108 APPLIED MECHANICS 
load on the tooth also varies during contact, and the investigation of the 
strength of the tooth is still further complicated by the variation in the 
‘form of the tooth, due to variations in the diameters of the pitch and 
rolling circles, and also to variations in the number of pairs of teeth in 
contact at one time. 
An approximate general solution is obtained by assuming that the 
load, Q (in Ibs.), is two-thirds of the load, P (in lbs.), at the pitch line due 
to the horse-power, H, transmitted when the velocity of the pitch line is 
V feet per minute, and that Q acts at the outer end of the tooth. 
PV 
~ 33000° 
There are two cases to consider: (1) where Q is distributed over the 
whole width of the tooth; (2) where Q acts at one corner of the tooth. 
It will also be assumed that ¢, the thickness of the tooth at the root, is 
the same as at the pitch line. 
Case I.—Load Q distributed over the whole width of the tooth 
(Fig. 146). This will obtain when the directions of the axes of the 
wheels are properly fixed and maintained, and the teeth are truly shaped. 
The greatest bending moment is at the root, and is equal to Qh. The 
moment of resistance to bending is 3622. Hence Qh=0b#f.  ¢ is gener- 
ally about 0°48 p, but allowing for wear ¢ may be taken at 0°4 p, where p 
is the pitch of the 
teeth. Takingh = 0°7p, 
b=np,and Q = $P,then 
P=/.np*f. Taking f 
at 3500 for cast-iron, 
the following simple 
formula is obtained, 
P= 200np?. 
Case Th hott Fig. 146. : Fia. 147. ; 
acts at one corner of the tooth (Fig. 147). This may result from 
inaccurate mounting of the shafts in the first instance, or through un- 
equal wear of the bearings, or from want of truth in the shape of the teeth. _ 
The tooth may break at a section ABCD, which makes an angle 0 
with the side of the wheel. EF being perpendicular to AB, EF =/ sin 0, 
F223 he ‘ _ih 2 
AB cere Then QA sin 0= amar *f, therefore 
6c ee 
~ 6sin@cos@ 3sin2 60° 
This shows that Q will be least when sin 26 is greatest, that is, 
when 0=45°. Hence if the tooth breaks at an oblique section, that 
2 
section will be inclined at 45° to the side of the wheel, and q-%. If 
Q =P, ¢=0°4p, then P=0-08p?f, and if f= 3500 for cast-iron, P = 280p”. 
If while Q acts at one corner the tooth breaks at the root, then 
2 
= pk and if the tendency to break at the root is the same as at the 
weakest oblique section, id = es or b=2h. This shows that if 0 is less 
