Solution of Maximum Uniformly Distributed Service Live Load That A Beam Can Support Based on Its Flexural Strength

Articles > Solution of Maximum Uniformly Distributed Service Live Load That A Beam Can Support Based on Its Flexural Strength

Question:

A cantilevered singly reinforced beam is subjected to a service dead load of 1.5 kip/ft, which includes the self-weight of the beam. The beam is reinforced with three no. 9 bars, and the yield stress of the tension steel is 60,000 psi. The beam’s span is 9.5 ft, and its capacity reduction factor is 0.9. The width of the beam’s compression edge is 16 in, and the beam’s effective depth is 22 in. The concrete’s specified compressive strength is 3000 psi. Determine the maximum uniformly distributed service live load that the beam can support based on its flexural strength.

Solution:

All the formulas used in this solution are obtained from Flexural Design of Reinforced Concrete Beams article.

From Section Properties of Reinforcing Bars & Cement Types, the cross-sectional area of one no. 9 bar is 1 in² .

The total cross-sectional area of the steel is

$A_s=n_{bars}A_b=3*1.00=3in^2$

The equivalent depth of the compression zone is

$a = frac{A_sf_y}{0.85f'_cb}=frac{300*60,000}{0.85*3000*16}=4.41in$

Then,

$phi M_n=phi A_sf_y(d-frac{a}{2})=0.9*3.0*(60,000)*(22-frac{4.41}{2})=3,207,000in-lbf$

converting to foot-kips,

$phi M_n=frac{3,207,000}{12*1000}=267ft-kip$

For a uniformly loaded cantilevered beam,

$M_u=frac{w_uL^2}{2}$

L is the span length in feet and wu is the factored uniformly distributed load in kips per foot. Using the strength requirement and solving for w_u gives

$frac{w_uL^2}{2}=phi M_n$

$w_u=frac{2phi M_n}{L^2}=frac{2*267}{9.5}=5.92kip/ft$

In terms of the service loads,

$w_u=1.2w_d+1.6w_l=1.2*1.5+1.6w_l=5.92kip/ft$

$w_l=2.58kip/ft$

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