A simply supported beam (B-2) supports a uniformly distributed load and is braced at six foot intervals (see Figure 8.48) in addition to a composite slab that is secured to the compression flange of the beam via shear studs. Assume that the beam is made of ASTM A992 grade with $F_{ y }$ = 50 ksi.

Select the lightest W shape to safely support the following loads:

DL = 100 psf; LL = 75 psf

Beams are spaced at 12 feet o.c.

Question

A simply supported beam (B-2) supports a uniformly distributed load and is braced at six foot intervals (see Figure 8.48) in addition to a composite slab that is secured to the compression flange of the beam via shear studs. Assume that the beam is made of ASTM A992 grade with $F_{ y }$ = 50 ksi.

Select the lightest W shape to safely support the following loads:

DL = 100 psf; LL = 75 psf

Beams are spaced at 12 feet o.c.

Select the lightest W shape to safely support the following loads:

DL = 100 psf; LL = 75 psf

Beams are spaced at 12 feet o.c.

Accepted Answer

[latex]ω_{ DL }[/latex] = 100 #/ft.² × 12' = 1,200 #/ft.

[latex]ω_{ LL }[/latex] = 75 #/ft.² × 12' = 900 #/ft.

Because this example represents a floor load condition, the two factored load considerations that apply are

1.4D = 1.4 × 1,200 #/ft. = 1680 #/ft. (for DL only)

1.2D + 1.6L = 1.2(1,200 #/ft.) = 1.6(900 #/ft.)

= 2,880 #/ft.

The factored load of 2,880 #/ft. governs the design for the floor beams.

[latex]M_{ u }=\frac{ω L^{2}}{8}=\frac{(2.88 k - ft .)\left(24^{\prime} ight)^{2}}{8}=207.4 k - ft.[/latex]

[latex]V_{ u }=\frac{ω L}{2}=\frac{(2.88 k - ft .)\left(24^{\prime} ight)}{2}=34.6 k[/latex]

[latex]Z_{ x }=\frac{12 M_{ u }}{\phi_{ b } F_{ y }}=\frac{12(207.4 k ext { -ft. })}{(0.90)\left(50 k / in .^{2} ight)}=55.3 in .{ }^{3}[/latex]

From Tables A11(a) and A11(b) in the Appendix, select the lightest trial size.

Try W18 × 35.

[latex]Z _{ x } [/latex] = 66.5 in.³, d = 17.7 in., h = 15.5 in.

[latex]\left.b_{ f }=6.00 ext { in., } t_{ f }=0.425 ext { in., } t_{ w }=0.30 ext { in. } ight)[/latex]

[latex]M_{ p }=F_{ y } Z_{ x }=\left(50 k -in.^{2} ight)\left(\frac{66.5 in .^{3}}{12 in . / ft } ight)=277 k - ft [/latex]

[latex]\phi M_{ p }[/latex]=(0.90)(277 k-ft.) = 249 k-ft.

This value can also be found directly in Table A11(b) in the Appendix.

Next, check the compactness of the beam section.

[latex]\frac{b_{ f }}{2 t_{ f }}=\frac{6.00^{\prime \prime}}{2\left(0.425^{\prime \prime} ight)}=7.06<9.19 [/latex] (the flange is compact)

[latex]\frac{h}{t_{ w }}=\frac{15.5^{\prime \prime}}{0.300^{\prime \prime}}=51.7<90.5[/latex] (the web is compact)

Shear check:

[latex]\frac{h}{t_{ w }} \leq \frac{418}{\sqrt{F_{ y }}}[/latex]

then use

[latex]V_{ u } \leq \phi_{ v } V_{ n }=(.90)\left(.6 F_{ y } t_{ w } d ight)[/latex]

[latex]\frac{h}{t_{ w }}=\frac{15.5^{\prime \prime}}{0.300^{\prime \prime}}=51.7<\frac{418}{\sqrt{F_{ y }}}=\frac{418}{\sqrt{50}}=59[/latex]

Therefore,

[latex]\phi_{ v } V_{ n }[/latex]=(0.90)(0.6)(50)(0.30 × 17.7") = 143.4 k

[latex]V_{ u }=34.6 k <\phi_{ v } V_{ n }=143.4 k[/latex]

[latex] herefore[/latex] the beam is OK in shear

The W18 × 35 is adequate and the lightest section to use as a floor beam.

Question 8.23: A simply supported beam (B-2) supports a uniformly distribut...

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