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MECHANICS. 295
Example 2.
Find by approximation the rectangular moment of inertia
for a surface, as Fig. 8, (the sectional area of an I beam) about
the axis xy.
When the beam is symmetrical, the neutral axis is at an
equal distance from the upper and lower side, and the moment
of inertia for the upper and lower half of the beam is equal
;
consequently, when calculating moment. of inertia for a surface
like Figs. 8 and 7, it is only nec-
essary to calculate the moment
of inertia for half the beam,
and multiply by 2 in order to
e X —
:
get the moment of the whole
beam.
Solution:
a = 3 X )4 X (2K)
2 = 11.34375
b — 3 X Yi. X (2X)
2 = 7.59375
c = 1 x % x (IK)’
2 = 1.53125
<t=i x y
2 x (ix)2 = 0.7S125
e = 1 X % X ( X)2 = 0:28125
/=i x X x( X)
2 = 0.03125
Moment of inertia = 21.5625
Moment of inertia = 21.5625
for upper half,
for lower half.
Moment of inertia = 43.125 for beam (approximately).
Area of cross-section of beam is 10 square inches.
Radius of gyration = \ ’ ’
* °
= 2.07 inches.
&J \ 10
Example 3.
Find approximately the moment of inertia of a surface, as
Fig. 9 (usual section for cast-iron beams), about the axis, x y,
passing through the center of gravity of the surface.
In shapes of this kind the axis through the center of gravity-
is not at an equal distance from the upper and lower side, but it
can be obtained experimentally by cutting a templet to the
exact shape and size of the surface and balancing it over a
knife’s edge, or it maybe calculated by the principle of moments,
as shown in this example. Divide the surfaces into three
rectangles, the upper flange, the web and the lower flange.
Assume some line as the axis, for instance, the line n m, which
is the center line through the lower flange; multiply the area of
each rectangle by the distance of its center of gravity from the
axis 11 m, and add the products. Divide this sum by the area of
the entire section, and the quotient is the distance between the
center of gravity of the section and the axis n m.
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