Simulation, Transfer Function

Transcription

1 Max Force (lb) Displacement (in) 1 ME313 Homework #12 Simulation, Transfer Function Last Updated November 17, 214. Repeat the car-crash problem from HW#6. Use the Matlab function lsim with ABCD format to perform the simulation. Note, that, for the car crash problem, the input u=. Before the simulation, compute the eigenvalues of [A], and the time constants and frequencies of oscillation. Set the time range for the simulation to be 3 times the longest time constant. Again consider the carcrash problem. Only now, write a Matlab script that will perform a simulation for a sequence of seat-belt stiffnesses k 2 ranging from k 2 =5 lb/ft to k 2 =5 klb/ft and seatbelt damping coefficients c 2 ranging from c 2 =2 lb s/ft to c 2 =2 lb s/ft. Use the function lsim to perform each simulation. After each simulation, extract the maximum force exerted by the seatbelt on the occupant, and the maximum relative displacement between the occupant and car. Plot both of these quantities versus the seat-belt stiffness k 2 and seatbelt damping ratio c 2 as a surface Your plots should look like those shown above Damping Coefficient (lbs/ft) Damping Coefficient (lbs/ft) Maximum Occupant Displacement Maximum Force on Occupant Stiffness (klb/ft 2 3 Stiffness (klb/ft

2 2 Esfandiari and Lu text, problem set 4.3, problems 1 and 2. Only compute the transfer function for the specified output. I.e., for problem 1, the specified output is x 2 (t). And, o for problem 1, determine T(j ), T(j ), arg[t(j )], and the steady-state output x 2 (t) if f(t)=5sin(2t). o For problem 2, determine T(j ), T(j ), arg[t(j )], and the steady-state output x 1 (t) if f(t)=.5sin(1.5t). Esfandiari and Lu text, problem set 4.3, problem 7. Defining q 1 (t) as the output o Determine the transfer function T(s) between Q 1 (s) as output and V(s) as input. o Plot the amplitude and phase response of q 1 (t) assuming that v(t)=3sin( t), and the driving frequency varies over the range < <1 rad/sec. o Use the amplitude and phase plot to determine the steady state response v(t) when =.5 rad/sec. It will be useful to you to use the command grid on; with your plot.

3 Displacement (m) 3 Handout problem Solve the problem by plotting the frequency response of the vertical motion of the car versus frequency, and identify the natural frequencies by locating the frequencies at which peaks in amplitude of car motion appear. Hint Suppose the road surface vertical variation is y(x) =.25sin ( 2π x) ft, where x (in ft) is the horizontal displacement of the vehicle along the road. This would model the road variation with a period of 2 ft as described in the problem statement. Then suppose the vehicle is travelling at a constant velocity v. Then, since v=x/t, the road surface variation as a function of time will be y(t) =.25sin ( 2π vt). 2 Again handout problem Use lsim and tf to simulate the motion of the car when driven over a speed bump. Perform the simulation over a time range of t 1 sec. Simulate the bump as an input of y(t)=.1m of t 1 sec, and y(t)=m of 1 t 1 sec. Repeat the simulation when a damper c=2 lb s/ft is added between the wheel and car (you will need to derive a new transfer function to include the damper). Plot both simulations on the same graph. Your result should look like: Car Without Damper Car With Damper Bump Time (sec)

4 Code for Car Crash Simulation clear all;close all;clc; HW 7 Linear Simulation Problem ICS x1= Vehicle position x2= Vehicle velocity x3= Occupant position (inertial) x4= Occupant velocity (inertial) xo(1)=; xo(2)=-(528/36)*5; xo(3)=; xo(4)=-(528/36)*5; k1=1e3; c1=5; m1=3/32.2; k2=5e3; c2=5 ; m2=15/32.2; A=[ 1 ; -(k1+k2)/m1 -(c1+c2)/m1 k2/m1 c2/m1 ; 1 ; k2/m2 c2/m2 -k2/m2 -c2/m2]; B=[ ; ; ; ]; C=[1-1 ]; D=; [E,L]=eig(A) wn1=abs(l(1,1));z1=-real(l(1,1))/wn1;wd1=wn1*sqrt(1-z1^2);td1=(2*pi)/wd1; tau1=1/(z1*wn1); wn2=abs(l(3,3));z2=-real(l(3,3))/wn2;wd2=wn2*sqrt(1-z2^2);td2=(2*pi)/wd2; tau2=1/(z2*wn2); Set up time range tend=5*max([tau1 tau2]); N=5; t=linspace(,tend,n); dt=t(2)-t(1); Perform simulation u=zeros(1,length(t)); sys=ss(a,b,c,d); [y,t,x]=lsim(sys,u,t,xo); y=x(:,1)-x(:,3); f=(-k2*(x(:,3)-x(:,1))-c2*(x(:,4)-x(:,2))); subplot(3,1,1); plot(t,x(:,1)*12,t,x(:,3)*12) title('inertial Coordinates of Car and Occupant') axis([ tend -1*12 1*12]); legend('x1','x3'); ylabel('displacement (in)');

5 Force (lb) Displacement (in) Displacement (in) subplot(3,1,2); plot(t,y*12) title('relative Displacement Between Car and Occupant') axis([ tend 1.1*min(y)*12 1.1*max(y)*12 ]); ylabel('displacement (in)'); subplot(3,1,3); plot(t,f) title('force of Seat-Belt on Occupant') axis([ tend 1.1*min(f) 1.1*max(f)]); ylabel('force (lb)'); xlabel('time'); Result of Simulation 1 Inertial Coordinates of Car and Occupant x1 x Relative Displacement Between Car and Occupant Force of Seat-Belt on Occupant Time

