640 MODERN GASWORKS PRACTICE
It will be noticed that the plane of the shearing force cuts the framing in two points (such as 0 and P, Fig. 391), consequently the shear on each of the horizontal planes will be half the total shear, or—
03
2 cq
I
>o
^8 CN (M
'I'
+
Q
'M
Cripps points out that sufficient accuracy will be obtained by taking this formula as—•
tt • , 1 K 24 L2 +D2 ■
Horizontal shear = — ---- tons.
10,000
(It will be noticed that the shear is spoken of as horizontal, for the reason that we are
now considering a horizontal cantilever as shown in Fig. 390. In the case of the gas-holder the cantilever is in a vertical position, therefore the shear is vertical shear.)
So far as computing the effect of the sliearing-stress is concemed, Cripps has shown that. although. the loading on the standards is in reality concentrated at each roller contact, it may be considered as a distributed load, for the stifEness of the standards converts the
forces given out opposite the rollers into a distributed load so far as the struts and ties are concemed.
THE STRESSES IN A BRACED CANTILEVER
Fig. 392.
The method of computing the stresses in the struts and ties of the guide-framing will readily be followed by considering an. ordinary braced cantilever (Fig.. 390) and the recipiocal stress diagram for the same. The equally distributed load will be apportioned amongst each. of the nodes, the loads AB, BC and CD, being equal (W tons) whilst the end node carries If, however, we assume that each node carries an equal weight (not a strictly accurate assumption, but nevertheless approved by Cripps) it is easily seen from the reciprocal diagram that the stress in the diagonal ties and vertical struts increases from the end to-wards the abutment in the proportion 1, 2, 3, 4. Therefore, if we have the total horizontal shear it is possible to com-pute the stress in the struts and ties. For instance, if
S, is the total shear resolved in the direction. of the ties, and Ss is the total