A steel beam (A572 Grade 50) is loaded as shown. Assuming a deflection requirement of $\Delta_{\text {total }}$ = L/240 and a depth restriction of 18 inches nominal, select the most economical section.

$F_{b}$ = 30 ksi; $F_{v}$ = 20 ksi; E = 30 × 10³ ksi

Question

A steel beam (A572 Grade 50) is loaded as shown. Assuming a deflection requirement of $\Delta_{\text {total }}$ = L/240 and a depth restriction of 18 inches nominal, select the most economical section.

$F_{b}$ = 30 ksi; $F_{v}$ = 20 ksi; E = 30 × 10³ ksi

[latex]F_{b}[/latex] = 30 ksi; [latex]F_{v}[/latex] = 20 ksi; E = 30 × 10³ ksi

Accepted Answer

[latex]V_{\max }[/latex]= 24 k

[latex]M_{\max }[/latex]=238 k-ft.

Steel beams are usually designed for bending. Once a trial section has been selected, shear and deflection are checked.

Bending

[latex]S_{ ext {req }^{\prime} d}=\frac{M}{F_{b}}=\frac{238 ext { k-ft. } imes 12 ext { in. } / ft. }{30 ksi }[/latex]=95.2 in.³

Try W18 ×55.

[latex]\left(S_{x}=98.3 in .^{3} ; I_{x}=890 in .^{4} ; t_{ w }=0.39^{\prime \prime} ; d=18.11^{\prime \prime} ight) [/latex]

[latex]M_{ ext {add. }}=\frac{\omega_{ ext {beam }} L^{2}}{8}=\frac{55 \# / ft.\left(28^{\prime} ight)^{2}}{8}=5,390 \#- ft.[/latex]

[latex]S_{ ext {add. }}=\frac{M_{ ext {add. }}}{F_{ b }}=\frac{5.39 k ext { -ft. } imes 12 ext { in. } / ft .}{30 ksi }=2.2 in .{ }^{3}[/latex]

[latex]S_{ ext {total }}[/latex] = 95.2 in.³ + 2.2 in.³ = 97.4 in.³ [latex]< [/latex] 98.3 in.³ [latex] herefore OK [/latex]

Shear Check

[latex]V_{ ext {add. }}=\frac{\omega_{ ext {beam }} L}{2}=\frac{55 \# / ft.\left(28^{\prime} ight)}{2}=770 \#[/latex]

[latex]f_{ v _{ average }}=\frac{V_{\max }}{t_{ w } d}=\frac{24,000 \#+770 \#}{\left(0.39^{\prime \prime} ight)\left(18.11^{\prime \prime} ight)}=3,510 psi[/latex]

[latex]f_{ v _{ ext {average }}}=3,510 ext { psi }<20,000 ext { psi } herefore ext { OK }[/latex]

Deflection Check

[latex]\Delta_{ ext {allow }}=\frac{L}{240}=\frac{28^{\prime} imes 12 ext { in. } / ft .}{240}=1.4^{\prime \prime}[/latex]

[latex]\Delta_{ ext {actual }}=\frac{P L^{3}}{48 E I}+\frac{5 \omega L^{4}}{384 E I}[/latex]

[latex]\Delta_{ ext {actual }}=\frac{20 k \left(28^{\prime} ight)^{3} 1,728}{48\left(30 imes 10^{3} ight)(890)}+\frac{5(1.06 k / ft .)\left(28^{\prime} ight)^{4} 1,728}{(384)\left(30 imes 10^{3} ight)(890)}[/latex]

[latex]\Delta_{ ext {actual }}=0.59^{\prime \prime}+0.55^{\prime \prime}=1.14^{\prime \prime}<1.4^{\prime \prime} herefore OK[/latex]

Use W18 × 55.

Question 8.16: A steel beam (A572 Grade 50) is loaded as shown. Assuming a ...

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