A partial plan of an office building is shown. All structural steel is A36 steel. Design a typical interior beam B1, and restrict the live load deflection to ΔLL

Question

A partial plan of an office building is shown. All structural steel is A36 steel. Design a typical interior beam B1, and restrict the live load deflection to [latex]\Delta_{ LL } [/latex] < L/360. Limit depth to 14 inches. Also design the spandrel beam, restricting its total deflection to [latex]\Delta_{ LL } [/latex] < L/240. Limit depth to 18 inches.

Loads:

Concrete floor: 150 pcf

1" finish wood floor: 2.5 psf

Suspended fire-resistant ceiling: 3.0 psf

LL: 70 psf*

Curtain wall: 400 #/ft.

*Occupancy for office building with moveable partitions

A36 steel:

[latex]F_{ b }[/latex] = 22 ksi

[latex]F_{ v }[/latex] = 14.5 ksi

E = 29 × 10³ ksi

Accepted Answer

Beam B1 Design

5" concrete floor: 62.5 psf

1" finish wood floor: 2.5 psf

Suspended ceiling: 3.0 psf

[latex]\space[/latex] [latex]_____[/latex]

Total DL 68 psf

Total LL = 70 psf × 8' (tributary width) = 560 #/ft.

Total DL + LL = 138 psf × 8' (trib. width) = 1,104 #/ft.

[latex]M_{\max }=\frac{\omega L^{2}}{8}=\frac{1.1 k / ft \cdot\left(28^{\prime} ight)^{2}}{8}=108 k - ft.[/latex]

[latex]S_{ ext {required }}=\frac{M_{\max }}{F_{ b }}=\frac{108 k - ft . imes 12 in . / ft. }{22 ksi }=59 in .^{3}[/latex]

Try W14 ×43 (S = 62.7 in.³; I = 428 in.[latex]^{4} [/latex] ). No check is necessary for [latex]S_{ ext {add. }}[/latex]

Deflection

[latex]\Delta_{ ext {allow }_{( LL )}}=\frac{L}{360}=\frac{28^{\prime} imes 12 in . / ft .}{22 ksi }=0.93^{\prime \prime}[/latex]

[latex]\Delta_{ ext {actual }_{( LL )}}=\frac{5 \omega_{ LL } L^{4}}{384 EI }=\frac{5(0.56 k / ft .)\left(28^{\prime} ight)\left(1,728 in \cdot^{3} / ft .^{3} ight)}{384\left(29 imes 10^{3} ksi ight)\left(428 in .^{4} ight)}=0.62^{\prime \prime}[/latex]

[latex]\Delta_{ ext {actual }_{( LL )}}=0.62^{\prime \prime}<\Delta_{ ext {allow }_{( LL )}}=0.93^{\prime \prime} herefore OK [/latex]

Use W14 × 43.

Beam Reaction onto Spandrel Beam

[latex]R=\frac{\omega_{ ext {total }} L}{2}=\frac{(1.1+0.043 k / ft .)\left(28^{\prime} ight)}{2}=16.1 k[/latex]

Spandrel Design

Curtain wall = 400 plf

Span length =32'

[latex]\Delta_{ ext {allow }_{( D + L )}}=\frac{L}{240}=\frac{32^{\prime} imes 12 in . / ft .}{240}=1.6^{\prime \prime}[/latex]

[latex]M_{\max }[/latex] = 309 k-ft.

[latex]S_{ ext {required }}=\frac{309 k ext { -ft. } imes 12 ext { in. } / ft .}{22 ksi }=168.5 in .{ }^{3}[/latex]

Try W18 × 97 [latex]\left(S_{x}=188 ext { in. }^{3} ; I_{x}=1,750 ext { in. }^{4} ight) [/latex].

[latex]M_{ ext {add. }}=\frac{0.097 k / ft \cdot\left(32^{\prime} ight)^{2}}{8}=12.4 k - ft.[/latex]

[latex]S_{ ext {add. }}=\frac{12.4 k - ft . imes 12 in . / ft .}{22 ksi }=6.8 in .^{3}[/latex]

[latex]S_{ ext {total }}=168.5+6.8=175.3 ext { in. }^{3}<188 ext { in. }^{3} herefore ext { OK }[/latex]

Deflection

[latex]\Delta_{ ext {total }_{( DL + LL )}}=\frac{5 \omega L^{4}}{384 E I}+\frac{P L^{3}}{20.1 E I}[/latex]

[latex]\Delta_{ ext {total }_{( DL + LL )}}=\frac{5(.5)\left(32^{\prime} ight)^{4}(1,728)}{384\left(29 imes 10^{3} ight)(1,750)}+\frac{16.1\left(32^{\prime} ight)^{3}(1,728)}{20.1\left(29 imes 10^{3} ight)(1,750)}[/latex]

[latex]\Delta_{ ext {actual }}=0.23^{\prime \prime}+0.91^{\prime \prime}=1.14^{\prime \prime}<\frac{L}{240}=1.6^{\prime \prime} herefore OK[/latex]

Shear Check

[latex]f_{ vaverage }=\frac{V}{t_{ w } h}=\frac{30.6 k +1.6 k }{\left(0.535^{\prime \prime} ight)\left(18.59^{\prime \prime} ight)}=3.2 ksi <14.5 ksi herefore OK [/latex]

Use W18 × 97.

Question 8.17: A partial plan of an office building is shown. All structura...

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