D = 55 lb/ft + (8 ft)(50 psf) = 455 lb/ft L = (8 ft)(80 psf) = 640 lb/ft Area of slab supported per foot of length by W24 × 35 ! 1 ft 0 in' Area supported by one beam (shaded) W24 x 55 w24 x 55 4 ft 0 in 4 ft 0 in w24 x 55 W24 x 55 8 ft 0 in 8 ft 0 in 8 ft 0 in 8 ft 0 in 8 ft 0 in
D = 55 lb/ft + (8 ft)(50 psf) = 455 lb/ft L = (8 ft)(80 psf) = 640 lb/ft Area of slab supported per foot of length by W24 × 35 ! 1 ft 0 in' Area supported by one beam (shaded) W24 x 55 w24 x 55 4 ft 0 in 4 ft 0 in w24 x 55 W24 x 55 8 ft 0 in 8 ft 0 in 8 ft 0 in 8 ft 0 in 8 ft 0 in
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Saeed Moaveni
Chapter7: Length And Length-related Variables In Engineering
Section: Chapter Questions
Problem 42P
Related questions
Question
The interior floor system shown in Figure 2.2 has W sections spaced 8 ft on
center and is supporting a floor dead load of 50 psf and a live floor load of 80 psf. Determine the governing load in lb/ft that each beam must support.

Transcribed Image Text:D = 55 lb/ft + (8 ft)(50 psf) = 455 lb/ft
L = (8 ft)(80 psf) = 640 lb/ft
Area of slab
supported per
foot of length
by W24 × 35 !
1 ft 0 in'
Area supported
by one beam
(shaded)
W24 x 55
w24 x 55 4 ft 0 in 4 ft 0 in w24 x 55
W24 x 55
8 ft 0 in
8 ft 0 in
8 ft 0 in
8 ft 0 in
8 ft 0 in
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