Question: ( Continuous foundation ) ( Inter or spread toot ng ) A simple light - framed wood building is subjected to the load conditions as

(Continuous foundation)
(Interor spread tootng)
A simple light-framed wood building is subjected to the load conditions as specified.
Along rafter length:
Roof pitch: 9:12
Dead load (roofing + sheathing + ceiling + rafters)=20 psf
Live load (snow)=40 psf
Ridge beam spans 16 feet from column support to column support
Please round to the nearest one-tenth (i.e.,0.1).
Determine the equivalent (horizontally projected) total load on the rafters = Answer 1 Question 1
lb/ft
Determine the load per foot on the bearing wall (at the bottom), including the wall weight (wall dead load =7 psf and wall height, H =8 ft)= Answer 2 Question 1
lb/ft
Determine the load per foot on the ridge beam = Answer 3 Question 1
lb/ft
Determine the load on the Interior column = Answer 4 Question 1
lb
Floor load conditions:
Floor dead load (flooring + subfloor + joists)=20 psf
Live load (occupancy)=50 psf
Determine the floor joist reactions into load per foot on the continuous foundation (one side)= Answer 5 Question 1
lb/ft
Determine the floor joist reactions into load per foot on the floor beam = Answer 6 Question 1
lb/ft
Foundation conditions:
Continuous foundation's stem wall is 8 inch thick and 2 ft tall
Footing base is 8 inch thick
Concrete density =150 pcf
Soil bearing capacity =1200 psf
Determine the minimum theoretical width of the continuous foundation = Answer 7 Question 1
ft (Round to the nearest tenth!)
Determine the minimum theoretical width of the interior spread footing for the critical center column = Answer 8 Question 1
ft (Round to the nearest tenth!)
( Continuous foundation ) ( Inter or spread toot

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