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2 9 4 MECHANICS.
The least rectangular moment of inertia is always equa.
to the area of surface of cross-section, multiplied by the square
of the radius of gyration, when the surface is assumed to rotate
about the neutral axis of the section.
Mathematicians calculate the moment of inertia by means
of the higher mathematics, but it may also be calculated
approximately by dividing the cross-section of the beam into
any convenient number of small strips and multiplying the area
of each strip by the square, of its distance from its center-line to
the neutral axis, and the sum of these products is the moment
of inertia, very nearly.
The narrower each strip is taken, the more exact the result
will be; but it will always be a trifle too small.
Example 1.
Find approximately the rectangular moment of inertia for
a surface (or section of :
a beam) 6" X 2", about its axis x y.
(See Fig 7,)
Divide the surface into narrow strips, as a, fi, c\ d< e, f, g,
A, i,j. k, /, and multiply each strip by the square of its distance
from the neutral axis, xy, and the sum of these products is the
moment of inertia of the surface.
h
2- *| FIG. 7.
a = 2 X % X (2%f
b = 2 x y
2 X (2X)’
2
r = 2 X % X dX)2
7.5625
5.0625
3.0625
d= 2 X y
2 X (IX)’2
= 1.5625
e = 2 X V
2 X ( X)’
2 — 0.5625
g — 2 X % X ( X)’
2
= 0.0625
//. — 2 X y
2 X ( % Y — 0.5625
z = 2 X % X (IX)’2
= 1.5625
j = 2 X y
2 X (IX)’2
= 3.0625
k = 2 X y
2 X (2X)’
2
= 5.0625
l—2Xy2 X (2X)
2
= 7.5625
Moment of inertia = 35.75 (approximately).
The correct value for the least rectangular moment of
inertia for such a surface is obtained by the formula,
(Depth)* X width
andfQrFi
12
-
12
7 will be
63 X 2
= 36. Thus, the
approximate rule gives results a trifle too small, but if the sur-
face had been divided into smaller strips, the result would have
been more correct.
Radius of gyration for this surface, when rotating about
the axis xy, is :
(moment of inertia
area > 12
= 1.73 inches
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