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296 MECHANICS.
FIG. 9.
SOLVING FOR CENTER OF GRAVITY :
(Area.) (Distance.)
Area of upper flange = 2 X 1 = 2 square inches X 5 =10
Area of web = 4 X 1 = 4 square inches X 2 l
/
2 = 10
Area of lower flange =4X1 = 4 square inches X =0
10 20
and 20 divided by 10 = 2’’ which is the distance from the center
of gravity of the lower flange to center of gravity of the section
of the beam, or the neutral axis x y.
SOLVING FOR MOMENT OF INERTIA :
« = 2x;4 X (3X)
2
= 10.56250
b = 2X yA X (2^)2
= 7.56250
c = \ X ,
<* X (2X)
2
= 2.53175
d=l X ,4 x(iX)2
= 1.53225
e = l X ,/2 x(iX)2
= 0.78125
/=ix;4 x ( U)2
= 0.28125
^=1 x )4 x ( X)
2
= 0.03125
h — 1 X ,* x ( X)
2
= 0.03125
z = l X ,
/
2 x ( %)
2
= 0.28125
; = ixj4 X(1X)2
= 0.78125
k = 4 X )4 X (IX)2
= 6.12500
/=4X,4 X (2X)
2
=
ia of beam =
10.12500
oment of inert 40.6266 (
Area of cross-section of beam = 10 square inches
Radius of gyration of beam = -\j
40.6266
10
2.015 inches.
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