“20 _ PHYSICS. 
axis to pass through the centre of gravity of the surface of fracture, then - 
o = %, and g = 3 h, which gives the relative strength of the parallelopipedon, 
bh? $ 
Q= 5 tan: The relative strengths, therefore, of parallelopipedal bodies 
of the same material are as their breadths, as the squares of their depths, 
and inversely as their lengths. If it be necessary to consider the weight, 
G, and if the centre of gravity be taken at half the length, we obtain 
ape 1m . bh? 
mein G. e- 
As an illustration of the application of this proportion, let fig. 10 represent 
a rectangular plate, with its longer edge, AF, walled in horizontally : suppose 
a weight, Q, to be suspended at E, and increased until fracture ensues 
Required the direction of the line of fracture, BD, and the magnitude of the 
weight, Q. Representing the height or depth of the plate, BF, by h, then 
D. he 
ame: TEE 
If the unknown angle, DBC, be represented by a, then 
BD os or if tgez = z, BD = BC V 1+ 22; alsoGC =BCsin. 2 —BC 
~ Oo 
x 
===, and these values substituted in the equation for Q, give 
wl ob e* ¥ 
1+2z? a : ee 
Q —1m. —h2. Finding from maxima and minima, the value of z, 
x 
1+2z? 
for which the factor, -, 1s @ minimum, we learn that this is the case 
when x = 1, whence tga —1, and « = 45°: Q is then 1m./h?. 
The strength of a beam, AB (fig. 12), exposed to fracture from a weight, 
@, acting in a direction perpendicular to its fibres, is as the product of the 
transverse section at the place where the weight is applied, and the distance 
from the centre of gravity of the same cross-section, to the point or line 
where the fracture terminates. In beams of square sections, the strengths 
are as the cubes of the sides; in cylindrical beams, as the cubes of the 
diameters; in two similar beams, as the cubes of the homologous 
sides. ) 
The strongest rectangular beam which can be cut from a given cylinder, 
is one in which the squares of the breadth, depth, and diameter of the 
cylinder are as 1:2:3. This beam may be found, according to pl. 17, fig. 
7,in the following manner :—Divide the diameter, AE, into three equal 
parts at G and H; erect GF and DH perpendicularly to these points, and 
produce them to the circle, BC; A, F, D, and E, will determine the four 
corners of the beam. Here the breadth of the beam is to its depth as 5:7, 
or more accurately as 12: 17. 
The strain to which beams are exposed, under different circumstances, is 
determined by very complicated calculation. Let L represent the length 
of leverage, from the neutral axis to the point of attachment of the weight, 
W the weight, and « the angle made by the above-mentioned leverage with 
the horizon at the instant of fracture; then the strain for the case repre- 
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