Problem 2.13
Uniformly Distributed Load
Item Hint Value Reference AT 26 ft * 16 ft given D 20 psf given L0 classroom 40 psf ASCE 7 table 4-1 page 2 or IBC table 1607.1 page 2 KLL interior beam 2 ASCE 7 table 4-2 or IBC table 1607.9.1 L supports only
one floor30.80 psf ASCE 7 section 4.8.1 or IBC section 1607.9.1
KLLAT = 832 ft2 > 400 ft2 so reduction is permitted,
check L > 0.50 L0 okwD+L 812.8 lb/ft wD+L = (D + L) (tributary width)
Vmax = 812.8 * 26 / 2 = 10,570 lb Mmax = 812.8 * 262 / 8 = 68,680 ft-lb
vmax = 5 * 812.8 * 264 / 384 EI = 4,836,000/EI
V&M diagrams
deflection
Alternative Concentrated Load
PL classroom 1000 lb ASCE 7 table 4-1 page 2 or IBC table 1607.1 page 2 wD 320 lb/ft wD = (D + L) (tributary width)
Vmax = 1000 + 320 * 26 / 2 = 5,160 lb Mmax = 1000 * 26 / 4 + 320 * 262 / 8 = 33,540 ft-lb
vmax = 1000 * 263 / 48 EI + 5 * 320 * 264 / 384 EI = 2,270,000/EI
V&M diagrams
deflection
Uniformly distributed load causes higher shear force, bending moment and deflection.