6 Maximum force and relative displacement for varying seatbelt stiffness and damping ratio. Code below to generate data shown in plots contained in assignment (although plot axis limits are included in the code). clear all; close all; Compute maximum force exerted by seatbelt on occupant for a range of seatbelt stiffnesses k1=1e3; c1=5; m1=3/32.2; k2=5e3; c2=5 ; m2=15/32.2; Nk2=2; k2t=linspace(5,5e3,nk2); Nc2=2; c2t=linspace(2,2,nc2); ICS x1= Vehicle position x2= Vehicle velocity x3= Occupant position (inertial) x4= Occupant velocity (inertial) xo(1)=; xo(2)=-(528/36)*5; xo(3)=; xo(4)=-(528/36)*5; for i1=1:nk2 for j1=1:nc2 k1=1e3; c1=5; m1=3/32.2; k2=5e3; c2=5 ; m2=15/32.2; k2=k2t(i1); c2=c2t(j1); A=[ 1 ; -(k1+k2)/m1 -(c1+c2)/m1 k2/m1 c2/m1 ; 1 ; k2/m2 c2/m2 -k2/m2 -c2/m2]; B=[ ; ; ; ]; C=[1-1 ]; D=; [E,L]=eig(A) wn1=abs(l(1,1));z1=-real(l(1,1))/wn1; tau1=1/(z1*wn1); wn2=abs(l(3,3));z2=-real(l(3,3))/wn2; tau2=1/(z2*wn2); tend=5*max([tau1 tau2]); N=5; t=linspace(,tend,n); u=zeros(1,length(t)); sys=ss(a,b,c,d); [y,t,x]=lsim(sys,u,t,xo); y1=x(:,1)-x(:,3); ymax(j1,i1)=max(abs(y1));

7 f=(-k2*(x(:,3)-x(:,1))-c2*(x(:,4)-x(:,2))); fmax(j1,i1)=max(abs(f)); end end subplot(2,1,1); surf(k2t/1,c2t,ymax*12) title('maximum Occupant Displacement') xlabel('stiffness (klb/ft'); ylabel('damping Coefficient (lbs/ft)'); zlabel('displacement (in)'); subplot(2,1,2); surf(k2t/1,c2t,fmax) title('maximum Force on Occupant') xlabel('stiffness (klb/ft'); ylabel('damping Coefficient (lbs/ft)'); zlabel('max Force (lb)');

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9 Code for Esfandiari and Lu text, set 4.3, problem 7 and accompanying instructions V=3; Nw=5; w=linspace(,1,nw); for i1=1:nw s=1i*w(i1); T(i1)=(2*s+4)/(2*s^3+6*s^2+9*s+2); end subplot(2,1,1); plot(w,v*abs(t)) ylabel('amplitude Q_1'); xlabel('frequency (rad/sec)'); grid on; subplot(2,1,2); plot(w,angle(t)*(18/pi)) ylabel('phase {\phi} (deg)'); xlabel('frequency (rad/sec)'); grid on; Resulting Output 6 Amplitude Q Frequency (rad/sec) Phase (deg) Frequency (rad/sec) At =5 rad/sec, the amplitude of q1(t) is approximately 3, and the phase is approximately -65 o. So, the steady state solution q 1(t) is q 1(t)=3sin(.5t-65o).

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12 Car Displacement Amplitude x3 (ft) Car Displacement Amplitude x3 (ft) Code for frequency response clear all;close all;clc ms=625/32.2; mu=19/32.2; ks=1e3; kt=8e3; bs=; Nv=5; v=linspace(1,1,nv); w=((2*pi)/2)*v; for i1=1:nv s=i*w(i1); N=bs*kt*s+ks*kt; D=ms*mu*s^4+(bs*ms+bs*mu)*s^3+(ks*ms+kt*ms+ks*mu)*s^2+bs*kt*s+ks*kt; T(i1)=N/D; end plot(v,abs(t)*.15) ylabel('car Displacement Amplitude x3 (ft)'); xlabel('velocity (ft/sec)'); title(['b_s=',num2str(bs)]); axis([ 1 1]); Frequency response for bs= and bs=4 1 b s = Velocity (ft/sec) 1 b s = Velocity (ft/sec)

13 Code for simulation of car over speed bump clear all; close all; clc Car Suspension Frequency Response ms=625/32.2; mu=19/32.2; ks=1e3; kt=8e3; bs=4; b=[bs*kt ks*kt]; a=[ms*mu bs*ms+bs*mu ks*ms+kt*ms+ks*mu bs*kt ks*kt]; TFsys=tf(b,a); Nt=5; t=linspace(,1,nt); x1=zeros(1,nt); for i1=1:nt if (t(i1) <= 1) x1(i1)=.1; else x1(i1)=; end end x3(1)=; x3(2)=; [x3,t]=lsim(tfsys,x1,t,x3); plot(t,x3,t,x1) ylabel('car Displacement x3 (ft)'); title(['shock Absorber b_s=',num2str(bs),' lbs/ft']); xlabel('time (sec)');

14 Car Displacement x3 (ft) Car Displacement x3 (ft).25 Shock Absorber b s = lbs/ft Time (sec).2 Shock Absorber b s =4 lbs/ft Time (sec)