248 On the Transverse Strain, 
side of the beam, hi another line perpendicu- 
lar to the side at h, and which, before flexure 
coincided with kl; hm and km will then be 
the increments of the outer fibre at h andk, 
and equal to one another. In the same man- 
ner In and in will be the corresponding de- 
crements, which will likewise be equal, and 
p the situation of the neutral line. But lm 
and km will be parallel to in and In respec- 
tively, and hence the quadrilaterals hmkp 
and inlp are similar: we have then on that 
account mp: pn :: hm + mk: In + ni, which 
is as the extension to the compression. 
In the actual bending of a body, its flexure 
is made up of an infinite number of these 
angles hpk, Ipi, each indefinitely small. 
Their supplements therefore, or the angles 
hmk and Ini will each be indefinitely near to 
two right angles; and the indefinitely small 
extensions and compressions hmk and Ini may 
be taken as straight lines; which we shall do 
in the following problem. 
20. To find the position of the neutral 
line: The ratio of the forces necessary to 
produce an equal extension and compression 
in a fibre being given, as s to c; and the 
alteration in its length in consequence of s 
or c being assumed unity. 
Since the sum of the forces of extension 
