Chapter 14: Airframe loads

Example 14.1 Pg.No.407

In [10]:
from __future__ import division
import math

W=45             #weight of aircraft (kN)
m=45/9.8          #mass of aircraft(k-kg)
a=3*9.8           #deceleration due to cable (ms^-2)
m1=4.5/9.8        #weight of aircraft after AA in Diagram (k-kg)
v0=25             #touch down speed (m/s)
alpha=10             #center line angle with ground (degree)

T=m*a/math.cos(math.radians(alpha))
print "tension in cable = %3.1f kN\n"%(T)

R=W+T*math.sin(math.radians(alpha))
print "load on each undercarriage strut = %2.1f kN\n"%(R/2/math.cos(math.radians(20)))

N=T+m1*9.8*math.sin(math.radians(alpha))-m1*a*math.cos(math.radians(alpha))
S=m1*a*math.sin(math.radians(alpha))+m1*9.8*math.cos(math.radians(alpha))
print "N and S forces are shown in Fig 14.4 N=%2.2f kN  S=%2.2f kN\n"%(N,S)

s=v0**2/2/a
print "length of deck covered = %2.2f m\n"%(s)

tension in cable = 137.1 kN

load on each undercarriage strut = 36.6 kN

N and S forces are shown in Fig 14.4 N=124.57 kN  S=6.78 kN

length of deck covered = 10.63 m

Example 14.2 Pg.No.409

In [20]:
from __future__ import division
import math

g=9.8
W=250           #weight of aircraft(kN)
m=250/g        #mass of aircraft (K-kg)
I_CG=5.65*10**8     #inertia about center of mass(N.s^2.mm)
v0=3.7            #vertical velocity of undercarriage(m/s)
R_h=400          #horizontal reaction (kN)
R_v=1200         #vertical reaction (kN)
l=1             # nose wheel distance from ground (m)
d=2.5           # distance of CG from ground (m)

#horizontal equilibrium
ax=R_h/m

#vertical equilibrium
ay=(R_v-W)/m


alpha=(R_v*l+R_h*d)*10**6/I_CG
print "angular acceleration of aircraft = %2.1f rad/s^2 \n"%(alpha)

#v=v0+ay*t
t=v0/ay
print "time taken for vertical velocity to become zero = %1.3f s\n"%(t)

#w=w0+a*t
w=a*t
print "angular velocity of aircraft = %1.2f rad/s \n"%(w)

angular acceleration of aircraft = 3.9 rad/s^2 

time taken for vertical velocity to become zero = 0.099 s

angular velocity of aircraft = 0.39 rad/s

Example 14.3 Pg.No.414

In [31]:
from __future__ import division
import math

W=8000             #weight of aircraft (N)
n=4.5              # wing loading
S=14.5              #wing area (m^2)
V=60                #speed (m/s)
rho=1.223           #density (kg/m^3)
alpha=13.75          #from Fig 14.8 (a)
C_Mcg=0.075           #from Fig 14.8 (a)
c=1.35                #mean chord (m)


L=n*W
C_L=L/(0.5*rho*V**2*S)
print "lift coefficient of aircraft = %1.3f \n"%(C_L)

#from Fig 14.8 (b)
l=4.18*math.cos(math.radians(alpha-2))+0.31*math.sin(math.radians(alpha-2))
print "length of tail arm = %1.3f m \n"%(l)


C_L=C_L-c/l*C_Mcg
print "lift coefficient =%1.3f \n"%(C_L)

alpha=13.3
l=4.18*math.cos(math.radians(alpha-2))+0.31*math.sin(math.radians(alpha-2))
print "Now tail arm length =  %2.3f m\n"%(l)

L=0.5*rho*V**2*S*C_L
print "Lift = %5.1f N\n"%(L)

P=n*W-35000
print "Tail Load = %5.1f N\n"%(P)

D=0.5*rho*V**2*S*0.0875
print "Drag = %5.1f N\n"%(D)
print "Forward inertia force = %5.1f N\n"%(D)  #eqn 14.13

lift coefficient of aircraft = 1.128 

length of tail arm = 4.156 m 

lift coefficient =1.103 

Now tail arm length =  4.160 m

Lift = 35222.3 N

Tail Load = 1000.0 N

Drag = 2793.0 N

Forward inertia force = 2793.0 N
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