282 On the Transverse Strain, 
ing them equal we have p+p’ or PP’ — 2 P. 
To find P; let any distance OR= 2, then 
RS= (r2—2*), and the distance of the 
centre of percussion of the semicircle 
HID from O te ri a zs (when 2 =r) 
fr—2Va x 
er+ 
PEt (Dealtry, Fluent 25) 
=— cr, wherec = 
16 
73 
3 
3°14159 &c.: P therefore = a cr; and p+p’ 
sep = - ee =x 3+14159.r — 1.17809 &e. 
Xr; and the above values of gand p+p’ being 
substituted in the formula give the strength 
2h3 
q + (a—r)A 
sX oo xX AX 1178097 &e. Xr 
ay aLC = 
1:178097 X rs 
aLC 
2 2 2 = co 9 2 + 
($2re-a ) +(a-r) 4), since 6 = (2ra-—a?)*, 
1:178097 &c. 5 [ 2b3 
Ss aaa Ga (a—r)A ) = 
Cor. If ar, or the section of tension 
equal to that of compression, strength 
= ic X section of tension x 4 diameter. 
Example 5th. Suppose it were required 
to find the strength of a hollow cylinder, the 
situation of whose neutral line had been pre- 
viously obtained, as before? 
