90 t To solve the oliear equatio, trasform it ito a iterative form: t = l + t l ( ) ( ) Assume t 90 = iitially, repetitively applies the above equatio t 90 overges to 3.89.

4 Problem 3. A first order system is exposed to a step hage of from 0 to 00 uits. After. se the istrumet idiates 80 uits. a. Estimate the time ostat of the istrumet (0 pts) b. Determie the 90% rise time of the system (5 pts) Solutio: a.) Respose of a first-order system with a step hage from y0 to y is: t y() t = y + ( y0 y ) exp, where τ is the time ostat of the first-order system. τ Hee,... y (.) = 80 = 00 + ( 0 00) exp exp = 0. = l ( 0.) τ τ τ Solve this equatio, τ =.3048 se. b.) To fid 90% rise time t t 90 = 00 + ( 0 00) exp exp = 0. t =.3048* l ( 0.) = So it takes about seods for the system to reah 90% of the step iput.

5 Problem 4. A weight m = 30 kg is oeted to a sprig with k = 000 N/m ad a visous damper with the dampig oeffiiet estimated to be v = 00 N-se/m. a. Determie the atural frequey ( ω ) of the system (5 pts) b. Determie the dampig ratio (ζ ) of the system (5 pts). Determie the damped frequey ( ω d ) of the system (5 pts) d. If the system is give a iitial displaemet of 0.05 m from the equilibrium x = positio l o ad the released suddely, determie the vibratio respose x( t) (0 pts) Solutio: The system a be desribed by the followig equatio of motio mx && + x& + kx = 0 a.) The atural frequey ω is determied by [ ] [ ] v Kg*m 000 se N 000 m k m ω = = = = rad/se m 30 Kg 30 Kg [ ] b.) To alulate dampig ratio, first determie the ritial dampig is ritially damped,. Whe the system 4mk = 0 k = mk = m = mω m Kg N*se m = * 30 Kg * rad/se = = = N*se m se se The dampig ratio is the alulated v 00 ζ = = = [ ] [ ] [ ]

6 .) Sie the system is uder-damped, the motio is periodi, ad the frequey of the damped motio, ω, is d ω = ω ζ = = d [ ] rad/se d.) To determie the vibratio respose, the equatio of motio eeds to be solved with the iitial oditios && x+ 00x& + 000x = 0 && x+ x& + x = the harateristi equatio ad the harateristi roots for above equatio are 0 00 λ + λ + = 0, λ = i, λ = i 3 3 Defie a = 3.333, b= 4.74 So, the geeral solutio of the system a be expressed i the followig form, at at x t = Ce si bt + C e os( bt) ( ) ( ) Take the first derivate of x(t) with respet to t, at at at at x& t = ac e si bt + bc e os bt + ac e os( bt) bc e si bt) () ( ) ( ) ( at ( ac bc ) e si ( bt) ( bc ac ) os( bt) = + + Use iitial oditios are used to determie C ad C, x 0 = 0.05 C = 0.05 ( ) a x& ( 0) = 0 bc+ ac = 0 C = C = 0.707C = b So the system time respose is x () t = *exp( t) *si ( 4.74t) exp( 3.333t) os( 4.74 t ) Survey Questio: What a we do to improve your learig experiee i this ourse? (Aswer briefly, ot aouted i your grades)