166 THE LONGITUDINAL STRENGTH OF RIGID AIRSHIPS. 
T’,=-T,,=3 oe P sec $ 
a+ 
T =T_=_T) =-T,,=4 x 866 ——— Psec¢ 
@ + 
T.=T),=-T),=-T: =I 1500 ; P seco 
T= T),=0 
(36) 
B= ey ee | 
a+2a 
Foe 2 =) 866, P. 
a+2a 
Fo =—F’ = .500 
a+ 2a 
IM = © 
& 
The suffix (1) is used for the shear wires, the suffix (2) for the counterwires. 
At the same time the shearing force }Q acting on frame (2) will cause tensions in the 
wires and end loads on the girders. 
The angles which the panels make with the horizon are: 
0, =O, 3 155 Simg— P2156 
0, = Os = as, Sil — 707 
(, =0.= 75; sin 0, = .966 
.". Z sin’ @ = 4 (sin’ 6, + sin’ 0, + sin’ 6.) = 6.0 
By substitution in (29”) and (30) we find: 
= f= ti — es = 4b 2043 cosec o 210 
hi eee = oe |COSCCIOI LO) 
nS aI, = 1, SS a nS ou CONSE GH DO 
F, = —F, = + x .086 cot ¢ Z Q (37) 
1, Sa aI, S30 NK AO COE CH 0) (OD) 
1D, Sal Sy SS OND EOE BD 
F, =o 
Combine now the effects of bending and shearing and'let: 
7A a 
+ cosecd 2 O=Q, Pe Pe 
