HOW FAILURE UNDER STRESS OCCURS IN TIMBER. 433 
or 
sf_ 8M 2:3 T+C_y 93 T+C 
Spetore agus i abet Cn” 
where F = a = shearing force on that section of the beam. 
Total difference of tension or longitudinal shear at neutral axis per unit of breadth 
Nie é' 
=>” from extreme fibre to neutral axis = S=shearing stress per square inch. 
ox 
Therefore, if fibre stress diagram is rectangular in form, 
dt 2°33 T+C ( C ) Fx 2:3 
=, ==), =, | =.) = SS .-- 
bx e bd C T+C “ bd 
t 
; 2 
But the area of stress diagram has been assumed to be . 
form, where K,=— ‘625; 
therefore 
times that of rectangular 
(2). 
To determine the condition that fracture may equally occur either by tearing or 
erushing or shearing, 
by equations (1) and (2), where & is a constant. If S and ¢ be known, then can be 
determined, and may be called the critical ratio. With a ratio of length to depth less 
than this, the presumption is that the fracture will take place by shear; with a ratio 
greater than this, that the fracture will be caused by tearing or crushing, beginning 
at the extreme fibres. 
The critical ratios, assuming the ultimate direct stresses for ¢ and S, are :— 
Oak. AsH. Boxwoop. 
9-38 10-2 
eeeeTAle At oa AD 
Ui te VAD _4 9 x 2°35 1-26 x 1:94 
d Sk 6x175 to to 
9-38 10-2 
= — o —— eS 15 
*6 x 2°35 as 8x 1:94 c 
In the following tables of the results of experiments, the values of ¢, c, and S 
have been worked out from the formule (1) and (2), and give a better idea of the 
relative stresses. 
TRANS. ROY. SOC. EDIN., VOL. XLVIII., PART Ii. (NO. 21). 65 
