Q74 On the Transverse Strain, 
The mean from all these is 4 to 5, as found 
for pine. 
And to divide any surface ABCD, (fig. 8) 
as required above:4 Let EFCD and 
EFBA be the sections of tension and com- 
pression respectively, and G and g their cen- 
tres of gravity. We shall then have EFCD 
x iG: EFBA X ig in a given ratio, which 
we will call m ton. But since, by the Cen- 
trobaryc Method, the solid, formed by the re- 
volution of any surface, as EFCD, round 
EF, is equal to the area of that surface mul- 
tiplied by the distance passed over by its cen- 
tre of gravity, or equal EFCD x iG X 
2p, (where p is = 3.1416); we have then, 
solid formed by revolution of EFCD : solid 
formed by revolution of EF BA (round EF) 
:: EFCD x iG X 2p: EFBA X ig x 2p:: 
EFCD x iG: EFBA X ig::m:n. The 
solids formed by the revolution of the sections 
round the neutral line are therefore in the 
given ratio; and this must be the case what- 
ever be the form of the beam. 
Example. Let it be required to find the 
situation of the neutral line in a beam, the 
form of which is the frustum ABCD (fig. 8) 
of an isosceles triangle, whose top DCH 
has been cut off parallel to AB. 
Call Hk, the height of the triangle, =a, 
hk, that of the frustum, = d, 
