from __future__ import division
from scipy import integrate
from sympy import *
import math
l1=100 #lengths shown in Fig(mm)
l2=200
F=100*10**3 #force applied (N)
y_bar=(2*l1*2*50+2*l2*2*l1+l2*2*l2)/(4*l1*2+4*l2*2)
Ixx=2*(2*l1**3/12+2*l1*25**2)+400*2*75**2+l2*2*125**2+2*(2*l2**3/12+2*l2*25**2)
s1=Symbol('s1')
q12=-round(10**4*F/Ixx)/10**4*(integrate(-50+2*s1,s1))
print "\nq12 = %s"%(q12)
s2=Symbol('s2')
q23=-round(10**4*F/Ixx)/10**4*(integrate(2*75,s2))-34.5
print "\nq23 = %s"%(q23)
s3=Symbol('s3')
q03=-round(10**4*F/Ixx)/10**4*(integrate(2*75,s3))
print "\nq03 = %s"%(q03)
s4=Symbol('s4')
q34=-round(10**4*F/Ixx)/10**4*(integrate(2*(75-s4),s4))-242.5
print "\nq34 = %s"%(q34)
s5=Symbol('s5')
q94=-round(10**4*F/Ixx)/10**4*(integrate(-2*125,s5))
print "\nq94 = %s\n"%(q94)
from __future__ import division
from scipy import integrate
from sympy import symbols
import math
A=20000 #nose cell area (mm^2)
L_w=900 #outer wall (mm)
L=300 #width of wall (mm)
length=600 #length of open section (mm)
G=25000 #shear modulus (N/mm^2)
T=10*10**6 #torque applied (kN.m)
t1=1.5 #thickness of closed section
t2=2 #thickness of open section
GJ_cl=4*A**2*G/(L_w+L)*t1
print "torsoinal rigidity of closed section = %2.2e N.mm^2 \n"%(GJ_cl)
GJ_op=G*(length+L)*t2**3/3
print "torsional rigidity of open section = %2.1e N.mm^2 \n"%(GJ_op)
GJ=GJ_cl+GJ_op
print "total torsional rigidity = %5.3e N.mm^2\n"%(GJ)
dO_by_dz=T/GJ
print "angle of twist per unit length = %1.4f rad/mm\n"%(dO_by_dz)
q_cl=GJ_cl/2/A*dO_by_dz
print "maximum shear stress in the closed section = %3.1f N/mm^2\n"%(q_cl/1.5)
#eqn 18.10 T_max=GtdO/dz
T_max=G*t2*dO_by_dz
print "maximum shear stress in the open section = %2.0f N/mm^2\n"%(T_max)