A four-story building has a structural steel beam-girder-column framing system. Columns are spaced at 20 feet o.c. in one direction and 30 feet o.c. in the perpendicular direction. An interior column supports a typical tributary floor area of 600 square feet. For a preliminary design, find an econo

Question

A four-story building has a structural steel beam-girder- column framing system. Columns are spaced at 20 feet o.c. in one direction and 30 feet o.c. in the perpendicular direction. An interior column supports a typical tributary floor area of 600 square feet.

For a preliminary design, find an economical W10 or W12 section for an interior first-floor column. Assume that the columns have unsupported lengths of 14 feet and K = 1.0. [latex]F_{y}[/latex] = 36 ksi.

Roof Loads: DL = 80 psf

SL = 30 psf

Floor Loads: DL = 100 psf

LL = 70 psf

Accepted Answer

Total roof loads: DL + SL =80 + 30 =

(110 psf) × (600 ft.² ) = 66,000# = 66 k

Total floor loads: DL + LL = 100 + 70 =

(170 psf) × (600 ft.² ) = 102,000# = 102 k per floor

The load at the top of the interior first-floor column is a result of the roof plus three floor loads.

Total load on the first-floor column = [latex]P_{ ext {actual }}[/latex]

[latex]P_{ ext {actual }}[/latex]= 66 k (roof) + 3(102 k) (floors) = 372 k

Try W10 × 60.

[latex]W 10 imes 60\left(A=17.6 in .^{2} ; r_{y}=2.57^{\prime} ight)[/latex]

The assumption being made is that the y axis is the critical buckling direction, because no weak axis bracing is provided.

[latex]\frac{K \ell}{r_{y}}=\frac{(1.0)\left(14^{\prime} imes 12 in . / ft . ight)}{2.57^{\prime \prime}}=65.4[/latex]

[latex] herefore F_{ a }=16.9 ksi[/latex]

[latex]P_{ a }=16.9 ksi imes 17.6 in .^{2}=297.4 k <372 k[/latex]

Therefore, this column section is overstressed. Select a larger section.

Try W10 × 77.

W10 × 77 (A = 22.6 in.², [latex] r_{y} [/latex] = 2.60")

[latex]\frac{K \ell}{r_{y}}=\frac{(1.0)\left(14^{\prime} imes 12 ext { in. } / ft . ight)}{2.60^{\prime \prime}}=64.6[/latex]

[latex] herefore F_{ a }=16.98 ksi[/latex]

[latex]P_{ a }=16.98 ksi imes 22.6 in .^{2}=383.7 k[/latex]

[latex]P_{ a }=383.7 k >372 k herefore OK[/latex]

[latex]\% ext { stress }=\frac{372}{383.7} imes 100 \%=97 \%[/latex]

Use W10 × 77.

A column design using a W12 section can be carried out using an identical procedure. The resulting W12 size would be

[latex]W 12 imes 72\left(P_{ a }=377 k ight)[/latex]

Both of these sections are adequate for stress and efficient in material. However, the W12 × 72 is more economical, because it is 5 pounds per foot less in weight. The final decision on selection will undoubtedly involve issues concerning dimensional coordination and construction.

Question 9.12: A four-story building has a structural steel beam-girder-col...

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