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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 Page 87 - Building stability - Moment of inertia of columns
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