# Chapter 20: Structural idealization¶

## Example 20.1 Pg.No.560¶

In [10]:
from __future__ import division
import math

sigma6=200
sigma1=200
sigma2=sigma5=150         #these are not real shear stress but taken
sigma3=sigma4=100          #proportional to length because we need just ratio
l16=400
t16=3
l12=l21=600
t12=t21=2                #thickness and lengths as shown in Fig 20.4
l23=600
t23=1.5
l25=300
t25=2.5
l34=200
t34=2
#eqn 20.1 B1=t_D*b/6*(2+sigma_2/sigma_1)
#eqn 20.2 B2=t_D*b/6*(2+sigma_1/sigma_2)
B1=B6=300+l16*t16/6*(2-sigma6/sigma1)+l12*t12/6*(2+sigma2/sigma1)
print "B1=B6=%5.2f mm^2\n"%(B1)

B2=B5=2*300+l12*t12/6*(2+sigma1/sigma2)+t25*l25/6*(2-sigma5/sigma2)+l23*t23/6*(2+sigma3/sigma2)
print "B2=B5=%5.2f mm^2\n"%(B2)

B3=300+l23*t23/6*(2+sigma2/sigma3)+l34*t34/6*(2-sigma4/sigma3)
print "B3=B4=%5.2f mm^2\n"%(B3)

B1=B6=1050.00 mm^2

B2=B5=1791.67 mm^2

B3=B4=891.67 mm^2



## Example 20.2 Pg.No.562¶

In [14]:
from __future__ import division
import math

Mx=100*10**6          #bending moment(N.mm)
y=[660,600,420,228,25,-204,-396,-502,-540]
B=[640,600,600,600,620,640,640,850,640]

print "direct stress in each boom in last column"
print "Boom\t y(mm)\t B(mm^2)\t delIxx=By^2\t sigma_z"
for i in range (0,9):
print "%1.0f \t %3.0f \t %3.0f \t \t%2.1e \t %2.1f"%(i+1,y[i],B[i],B[i]*y[i]**2,Mx*y[i]/(1854*10**6))

direct stress in each boom in last column
Boom	 y(mm)	 B(mm^2)	 delIxx=By^2	 sigma_z
1 	 660 	 640 	 	2.8e+08 	 35.6
2 	 600 	 600 	 	2.2e+08 	 32.4
3 	 420 	 600 	 	1.1e+08 	 22.7
4 	 228 	 600 	 	3.1e+07 	 12.3
5 	  25 	 620 	 	3.9e+05 	 1.3
6 	 -204 	 640 	 	2.7e+07 	 -11.0
7 	 -396 	 640 	 	1.0e+08 	 -21.4
8 	 -502 	 850 	 	2.1e+08 	 -27.1
9 	 -540 	 640 	 	1.9e+08 	 -29.1


## Example 20.3 Pg.No.566¶

In [17]:
from __future__ import division
import math

Ixx=48*10**6
Sy=4.8*10**3
B=300

q12=-Sy/Ixx*B*200       #until point 2
q23=q12-Sy/Ixx*B*200
q34=q23-Sy/Ixx*B*(-200)
print "shear flow in flange 12 = %2.0f N/mm\n"%(q12)
print "shear flow in web 23 = %2.0f N/mm\n"%(q23)
print "shear flow in flange 34 = %2.0f N/mm\n"%(q34)

shear flow in flange 12 = -6 N/mm

shear flow in web 23 = -12 N/mm

shear flow in flange 34 = -6 N/mm



## Example 20.4 Pg.No.569¶

In [28]:
from __future__ import division
import math

B=[200,250,400,100,100,400,250,200]
Ixx=13.86*10**6
Sy=10*10**3
qb23=0
qb34=qb23-Sy/Ixx*B[2]*100
qb45=qb34-Sy/Ixx*B[3]*50
qb56=qb34
qb67=qb23
qb21=qb67-Sy/Ixx*(B[1]*100)
qb18=qb21-Sy/Ixx*B[7]*30
qb87=qb21
qs0=-5.4
print "Distribution of shear flow :"
print "q23 = %2.1f N/mm"%(qb23+qs0)
print "q21 = %2.1f N/mm"%(qb21-qs0)
print "q34 = %2.1f N/mm"%(qb34-qs0)
print "q45 = %2.1f N/mm"%(qb45-qs0)
print "q56 = %2.1f N/mm"%(qb56-qs0)
print "q67 = %2.1f N/mm"%(qb67+qs0)
print "q18 = %2.1f N/mm"%(qb18-qs0)
print "q87 = %2.1f N/mm\n"%(qb87-qs0)

Distribution of shear flow :
q23 = -5.4 N/mm
q21 = -12.6 N/mm
q34 = -23.5 N/mm
q45 = -27.1 N/mm
q56 = -23.5 N/mm
q67 = -5.4 N/mm
q18 = -17.0 N/mm
q87 = -12.6 N/mm



## Example 20.5 Pg.No.575¶

In [36]:
from __future__ import division
import math
from sympy import symbols, integrate
z=symbols('z')

E=70000              #youngs modulus (N/mm^2)
G=30000              #shear modulus (N/mm^2)
P=4.8*10**3          #applied force (N)
L=2000               #cantilever length(mm)
Sy=P
Ixx=48*10**6         #second moment of area
t=1                   #actual thickness (mm)

Mx0=-Sy*(L-z)
Mx1=-(L-z)

del_M=integrate(Mx0*Mx1/E/Ixx,(z,0,L))
del_S=integrate((1/G/t/Sy*(6**2*200+12**2*400+6**2*200)),(z,0,L))
print "total deflection in vertical direction = %1.2f mm\n"%(del_M+del_S)

total deflection in vertical direction = 4.81 mm


In [35]: