Chapter 14 Nonferrous Alloy

Exaple 14_1 pgno:545

In [1]:
from math import pi,sqrt
# Initialisation of Variables
d1=0.5;#Diameter of a steel Cable in in.
rhoy=70000.;#Yield Strength of Steel Cable in psi
rhoa1=36000.;#Yield Strength of Aluminum in psi
rhos=0.284;#Density of Steel in lb/in**3
rhoa2=0.097;#Density of Aluminum in lb/in**3
F=rhoy*((pi/4)*(d1**2));#Load applied on Aluminum in lb
d2=sqrt((F/rhoa1)*(4/(pi)));#Diameter of Aluminum in in.
Ws=(pi/4)*(d1**2)*12*rhos;#Weight of Steel  in lb/ft
Wa=(pi/4)*(d2**2)*12*rhoa2;#Weight of        Aluminum in lb/ft
print "a. Load applied on Aluminum in lb:",F
print "c.  Weight of Steel  in lb/ft:",Ws
print "Weight of Aluminum in lb/ft:",round(Wa,3)
print "b. Diameter of Aluminum in in.: ",round(d2,3)
a. Load applied on Aluminum in lb: 13744.4678595
c.  Weight of Steel  in lb/ft: 0.669159235215
Weight of Aluminum in lb/ft: 0.444
b. Diameter of Aluminum in in.:  0.697