1 / 7

Solutions of system of nonlinear equations:

Solutions of system of nonlinear equations:. L 3. θ 3. L 4. L 2. θ 2. θ 4. s 1. 1. 4. 2. 3. Newton-Raphson Example 4:. The kinematic equations for a Four-Bar mechanism can be written as (5th semester, Mechanisms Course). L 2 =0.15 m L 3 =0.45 m L 4 =0.28 m s 1 =0.2 m.

rupali
Télécharger la présentation

Solutions of system of nonlinear equations:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solutions of system of nonlinear equations: L3 θ3 L4 L2 θ2 θ4 s1 1 4 2 3 Newton-Raphson Example 4: The kinematic equations for a Four-Bar mechanism can be written as (5th semester, Mechanisms Course) L2=0.15 mL3=0.45 mL4=0.28 ms1=0.2 m Where link 2 is the input member. How do you calculate θ3 and θ4 when θ2=120°. -0.075 0.13

  2. Solutions of system of nonlinear equations: 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) Following changes are made in the computer program. • Sub newtonrn_Click() • - - - • 40 n=2 • 41 xb(1)=0.5:xb(2)=1:xh(1)=.001:xh(2)=.001 • - - • 45 ‘…Error equations… • a(1,1)=-0.45*Sin(xb(1)):a(1,2)=0.28*Sin(xb(2)) • a(2,1)=0.45*Cos(xb(1)):a(2,2)=-0.28*Cos(xb(2)) • b(1)=-(0.45*Cos(xb(1))-0.28*cos(xb(2))-0.275) • b(2)=-(0.13+0.45*Sin(xb(1))-0.28*Sin(xb(2))) • 46 ‘... • - - • End sub (Initial angle values must be given in RADIAN) ANSWER: θ3=0.216 rad (12.37°) θ4=0.942 rad (53.97°)

  3. 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 mL3=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

  4. Solutions of system of nonlinear equations: Following changes are made in the computer program. • Sub newtonrn_Click() • - - - • 40 n=2 • 41 xb(1)=-1:xb(2)=0.8:xh(1)=.001:xh(2)=.001 • - - • 45 ‘…Error equations… • a(1,1)=-0.6*Sin(xb(1)):a(1,2)=-1 • a(2,1)=0.6*Cos(xb(1)):a(2,2)=0 • b(1)=-(0.075+0.6*Cos(xb(1))-xb(2)) • b(2)=-(0.1299+0.6*Sin(xb(1))) • 46 ‘... • - - • End sub ANSWER: θ3=-0.2182 rad (-12.5°) s=0.6607 m 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)

  5. Solutions of system of nonlinear equations: Solution with MATLAB clc;clear t=solve('t^3-t^2-4*t+3=0'); vpa(t,6) 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? Sub newtonr1_Click () ' CHANGE LINES 30, 35 AND 37 FOR DIFFERENT PROBLEMS 30 x = 1: AERROR = .0001 niter1 = 5: niter2 = 20: ir = 0: Call cls1 32 xp = x 35 f = x ^ 3 - x ^ 2 - 4 * x + 3 37 f1 = 3 * x ^ 2 - 2 * x - 4 … End Sub ANSWER T=0.713 s t=2.198 s Using roots command in MATLAB a=[ 1 -1 -4 3];roots(a)

  6. Graph Plotting: Given: z=0.36 ω0=24*2*π (rad/s) A=1.2 Φ=-42*π/180 (rad)=-0.73 rad ω ω0=150.796 rad/s α -σ 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.

  7. 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)');

More Related