714 
MR. FAIRBAIRN’S EXPERIMENTAL INQUIRY 
Substituting these values and reducing, we find 
1,=^ \Aee\—{e,—Q\e—Q — {e—t){e,—t,){{e, — t,f-\-{e,—Q{‘le, — t,-t^)]']. . . (3.) 
Also (Moseley’s Engineering, p. 82) we have 
I = I.-AX^ (4.) 
Moreover, by the formula of rupture, 
^/_SI 
4 “X’ 
Taking the data of Table XVI., we have 
ez=2-5, ei = 2-6, #=-32, ^i = -35, t^=’A2 ; 
therefore, by equation (2.), 
A=(-32+'25)(2'6— •42) + (-35 + '42)(2'5 — •32)+2X-32X'42 = 3-19. 
By equation (1.), 
X= {3 X 2-5 X 2-6"- (2-6--42)"(7-5 - •64-'25) - (2-5--32) 
X (•42--35)(7'8--84- -35)} ^3X3*19= 1-91, 
which is the distance of the neutral axis from the edge ne of the beam. 
By equation (3.), 
(5.) 
I,=i[4X2-5X2-6*— (2-6 — •42)"(2-5-*25)-(2-5--32)(2-6--35) 
X {(2-6 — •35)"4-(2'6-'42)(5'2--35— -42)}] = 13-375. 
By equation (4.), 
I=13-375-3-19X ]-9P=l-738. 
By equation (5.), 
S=W/x 4 X 1 - 738 1*^8.= W/x tons. 
In experiment 1, W=3409, 1=7 X 12 = 84, 
S= 
3409 X 84 X 1-91 
15568 
= 35 tons. 
Let Xi= the distance of the neutral axis from the edges AB, and Si= the force 
per square inch opposed to extension or compression, as the case may be, at the edge 
AB, then 
and 
X, = 2-6- 1-91 =-69 
•69x35 
1-91 
s,=^.s=- 
= 12-6 tons. 
In experiment 2, W = 773fi + 18 = 77fi3, and 1=27 
^ 7753x27x 1-91 
15568 —25 6 tons, 
^ •69x25-6 
S, = =9-3 tons. 
and 
1-91 
