INTO THE STRENGTH OF WROUGHT-IRON PLATES. 
713 
and exact proportions of the parts, in order to obtain the necessary formula for cal- 
culating the strength of beams and frames of this description. 
FORMULA RELATIVE TO THE BEAMS IN THE FOREGOING EXPERIMENTS. 
Beams with a Single Flanch. 
Let ABF be a section of a beam having a single flanch ABkh, 
the material being symmetrically distributed with respect to the 
vertical line EF. Let hacd and DFC be parallel to AB ; cr and 
hdC parallel to EF. Put EB=EA=e, CB=DA=ei, ac—¥r=t, 
BA“Ag=^i, Ea=B6?=^2, Fe=F/z = ^3, area section ABcen/?g=A, 
and the distance of the centre of gravity of this section from the 
edge ne='K. 
A. jB 
To Jind the Position of the Neutral Axis. 
Assuming the material to be perfectly elastic, the neutral axis will be in the centre 
of gravity of the section. Hence we have, by calculating the moments with respect 
to the line DC, 
AxX=eef- (Ci — C)'(e— 0-§(ei — C) — (e— 0(^2— C)(ei — 
X: 
3A 
( 1 .) 
which expresses the distance of the neutral axis from the edge ne, where 
A — {tft^{ei — t^-\-{ti-\-t.2){e — t)-\-2tt2 (2.) 
To find the Moments of Rupture. 
Let 1 = the moment of inertia of the section about the neutral axis. 
I,= the moment of inertia of the section about DC. 
W= the breaking weight of the beam. 
I— the distance between the supports. 
S= the force per square inch of the material opposed to extension or com- 
pression, as the case may be, at the thin edge of the beam. 
Taking DC as the axis, — 
Ii= moment inertia ABCD — 2 moment inertia rQdc —2 moment inertia ere— 2 
moment inertia cdk. 
2 
Now, moment inertia ABCD= -^eeX, 
2 
2 moment inertia rCdc= ^ 
2 moment inertia erc—^{t—t^{ei—tf^, 
2 moment inertia c<;?A= |(e— ^){(ei — ^i)^— 3 (ei— — 
4 Y 
MDCCCL. 
