104 APPLIED MECHANICS 
Consider an indefinitely thin layer of material LN parallel to the 
neutral surface, and at a distance y from it. When unstrained, LN = HK, 
' but in the bent beam LN becomes L’N’. Now LN’ _R+y there- 
HK R 
(R+ + as 
fore L’N’ = and the strain produced in LN is 
UN'—IN, (R+YHK x) : ey 
pi sHK=4. 
But E= et hence if 7 is the stress produced in the layer LN in the 
strain 
process of bending the beam, Z = | that is to say, f is proportional to y. 
The distribution of the stress on a cross section will therefore evidently be 
as shown in Fig. 142, f, being the maximum tensile stress, and f, the 
maximum compressive stress. 
The line in which the neutral surface cuts a transverse section of a 
beam is called the neutral axis of that section. 
110. Moment of Resistance to Bending—Position of Neutral 
Axis.—The resultant of the external forces which produce pure bending 
is a couple, and this external couple is balanced by internal forces in the 
beam, and the resultant of these internal forces must therefore be a couple, 
because only a couple will balance a couple. The two forces which form 
the internal couple are the resultants of the tensile and compressive 
stresses, therefore these resultants must be equal and parallel. 
fa | 
1 
Fig, 142. Fig. 143. 
Let YY (Fig. 143) represent a face view and Y’Y’ an edge view of a 
transverse section of a beam, and let XX be the neutral axis. Consider 
an indefinitely narrow strip ss of the section parallel to XX and at a 
distance y from it. Leta ey the ared of the strip ss. The stress f 
on the strip ss is such that 2 af “1, therefore / =f, The resultant of 
Wy. 
the stress on ss is fa =n, = resultant R, of the stress on the part 
1 
of the section below XX must be =fa= = 2A =A “1¥ay. In like manner 
1 
R,= f ae Therefore if R, = R,, f faa 2zay. But ds i a, therefore 
YW Y2 
Zay far the area below XX must be ker to Lay for the area above 
