432 MR ANGUS R. FULTON ON EXPERIMENTS TO SHOW 
The neutral axis, if the extreme limit of a rectangular area of stress was reached, 
would divide the beam inversely as the maximum stresses, 7.e. ae and 
Cr "5 
O Nea) (Coe eee a6: 
7 cae therefore Omar a 
With the stress areas assumed, the position of the neutral axis and the distance 
between the centres of gravity of the two component stress diagrams will not differ 
appreciably from that got with a purely rectangular stress diagram. 
Therefore, 
9 
Tension area = hd SCs as K 
t 
TLAIC B t 
and similarly 
2 4h 2+K 
C i, ce, 
ompression area bd aa x 3 c 
The moment of the internal stresses resisting the external bending moment at the 
instant of fracture will be : 
' d , d 
Tension stress area x 9 Compression stress area x ie M, 
C 2K, a i We Kew 
se ae 3 Om are * 3 ex5=M, 
where ¢ and c are the extreme fibre stresses. For timber, K,=K,=°625 approx. 
Therefore 
_2:3M T+C _ 23M T+C 
St ae and C= aa : : ANN 
Now, extreme direct stresses for 
Oak. ASH. Boxwoop. 
T= T45 9°38 10:2 
C= 3°6 3:0 4:25 
WL 2°3 11:05 WL WL 2:3 11:05 WL 
O ————— =o = eee = 845 —_ 
z £ ba? 36 elder 4 ba 7-45 ba? 
12°38 12°38 
ASH t ” 3 35 ” ¢ ” 9:38 i ” 
14°45 14°45 
B = ay: = Coe 
a 2° 25 Loe? > 109 
Maximum SHEAR STRESS. 
To discover the maximum shear stress, let ¢ and (t+ dt) be the extreme fibre stresses 
un two transverse sections of the beam at a distance dx apart measured along the axis 
of the beam. ‘Then 
TOw tie WDC) Nese 
t= igen and Do ae eis SP Bye 
therefore 
teliC, 3273 
ot= T+ pe 6M per oz length, 
