under endlong compression. 
195 
siderable. A Table is given showing the values of P and W 
measured at the breakdown point : — 
Table I. 
TV lbs. 
1 
P lbs. 
Mean P lbs. 
TV lbs. 
P lbs. 
Mean P lbs. 
0 
0 
00 
104*5 
12 
70 1 
70 
0 
101 J 
12 
70 J 
5 
88 1 
89 
20 
46 1 
47 
5 
90 J 
20 
48 J 
10 
73 1 
74 
30 
27 1 
26-5 
10 
75 j 
30 
26 J 
From this table it will be seen that when the value of W was 
30 lbs., a vertical load equal to 26 lbs. was sufficient to break the 
strut-beam ; with this lateral loading however an endlong load of 
25 lbs. was withstood, and it was then noted that the central 
deflection was about 2 inches, though the calculated value on 
the usual elastic hypothesis was only 1*7 inch. The formula used 
to calculate the central deflection is 1 
Vi = 
W 
2P |yp 
U> i - pr i 
-t= tan VP — 
l l 
2a 2 
where y 1 = central deflection, l = length of strut, a = ^lEI, 
E = Young’s Modulus and / = Moment of Inertia of section about 
the neutral axis. 
1 The differential equation to be solved is of the form 
py 
dx 2 
P W 
my + 2Er x=0 ’ 
where y is deflection at any point and x is distance measured from one end of the 
strut. 
This, on solution, gives 
maximum deflection 
P 
El 
and maximum 
-5-k /¥~ W _ 
stteEB ^X J 
where 2 is half the thickness of the rod. 
VOL. XI. PT. III. 
15 
