# Chapter 19: Combined open and closed section beams¶

## Example 19.1 Pg.No. 552¶

In [35]:
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)

q12 = -0.0069*s1**2 + 0.345*s1

q23 = -1.035*s2 - 34.5

q03 = -1.035*s3

q34 = 0.0069*s4**2 - 1.035*s4 - 242.5

q94 = 1.725*s5



## Example 19.2 Pg.No.555¶

In [31]:
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)

torsoinal rigidity of closed section = 5.00e+10 N.mm^2

torsional rigidity of open section = 6.0e+07 N.mm^2

total torsional rigidity = 5.006e+10 N.mm^2

angle of twist per unit length = 0.0002 rad/mm

maximum shear stress in the closed section = 166.5 N/mm^2

maximum shear stress in the open section = 10 N/mm^2