Observe the four bar mechanism given below This is a quick r

Observe the four bar mechanism given below. This is a quick return mechanism, used for converting rotary motion into oscillatory angular motion. Using vector loops, find symbolic functions for the motion of bar C as a function of phi and its derivatives. Use L_a for the length of bar A, L_b for the length of bar B, L_c for the length of bar C, and d for the separation between the mounting brackets. Both phi and theta are measured positively CCW from the positive horizontal axis. Give functions for theta, theta, theta in terms of unresolved dot products. Do not resolve the dot products in this step. Write out the vector frame transforms for all of the frames in your analysis. Make a list of the unresolved dot products found in your functions in step 2, and use your transforms to fully resolve those dot products. You don\'t need to rewrite the functions for 0 and its derivatives, just the resolution of the dot products found therein. EG: g_1.h_1 = Cos(alpha)Sin(gamma). Only make a list. Don\'t plug these dot products into your functions for theta and its derivatives until Step 4. Use a computer to generate plots of theta, theta, and theta using the following measurements: Bar A has a length of 3 inches, bar B has a length of 5 inches, and bar C has a length of 6 inches. The mounting brackets are separated by 5 inches. There is a motor on the left bracket that drives bar A at constant CCW angular velocity of 45 RPM. Let phi (0) = 90 degree. Run your plots from t = 0 to t = 3 seconds. Comment on the shape of the graphs in light of the function described in Part 1. Extra Credit: Make a parametric position plot for a point located at the halfway point of Bar B.

Solution

solution:

1)here for given four bar mechanism loop closure equation is for ABCD is given as in cloclwise numbering is as follows

DA\'=AB\'+BC\'+CD\'

in vector form it can bewritten as product of magnitude and unit vector along direction

DA\'=Ld.Ld^

AB\'=La.La^

BC\'=Lb.Lb^

CD=Lc.Lc^

on differentiating all constant term would vanished and only varying terms left as follows we get velocity of CD link as follows

LaWa(La^*K^)+LbWb(Lb^*K^)+LcWc(Lc^*K^)=0

here La^=unit vector along La\'

K^=unit vector along z direction

i^=unit vector along x direction

j^=unit vector along y direction

where as both velocity Wb andWc are unkown hence multiplied by (Lb^*K^) and (Lc^*K^),we get both velocity as follow equation

Wc=-[La*Wa*(La^)*(Lb^)*K^]/[Lc*(Lc^)*(Lb^)*K^]

for Wb we get

Wc=-[La*Wa*(La^)*(Lc^)*K^]/[Lb*(Lb^)*(Lc^)*K^]

where accelaration obtain by differentiating velocity equation and on multiplying by (Lb^*K^)

ac=[[La*Wa^2*(La^)*(Lb^)*K^+Lc*Wc^2*(Lc^)*(Lb^)*K^]/[Lc*(Lc^)*(Lb^)*K^]]

2)for getting angle beta b and theta we have solve by chace solution here we get

DA\'-AB\'=BC\'+CD\'

5i-3j=BC\"+CD\'=R\'

here by chace method angle is given from point of intersection of two circle

y=R^2+BC^2-CD^2/2*R=5.8309^2+5^2-6^2/2*5.8309=1.9720

x=4.594

from

x^2+y^2=BC^2

so BC\' and CD\' is given as

BC\'=x(R^)+y(R^)

BC\'=3.937i-3.374j

BC^=.759i-.65j

from above relation

R\'=BC\'+CD\',we get CD\'

CD\'=1.063i+.374j

CD^=.837i+.2945j

5)in this way angular velocity for input angle 90 and velocity

Wa=4.712 rad/s

is as follows

Wc=-3*4.712*cos(49.42+90)/6cos(49.42-19.38)=2.067 rad/s

Wb=-3*4.712*cos(-70.61+90)/5cos(-70.61+40.57)=-3.080 rad/s

where accelaration is given as

ac=3*4.712^2*cos(49.42+90)+6*2.067^2*cos(49.42-19.38)/6cos(49.42-19.38)=-5.467 rad/s2

6)in this way by using input angle for various time we can calculate particular velocity and accelaration from given equation by vector method

in this way for time instant say t=1 sec

input angle

Wa=phi-90/time=phi-(pi/2)/1

for 1 sec phi1=6.2827*180/pi=359.97=360 degree=0 degree,Wc=-1.770 rad/s,ac=14.60 rad/s2

for 2 sec phi2=449.95=450 degree=90 degree as velocity and accelaration obtained above

for 3 sec phi3=899.93=900 degree=180 degree ,Wc=.4279 rad/s,ac=-8.15 rad/s2

7)parametric property for link BC at hakfway in vector form as is given by

BC\"=2.5.BC^