Reviewing W/D Ratio for Concrete Encased Steel

Daily Discussion (today's anonymously submitted question):

Reviewing the Example 2 on SFPE (4th) page 4-307, could not understand how it is W/D = 0.67 lb/ft-in.

Here is the effort:

• Found total cubic inches of the concrete encased W8x28 = (added 1.25 inches on all sides of the column) = [(8.06 + 2.5) x (6.535 + 2.5) x 12] = 1145 cu. in.

• Subtracted W8x28 cubic inches = (A x 12) = (8.3 x 12) = 99.6 cu. in.
• Got net lightweight concrete cubic inches = 1145 cu. in. - 99.6 cu. in. = 1045.4 cu. in.
• Interpolate with 110 lbs per 1728 cubic inches
• Got lightweight concrete only weight = 66.55 lbs per foot of the column
• Added 28 lbs per foot of the beam to 66.55 = 94.55 lbs
• So W = 94.55 lbs total weight of the linear foot of the column

D: As per figure 4-11.10 on page 4-305
D = 2(a + b)
So, Added 2.5 inches to beam flange, a becomes 9.035, and b becomes 10.56
D = 2(9.035 + 10.56)
D = 2(19.595)
D = 39.19 W/D = 94.55 ÷ 39.19 = 2.413

What is not right?

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