HOW FAILURE UNDER STRESS OCCURS IN TIMBER. 437 
show that the assumptions upon which the formula is based are approximately correct. 
This is illustrated if the shear values s= ul are plotted on a diagram as ordinates 
bd 
with the ratios of “ as abscissee, as has been done in fig. 39. A fair curve to pass 
through the mean of these points is a hyperbola with a constant which, in the case of 
Oak, is equal to 3°73. 
Thus 
WwW — C=3'73= L ae tensile stress) 
If the average value of ¢ is inserted here, say 6°53 tons, then 
poe i, 
which was the value of the calculated constant on the assumption of an approximately 
rectangular stress diagram. 
It is difficult to account for the very high values which are apparently reached in 
some cases by the shear stresses, though the average values are in practical agreement 
with those in direct shear. The cases in which this excessive calculated value is reached 
occur only when the span is very short relative to the depth, and where, consequently, 
the central load is not far removed from the line of the reaction. The material is thus 
subjected to a compressive bearing stress which tends to increase the value of the 
resistance of the material to horizontal shear. As the load becomes more remote from 
the reaction, this effect grows less, and is not observable at that ratio of length of span 
to depth of beam at which the material will give either by tension, compression, or shear. 
That there is some raising of the shear value is evidenced by shearing taking place 
habitually in beams at what may be called the normal shearing value when L is near 
d 
the critical ratio, as well as at the more inflated value when b is very small. Fig. 32 
d 
shows some typical shearing fractures of Oak, but these fractures never occur above the 
critical ratio. With horizontal shear tangential, 4 was the highest value of L attained 
d 
when shear was the direct cause of failure, and with horizontal shear radial, 5. 
The only condition under which so high a value of L as 10 could be reached in a beam 
d 
fractured by horizontal shear would be where a shake, situated near the neutral axis, 
already existed in the timber, which would, of course, offer little resistance to shear. 
CoNCLUSIONS. 
1. The initial cause of fracture in timbers lies in the medullary rays. 
2. Rectangular beams when laid on a tangential face are stiffer and have a higher 
fracture value than when on a radial face. 
