Newton-Raphson Example 4: Solutions of system of nonlinear equations: Newton-Raphson Example 4: The kinematic equations for a Four-Bar mechanism can be written as (5th semester, Mechanisms Course) L2=0.15 m L3=0.45 m L4=0.28 m s1=0.2 m s1 L2 L3 L4 θ2 θ3 θ4 3 4 2 1 Where link 2 is the input member. How do you calculate θ3 and θ4 when θ2=120°. -0.075 0.13

Following changes are made in the computer program. Solutions of system of nonlinear equations: Following changes are made in the computer program. clc, clear x=[0.5 1] ; err=[0.01 0.01]; niter1=10;niter2=50; err=transpose(abs(err)); for n=1:niter2 x %Error Equations--------------------------- a(1,1)=-0.45*sin(x(1));a(1,2)=0.28*sin(x(2)); a(2,1)=0.45*cos(x(1));a(2,2)=-0.28*cos(x(2)); b(1)=-(0.45*cos(x(1))-0.28*cos(x(2))-0.275); b(2)=-(0.13+0.45*sin(x(1))-0.28*sin(x(2))); %---------------------------------------------- bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps); if n>niter1 if abs(eps)<err break else display ('Roots are not found') end (Initial angle values must be given in RADIAN) ANSWER: θ3=0.216 rad (12.37°) θ4=0.942 rad (53.97°) Alternative solution with MATLAB clc;clear [x,y]=solve('0.45*cos(x)-0.28*cos(y)=0.275','0.13+0.45*sin(x)-0.28*sin(y)=0'); vpa(x,6) vpa(y,6)

Newton-Raphson Example 5: Solutions of system of nonlinear equations: Newton-Raphson Example 5: Kinematic equations for a crank mechanism are given below (5th semester Mechanisms Course) L2=0.15 m L3=0.6 m θ3 L2 L3 θ2 s Where link 2 (crank) is the input member. How dou you calculate θ3 and s with computer when θ2=60°. 0.075 0.1299

Following changes are made in the computer program. Solutions of system of nonlinear equations: Following changes are made in the computer program. clc, clear x=[-1 0.8] ; err=[0.01 0.01]; niter1=10;niter2=50; err=transpose(abs(err)); for n=1:niter2 x %Error Equations--------------------------- a(1,1)=-0.6*sin(x(1));a(1,2)=-1; a(2,1)=0.6*cos(x(1));a(2,2)=0; b(1)=-(0.075+0.6*cos(x(1))-x(2)); b(2)=-(0.1299+0.6*sin(x(1))); %---------------------------------------------- bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps); if n>niter1 if abs(eps)<err break else display ('Roots are not found') end ANSWER: θ3=-0.2182 rad (-12.5°) s=0.6607 m Alternative solution with MATLAB clc;clear [x,y]=solve('0.075+0.6*cos(x)-y=0','0.1299+0.6*sin(x)=0'); vpa(x,6) vpa(y,6)

Newton-Raphson Example 6: Solutions of system of nonlinear equations: Newton-Raphson Example 6: The time-dependent locations of two cars denoted by A and B are given as At which time t, two cars meet?

Newton-Raphson Example 6: Solutions of system of nonlinear equations: Newton-Raphson Example 6: clc;clear t=solve('t^3-t^2-4*t+3=0'); vpa(t,6) Alternative Solutions with MATLAB clc, clear x=1;err=0.001; niter=20; %---------------------------------------------- for n=1:niter f=x^3-x^2-4*x+3; df=3*x^2-2*x-4; eps=-f/df; x =x+eps; if abs(f)<err break end display('Answer is='),x Using roots command in MATLAB a=[ 1 -1 -4 3]; roots(a) ANSWER t=0.713 s t=2.198 s

Graph Plotting Example 7: From a vibration measurement on a machine, the damping ratio and undamped vibration frequency are calculated as 0.36 and 24 Hz, respectively. Vibration magnitude is 1.2 and phase angle is -42o. Write the MATLAB code to plot the graph of the vibration signal. Given: z=0.36 ω0=24*2*π (rad/s) A=1.2 Φ=-42*π/180 (rad)=-0.73 rad ω0=150.796 rad/s ω -σ α

yt=1.2*exp(-54.3*t).*cos(140.7*t+0.73); plot(t,yt) xlabel(‘Time (s)'); Graph Plotting: clc;clear t=0:0.002:0.1155; yt=1.2*exp(-54.3*t).*cos(140.7*t+0.73); plot(t,yt) xlabel(‘Time (s)'); ylabel(‘Displacement (mm)');