An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft. Dead load is 20 psf. Find: a. Basic floor live load Lo in psf
b. Reduced floor live load L in psf
c. Uniformly distributed total load to the beam in lb/ft.
d. Compare the loading in part c with the alternate concentrated load requried by the Code. Which loading is more critical for bending, shear, and deflection.?

a.  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load,  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

For the school, = 2

Therefore, we have;

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, = b × =

∴   =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, = 812.8 ft./lb

d. For the uniformly distributed load, we have;

= 812.8 × 26/2 = 10566.4 lbs

=  812.8 × 26²/8 = 68,681.6 ft-lbs

= 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

= 1000 lb

= 20 × 16 = 320 lb/ft.

= 1,000 + 320 × 26/2 = 5,160 lbs

=  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

= 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

1

explanation: