The sun deck is to be framed using Hem-Fir No. 2 grade timber. Joists are spaced at two feet o.c. with a span length of 10 feet. One end of the joist is supported by a concrete foundation wall and the other end by a beam. The supporting beam is actually made of two beams, spliced at the center of t

Question

The sun deck is to be framed using Hem-Fir No. 2 grade timber. Joists are spaced at two feet o.c. with a span length of 10 feet. One end of the joist is supported by a concrete foundation wall and the other end by a beam. The supporting beam is actually made of two beams, spliced at the center of the span. Joists and beams are to be considered as simply supported.

Loads:

2" plank deck = 5 psf

LL = 60 psf

For Joists:

[latex]Δ_{ ext {total }} [/latex] < L/240

[latex]F_{ b } [/latex]= 1,150 psi

[latex]F_{ v }[/latex]=75 psi

E = 1.4 × [latex]10^{6} [/latex] psi

For Beams:

[latex]Δ_{ ext {total }} [/latex] < L/240

[latex]F_{ b } [/latex] = 1,000 psi

[latex]F_{ v } [/latex]=75 psi

E = 1.4 × [latex]10^{6} [/latex] psi

Accepted Answer

Joist Design

DL—2" deck =5 psf

LL = 60 psf

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

Total DL+ LL = 65 psf

[latex]ω_{ DL + LL }[/latex] = 65 psf × 2'= 130 #/ft.

[latex]V_{\max }=\frac{ω L}{2}=\frac{130 \# / ft.\left(10^{\prime} ight)}{2}=650 \#[/latex]

[latex]M_{\max }=\frac{ω L^{2}}{8}=\frac{130 \# / ft.\left(10^{\prime} ight)^{2}}{8}=1,625 \# ext { -ft. }[/latex]

[latex]A_{ ext {required }}=\frac{1.5 V }{F_{ v }}=\frac{1.5(650 \#)}{75 \# / in .^{2}}=13 in .{ }^{2}[/latex]

[latex]S_{ ext {required }}=\frac{M}{F_{ b }}=\frac{1,625 \# ext { -ft. } imes 12 in . / ft. }{1,150 psi }=17 in .{ }^{3}[/latex]

Try 2 × 10.

[latex]\left(A=13.88 in. ^{2} ; S_{x}=21.4 in.^{3} ; I_{x}=98.9 in.^{4} ; \omega_{ ext {beam }}=3.5 \# / ft . ight)[/latex]

Note: Hem-Fir has a density of 30 pcf; hence, the conversion factor is [latex]ω_{ ext {beam }}[/latex] = 0.22 × cross-sectional area of the beam.

Bending and Shear

[latex]V_{ ext {add. }}=\frac{3.5 \# / ext { ft. }\left(10^{\prime} ight)}{2}=17.5 \#[/latex]

[latex]A_{ ext {add. }}=\frac{1.5\left(17.5 in.^{3} ight)}{75 ext { psi }}=0.35 in.^{2}[/latex]

[latex]A_{ ext {total }}=13.4 ext { in. }^{2} herefore O K[/latex]

[latex]M_{ ext {add. }}=\frac{3.5 \# ext { -ft. }\left(10^{\prime} ight)^{2}}{8}=43.75 \# ext { -ft. }[/latex]

[latex]S_{ ext {add. }}=\frac{43.75 \#- ft . imes 12 ext { in./ft. }}{1,150 ext { psi }}=0.5 in.^{3} [/latex]

[latex]S_{ ext {total }}=17.5 in .^{3} herefore OK[/latex]

Deflection

[latex]\Delta_{ ext {allow }_{(D L+L L)}}=\frac{L}{240}=\frac{10^{\prime} imes 12 ext { in. } / ext { ft }}{240}=0.5^{\prime \prime}[/latex]

[latex]\Delta_{ ext {actual }_{( DL + LL )}}=\frac{5 \omega L^{4}}{384 E I}=\frac{5(130+3.5)\left(10^{\prime} ight)^{4}(1,728)}{384\left(1.4 imes 10^{6} ight)(98.9)}[/latex]

[latex]\Delta_{ ext {actual }_{( DL + LL )}}=0.22^{\prime \prime}<0.5^{\prime \prime} herefore OK[/latex]

Use 2 × 10 joists at two feet o.c. (equivalent to 2 psf)

Beam Design

Because joists are spaced uniformly and occur at a relatively close spacing, assume loads to be uniformly distributed on the beam.

Loads:

2" plank deck = 5 psf

2 × 10 @ 2' o.c. = 2 psf

LL = 60 psf

ω = 67 psf × 5' = 335 #/ft.

[latex]V_{\max }=\frac{ω L}{2}=\frac{335 \# / ft \cdot\left(8^{\prime} ight)}{2}=1,340 \#[/latex]

[latex]M_{\max }=\frac{ω L^{2}}{8}=\frac{335 \# / ft .\left(8^{\prime} ight)^{2}}{8}=2,680 \#- ft[/latex]

[latex]A_{ ext {required }}=\frac{1.5 V }{F_{ v }}=\frac{1.5(1,340 \#)}{75 psi }=26.8 in .^{2}[/latex]

[latex]S_{ ext {required }}=\frac{M}{F_{ b }}=\frac{2,680 \#- ft . imes 12 in \cdot / ft .}{1,000 psi }=32.2 in .^{3}[/latex]

[latex] ext { Try } 4 imes 10 .\left(A=32.38 in .^{2} ; S_{x}=49.91 in .{ }^{3} ; I_{x}=230.84 in .^{4} ; \omega_{ ext {beam }}=8 \# / ft . ight)[/latex]

[latex]A_{ ext {add. }} ext { and } S_{ ext {add. }}[/latex]should not be critical here.

[latex]\Delta_{ ext {allow }}=\frac{L}{240}=\frac{8^{\prime} imes 12 ext { in. } / ft .}{240}=0.4^{\prime \prime}[/latex]

[latex]\Delta_{ ext {actual }}=\frac{5 \omega L^{4}}{384 E I}=\frac{5(335 \# / ft .)\left(8^{\prime} ight)^{4}(1,728)}{384\left(1.4 imes 10^{6} ight)\left(231 in .^{4} ight)}[/latex]

[latex]\Delta_{ ext {actual }}=0.1^{\prime \prime}<0.4^{\prime \prime} herefore OK[/latex]

Use 4 × 10 S4S beam.

Question 8.18: The sun deck is to be framed using Hem-Fir No. 2 grade timbe...

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