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 floor	30.80 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 ok
wD+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

 

ASCE 7 section 4.3 or IBC section 1607.4

 

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.