IMPACT UPON BEAMS. 
101 
Cor. 2. Since 
/ / 
b — W \/ —7— 
V p (w 
h 
p (w + r) y 
2 he w 2 
P 
b 2 
— w . 
Whence we may obtain the inertia of beams, by substituting for 
these values in the results of our experiments. Taking them, 
then, from each of the seven experiments upon the third beam, 
these being the most varied, we have : 
From the lbs. balls 
From the 4^ lbs. . . . . . . . r = 5T3 lbs. 
. r = 7*86 lbs. 
. r = 5T3 lbs. 
From the 9 oz. 7 drs.. r = 6*39 lbs. 
Mean 
'6-46 lbs. 
The weight of the beam between the supports was 13*75 lbs. 
646 
Whence-—— = '47, the coefficient by which to multiply the 
lO / d 
weight between the supports of an uniform beam to obtain a 
weight equal to its inertia. The inertia appears therefore to 
differ but little from half the weight of the beam between the 
supports ; and it will be assumed as just half in all our calcula¬ 
tions for comparison with experiment; and especially as half 
the weight is the pressure which a prop under the middle, or 
any other part, of an uniform beam would sustain through the 
weight of the beam, were it cut in two in that part. Mr. Tred- 
gold assumes it as half without proof, and calculates the effect 
of vertical impact as if it were horizontal. 
Prob. 3. Required the height from which a body must fall 
upon the middle of a given beam, supported at the ends, to de¬ 
flect it through an assigned distance; the weight of the beam 
being considered, and the striking body assumed as inelastic. 
Put e = C E the assigned deflec¬ 
tion of the beam; p — the pressure 
which, applied in the middle, would 
have produced it; e' — C D the de¬ 
flection of the beam from its own 
weight, x — any other deflection, q 
= the pressure in the middle of the 
beam from its weight (= half the 
weight of the beam if uniform), h = G D the height fallen through 
before impact, n = the power of the deflection to which the pres¬ 
sure of the beam is proportional, and the rest as before. 
Then, since e n : x n : :p the resistance of the beam at any 
a 
