THE LONGITUDINAL STRENGTH OF RIGID AIRSHIPS. 165 
For the other joints we must, in calculating the moment of inertia, take the mean value 
for each set of shear wires or counterwires meeting at each joint, one belonging to a panel 
above and the other to a panel below the joint. 
to the left with an area: 
For instance, at C, we find a shear wire a 
— JG 5, 88) 
OO 
Hie os .866 
and a shear wire c, to the right with an area 
ee You _ -683 | 
Vc nCGom 
The mean value of these is: 
eens 1.616 = 083 « 
DREGE 288 
For the counterwires meeting at C, we find likewise a-, = .933 a, 
Now it is found that in the symmetrical ship here under consideration all the so cor- 
rected a values become .933 a, but we shall for the sake of simplicity retain the notation a 
in the following calculations, bearing in mind that actually in this case 
a = .933 Eu a, cos? @. 
Ep 
Since all the joints have the same effective area, the geometrical neutral axis of the 
section must pass through the axis of the ship on the level of the G-girder. 
The moment of inertia is: 
[=6R(a+214). 
Apply now a bending moment M, to section (2) and find the stresses at the joints 
by the theory of bending. 
_4R 
so also 
Va RES pn= 
Be = —pr= -500 pa, 
The horizontal or P-forces acting at the joints of (2) are: 
P= Py =p, (a 4 2) a) 
which we shall denote by P for convenience. 
The remaining forces are: 
Pas = eS 66 I, 
M, .866 R _ B66 
Ht 
pco= O. 
P; = —P;=.500 P, Pe=o0. 
The P-forces are transmitted to frame (1) through the wires and longitudinals between 
(1) and (2), in which members they call forth the following forces: 
