Question

PLEASE HELP ME INPUT THIS INTO MATLAB, this is solutions manual for DESIGN of MACHINERY 5th Edition number 11-12

PROBLEM 11-12 Statement: Figure P11-5b shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections of 60 mm diameter. In the instantaneous position shown, the crank 0,4 has a = -10 rad/sec and a = 10 rad/sec2. There is a honizontal force at P of F = 500 N. Find all pin forces and the torque needed to drive the crank at this instant. Given: Link lengths: Link 2 (0 to 4) a = 0.86-m Link 3 (A to B) b = 1.85-m Link 4 (B to 01) c = 0.86-m Link 1 (0 to 01) d:= 2.22-m Coupler point: Rpa = 1.33- 63:= 0.deg F = 500-N T4 = 0-N-m = -10-rad-sec 2 = 10-rad.sec Crank angle and motion: 0 = -36-degw Link cross-section dims: dink = 60 mm Material specific weight: = 0.3-16f-in Solution: See Figure P11-56 and Mathcad file P1112. 1. Use program FOURBAR to determine the position, velocity, and acceleration of links 3 and 4. 03 = 46.028-deg W3 = 3.285-rad-sec. ay = -109.287-rad-see - 2 24 = 106.189-deg w4 = 11.417-rad-sec 1 4 -43.426-rad-sec-2 2. Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4: RCG2 = 0.5-a RCG2 = 0.430-m RCG4 = 0.5-c RoG4 = 0.430-m Link 3: RG 0.5.b RG3 = 0.925-m 3. Determine the mass and moment of inertia of each link. dont T-dink -dink m2 4 m44 m2 = 20.192- kg m3 = 43.436 kg m4 = 20.192 kg 1G2 = 1.249-kg-m

Answer 1

Matlab Code:

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clear;clc;

a = 0.86;b = 1.85;c = 0.86;d = 2.22;
R_pa = 1.33;del3 = 0;F = 500;T4 = 0;
q2 = -36;w2 = -10;alpha2 = 10; d_link = 0.060;
gamma_s = 8303.97*9.81;g = 9.81;

%1
q3 = 46.028; q4 = 106.189;
w3 = 3.285;w4 = 11.417;
alpha3 = -109.287; alpha4 = -43.426;
%2
R_CG2 = 0.5*a;R_CG4 = 0.5*c;R_CG3 = 0.5*b;
%3
m2 = (pi*d_link^2 * a* gamma_s)/(4*g);
m3 = (pi*d_link^2 * b* gamma_s)/(4*g);
m4 = (pi*d_link^2 * c* gamma_s)/(4*g);
IG2 = (m2/12)*(0.75*d_link^2+a^2);
IG3 = (m3/12)*(0.75*d_link^2+b^2);
IG4 = (m4/12)*(0.75*d_link^2+c^2);
%5
R12x = R_CG2*cosd(q2+180);
R12y = R_CG2*sind(q2+180);
R32x = R_CG2*cosd(q2);
R32y = R_CG2*sind(q2);
R23x = R_CG3*cosd(q3+180);
R23y = R_CG3*sind(q3+180);
R43x = (R_CG3-b)*cosd(q3+180);
R43y = (R_CG3-b)*sind(q3+180);
R34x = R_CG2*cosd(q4);
R34y = R_CG2*sind(q4);
R14x = R_CG2*cosd(q4+180);
R14y = R_CG2*sind(q4+180);
Rpx = (R_pa- R_CG3)*cosd(q3);
Rpy = (R_pa- R_CG3)*sind(q3);

%6
a_G2 = R_CG2*alpha2*(-sind(q2)+1i*cosd(q2)) - a*w2^2*(cosd(q2)+1i*sind(q2));
a_G2x = real(a_G2);
a_G2y = imag(a_G2);
a_A = a*alpha2*(-sind(q2)+1i*cosd(q2)) - a*w2^2*(cosd(q2)+1i*sind(q2));
a_CG3A = R_CG3*alpha3*(-sind(q3)+1i*cosd(q3))...
-R_CG3*w3^2*(cosd(q3)+1i*sind(q3));
a_G3 = a_A + a_CG3A;
a_G3x = real(a_G3);
a_G3y = imag(a_G3);
a_G4 = R_CG4*alpha4*(-sind(q4)+1i*cosd(q4))...
-c*w4^2*(cosd(q4)+1i*sind(q4));
a_G4x = real(a_G4);
a_G4y = imag(a_G4);
%7
Fpx = F;Fpy = 0;
%8
C = [1 0 -R12y 0 0 0 0 0 0;
0 1 R12x 0 0 0 0 0 0;
1 0 -R32y -1 0 R23y 0 0 0;
0 1 R32x 0 -1 -R23x 0 0 0;
0 0 0 1 0 -R43y -1 0 R34y
0 0 0 0 1 R43x 0 -1 -R34x
0 0 0 0 0 0 1 0 -R14y;
0 0 0 0 0 0 0 1 R14x;
0 0 1 0 0 0 0 0 0;]';

F = [m2*a_G2x m2*a_G2y IG2*alpha2 (m3*a_G3x - Fpx)...
(m3*a_G3y - Fpy) (IG3*alpha3 - Rpx*Fpy + Rpy*Fpx) m4*a_G4x...
m4*a_G4y (IG4*alpha4 - T4)]';

R = inv(C)*F

R = -1851 1315.4 497.15 -224.51 53.694 -1087.3 1046.9 -3156 -45.28