Design a Southern Pine No. 1 beam to carry the loads shown (roof beam, no plaster). Assume the beam is supported at each end by an eight inch block wall.

$F_{ b }$ = 1,550 psi; $F_{v }$ = 110 psi; E = 1.6 × $10^{6}$ psi

Question

Design a Southern Pine No. 1 beam to carry the loads shown (roof beam, no plaster). Assume the beam is supported at each end by an eight inch block wall.

[latex]F_{ b }[/latex]= 1,550 psi; [latex]F_{v }[/latex]= 110 psi; E = 1.6 × [latex]10^{6}[/latex] psi

Accepted Answer

Bending

[latex]S_{ ext {required }}=\frac{M_{\max }}{F_{ b }}=\frac{12.8 k - ft . imes 12 in. / ft. }{1.55 ksi }=99.1 in .^{3}[/latex]

Shear

[latex]A_{ ext {required }}=\frac{1.5 V_{\max }}{F_{ v }}=\frac{1.5 imes(2,750 \#)}{110 psi }=37.5 in .^2[/latex]

Deflection (Allowable)

[latex]Δ_{ ext {allow }}=\frac{L}{240}=\frac{15^{\prime} imes 12 ext { in. } / ft }{240}=0.75^{\prime \prime}[/latex]

Try 6 × 12 S4S.

(A = 63.3 in.²; [latex]S_{x}[/latex] = 121 in.³; [latex]I_{x}=697 ext { in. }^{4}[/latex] )

[latex]ω_{ ext {beam }} \approx[/latex] 0.252 × 63.3 = 16 #/ft.

Bending

[latex]M_{ ext {add. }}=\frac{\omega_{ ext {beam }} L^{2}}{8}=\frac{16 \# / ft \cdot\left(15^{\prime} ight)^{2}}{8}=450 \#- ft.[/latex]

[latex]S_{ ext {add. }}=\frac{M_{ ext {add. }}}{F_{ b }}=\frac{450 \# ext { -ft. } imes 12 in . / ft. }{1,550 psi }=3.5 in .^{3}[/latex]

[latex]S_{ ext {total }}=99.1 in .^{3}+3.5 in .^{3}=102.5 in .{ }^{3}<121 in .^{3} herefore OK [/latex]

Note: [latex]S_{ ext {add. }} [/latex] is usually approximately 2% to 5% of [latex]S_{ ext {required }}[/latex].

Shear

[latex]V_{ ext {add. }}=\frac{ω_{ ext {beam }} L}{2}=\frac{16 \# / ext { ft. }\left(15^{\prime} ight)}{2}=120 \#[/latex]

[latex]A_{ ext {add. }}=\frac{1.5 V_{ ext {add. }}}{F_{ v }}=\frac{1.5 imes 120 \#}{110 psi }=1.6 in .{ }^{2}[/latex]

[latex]A_{ ext {total }}[/latex]= 37.5 in.² + 1.6 in.² = 39.1 in.² [latex]<[/latex] 63.3 in.² [latex] herefore OK[/latex]

Actual Deflection

Using superposition (the combination, or superimposing, of one load condition onto another),

[latex]\Delta_{ ext {actual }}=\frac{5 \omega_{ LL } L^{4}}{384 E I}+\frac{23 P L^{3}}{648 EI }( ext { @ centerline })[/latex]

[latex] herefore \Delta_{ ext {actual }}=\frac{5(100+16)\left(15^{\prime} ight)^{4}(1,728)}{384\left(1.6 imes 10^{6} ight)\left(697 in .^{4} ight)}+\frac{23(2,000 \#)\left(15^{\prime} ight)^{3}(1,728)}{684\left(1.6 imes 10^{6} ight)\left(697 in .^{4} ight)}[/latex]

[latex]\Delta_{ ext {actual }}[/latex]= 0.12" + 0.35" = 0.47" [latex]<[/latex] 0.75" [latex] herefore OK[/latex]

Check the bearing stress between the beam and the block wall support.

[latex]f_{p}=\frac{P}{A_{ ext {bearing }}}=\frac{2,870 \#}{44 in.^{2}}=65.2 psi[/latex]

The allowable bearing stress perpendicular to the grain for Southern Pine No. 1 is:

[latex]F_{ c \perp}=440 psi herefore OK[/latex]

Use 6 × 12 S4S.

Question 8.15: Design a Southern Pine No. 1 beam to carry the loads shown (...

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