Problem 1: The drawing shows the beam and girder layout for the second floor of
ID: 1712171 • Letter: P
Question
Problem 1: The drawing shows the beam and girder layout for the second floor of a commercial building. The live load (LL) for the flow is 60 PSF and the dead load (DL), which consists of a 4" thick concrete slab and 5/8" thick terrazzo, is 42 PSF. 30' 20 81 832 81 8 3-2 B-3 a) Compute the distributed loads applied beam B-3. Show a FBD of these loads and compute the reactions on each side of the beam. Are these the same values for B-2 and B-4? Why or why not? b) Repeat part a, but now analyze beam B-1. Observe that your values for B-1 are the same as those for B-5. Problem 2: Complete the second portion of the in-class knee-bracing example by solving for the force in member FH and the reactions in pin G (Gx, Gy).Explanation / Answer
1a) Tributary width of beam B-3 = 8'
Dead load =42 psf
Uniform distributed load on beam due to dead load = 8*42=336plf
live load = 60 psf
Uniform distributed load on beam due to live load = 8*60=480 plf
beam span=30'
Reaction on each side of beam due to dead load = (336*30/2)/1000=5.04kips
Reaction on each side of beam due to live load = (480*30/2)/1000=7.2 kips
Yes,these values vill be same for beams B-2 and B-4,because they have same tributary width,loading and span as B-3
b)Tributary width of beam B-1 = 8/2=4'
Dead load =42 psf
Uniform distributed load on beam due to dead load = 4*42=168plf
live load = 60 psf
Uniform distributed load on beam due to live load =4*60=240 plf
beam span=30'
Reaction on each side of beam due to dead load = (168*30/2)/1000=2.52kips
Reaction on each side of beam due to live load = (240*30/2)/1000=3.6 kips
Yes,these values vill be same for beam B-5,because it has same tributary width,loading and span as B-1