# Chapter 24: Fuselage frames and wing ribs¶

## Example 24.1 Pg.No.638¶

In [2]:
from __future__ import division
import numpy
import math
#reference Fig 42.2
l1=250               #length of each section(mm)
l2=200                #length of DK(mm)
l3=100                 #length of KH(mm)
F1=4000
F2=5000                #forces as shown in Fig 24.1
theta=60               #force angle from DH (degree)

#solve (i) and (ii) equation
a=numpy.array([[1,-1],[200,100]])
b=numpy.array([13.8564,2000])
q=numpy.linalg.solve(a,b)
q1=q[0]
q2=q[1]
P_A=l1*q1+l1*q3+l1*q4
P_E=-l1*q2-l1*q3-l1*q4
P=3464.1
M_AE=F2*l1+F1/2*3*l1-P*50

P_A=M_AE/(l2+l3)+P/2
P_E=-M_AE/(l2+l3)+P/2

print "shear flow as shown in Fig 24.1\n"
print "q1=%2.2f N/mm\n"%(q1)
print "q2=%2.2f N/mm\n"%(q2)
print "q3=%2.2f N/mm\n"%(q3)
print "q4=%2.2f N/mm\n"%(q4)

print "Stiffener load at point A = %2.2f N/mm(tension)\n"%(P_A)
print "Stiffener load at point E = %2.2f N/mm(compression)\n"%(P_E)  #in book another method is also explained

shear flow as shown in Fig 24.1

q1=11.29 N/mm

q2=-2.57 N/mm

q3=6.67 N/mm

q4=23.33 N/mm

Stiffener load at point A = 10321.37 N/mm(tension)

Stiffener load at point E = -6857.27 N/mm(compression)



## Example 24.2 Pg.No.645¶

In [4]:
from __future__ import division
import numpy as np
import math

A1=50000
A2=95000        #area shown in Fig 24.13(mm^2)
A3=95000

A4=46000
A5=49000         #area shown in Fig 24.11 (mm^2)

F1=12000
F2=15000          #forces shown in Fig 24.9 (N)

l1=l3=300
l2=320           #lengths and angle shown in Fig 24.9
alpha=15

#solve equation (i)&(iii)
a=np.array([[600,-600],[190000,290000]])
b=np.array([12000,440000])
q=np.linalg.solve(a,b)
q12=q[0]
q23=q[1]
q31=(F2+l1*q23)/l1
print "shear flows are shown in Fig 24.9 (flanges)"
print "q12=%2.2f N/mm"%(q12)
print "q23=%2.2f N/mm"%(q23)
print "q31=%2.2f N/mm\n"%(q31)

Sy_1=7*l3
Px4=Px2=2*A1*7/l1
q1=(2100-2*625.2)/l1
P2=P4=(Px4**2+Py4**2)**0.5
P5=P6=2*((A1+A4)*7-A5*13)/l2
q2=(7*l1+7*10-13*10)/l2
q3=(6.4*l2+F2)/l2

M3=2*((A1+A2)*7-A2*13)+F2*l1
Px1=Px3=M3/l1
Py1=Py3=3626.2
P1=P3=(Px1**2+Py1**2)**.5
q3=(17100-2*Py1)/l1

print "P4=P2=%2.2f N"%(P2)
print "P6=P5=%2.2f N"%(P6)
print "P3=P1=%2.2f N\n"%(P1)

print "shear flow in webs"
print "q1=%2.2f N/mm"%(q1)
print "q2=%2.2f N/mm"%(q2)
print "q3=%2.2f N/mm (this value is given at the end of example)"%(q3)

shear flows are shown in Fig 24.9 (flanges)
q12=13.00 N/mm
q23=-7.00 N/mm
q31=43.00 N/mm