THE LONGITUDINAL STRENGTH OF RIGID AIRSHIPS. 163 
the members on either side of the joint is the same as would exist in a bar of the sectional 
area a + qa under the action of P. 
Where both the shear wires and the counterwires are taut and where they are fitted sym- 
metrically on both sides of a joint, as indicated in Fig. 13 (0), the effective area is a+ 2a. 
Where a diagonal wire is found on one side of a joint and not on the other, the aggre- 
gate elongation of the longitudinal reckoned between half-frame stations just forward and 
aft of the joint is approximately the same as if the girder had an area a + teas whence this 
2 
is reckoned to be the effective area at the joint (Fig. 13 (c)). 
This method applies to all wires so long as they are taut, whether the tension is in- 
creased as by a pull P or reduced as by a thrust P. It is strictly correct where the two 
longitudinals that are connected by a wire are subject to the same P-force, since in that case 
the horizontal pull exerted by either end of the wire is actually pa = T cos @, where p 
is the stress at the joint. But in an airship, subject to bending, the stresses at the joints 
are proportional to y, the distance from the neutral axis. If now we reckon the equivalent 
area to be equal to a at either end, we shall have a different total tension at the ends of the 
wire, which is impossible. We must, therefore, introduce a modification of the equivalent 
area. 
FIG. 14. 
Referring to Fig. 14, consider a panel between two frames (1) and (2) and two longi- 
tudinals c and e. Under pure bending the stresses, assumed to be tensile, at the joints are 
pc at C, and C, and peat FE, and E,. The distances of the joints from the neutral axis are 
‘yc, and yz respectively and let 
Vot VE 
= Vee. 
2 
The tension in the shear wire is 7, which is determined by the mean extension of the lon- 
gitudinal bars. We may, therefore, substitute for T a fictitious horizontal bar of area a 
midway between the longitudinals, in which case we obtain the correct pull exerted by the bar, 
ap, where 
BS Veni 
I Seer ae (34) 
In reality the pull acts at C, and £,, and it is not strictly correct, as proposed by some, 
to place a midway between the C and E girders in calculating the moment of inertia unless 
both shear and counterwires are active. It is desirable, when the counterwire is slack, to 
substitute for a a value a, at C, such that 
Bb 
AchcVc= 4 PYc OF Ac == 
Bc 
