a)
model of 2 x 4 

b)
model of 4 x 4 


I
= 4 in. ( 2 in.)^{3} 
I
= 4 in. ( 4 in.)^{3} 
12 
12 
=
2.67 in.^{4} 
=
21.3 in.^{4} 
. 
Fig. 130  Moments of inertia for columns with different
cross sections (visualization
models) 

. 
Notice
that doubling the thickness of the 2 x 4 to 4 x 4 inches increases the
moment of 
inertia
by a factor of eight. The 4 x 4 is much stiffer than the 2 x 4
because more of its mass 
is
concentrated farther away from the center of its cross section.
Since the critical buckling 
factor
of a column is directly proportional to its moment of inertia, doubling
the thickness 
of a
column increases its load bearing capacity, not by a factor of two as
might be thought, 
but by a
factor of eight! 

Consider
another example. Column c) is a solid steel column 2 in. x 2 in.
square. Column 
d) is a
hollow steel column with the same cross sectional area of mass as column
c). 
. 
c) model of a 

d) 2.828 x 2.828 with 2 x 2 hole 
solid 2 x 2 

Area = 4 in.^{2} 
Area = 4 in.^{2} 
I
= 2 in. ( 2 in.)^{3} 
I
= I
(large square)  Ι
(small square) 
12 
I
= 2.828 in. ( 2.828 in.)^{3}_ 2 in. (2 in.)^{3} 
= 1.33 in.^{4} 
12
12 


= 4 in.^{4} 
. 
Fig. 131  Moments of inertia for columns
with equal cross sectional areas of mass 

. 
Despite
the fact that both columns have the same cross sectional area of mass, and
thus the 
same
weight of material per linear foot of column, the moment of inertia of the
hollow 
column is three times that of the solid column. Therefore the hollow
column will support 
three
times the weight for the same cost of materials. 
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Knowhere 

Page 87 
Building stability  Moment of inertia of columns 

