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clc;
clear;
mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-10.1 Page No.195\n');
P=100; //[lb/in^2] Hydraulic pressure
F=450; //[lb] Extension force
Fr=400; //[lb] Retraction force
A=F/P; //[in^2] Cross section area
D=sqrt(4*A/%pi); //[in] Bore of cylinder
mprintf('\n The bore of cylinder is %f in.',D);
//Use 2.5in bore cylinder
Dm=2.5; //[in] Bore of cylinder
Dr=1; //[in] Diameter of rod
A2=%pi*Dm^2/4-%pi*Dr^2/4; //[in^2]
F2=P*A2; //[lb] Force
if F2>=Fr then
mprintf('\n The diameter of rod is %f in.',Dr);
else
mprintf('\n This would not meet requirement');
end
//This would meet requirement
Ab=%pi*Dm^2/4; //[in^2] Cross section area
//Note-In the book V=180.7 is used instead of V=180.64158
d=20; //[in] stroke
V=Ab*d+A2*d; //[in^3] Volume per cycle
t=2; //[s] Cycle time
FR=V/t; //[in^3/s] Flowrate
FR=FR*7.48*60/1728; //[gal/min] Flowrate
mprintf('\n Flow rate required is %f gal/min.',FR);
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