The floor system of an apartment building consists of a 4-in.-thick reinforced c
ID: 1711500 • Letter: T
Question
The floor system of an apartment building consists of a 4-in.-thick reinforced concrete slab resting on three steel floor beams, which in turn are supported by two steel girders, as shown in the figure. The areas of cross section of the floor beams and the girders are 18.3 in^2 and 32.7 in^2 respectively. If a 6-in.-thick brick wall, which is 7 ft high and 25 ft long, bears directly on the top of beam CD Determine the dead loads acting on the beam CD and the girder AE. Unit weight of concrete 150 lb/ft^3. Unit weight of masonry is 120 lb/ft^3. Self- weight of steel beam and girder 490 lb/ft^3.Explanation / Answer
The slab will be spanning one way in North-South direction
self weight of beam CD = 490*18.3/144 = 62.27 plf
self weight of 4" thick slab = 150*4/12 = 50psf
Tributary width of beam CD = 12ft
Uniformly distributed load on beam due to self weight of slab = 50*12 = 600 plf
Uniformly distributed load of wall on beam = 120*7*(6/12)=420plf
Total uniformly distributed dead load on beam CD = 62.27+600+420 = 1082.27 plf
Total dead load on beam CD = 1082.27*25 = 27056 lb =27 kips
Load on girder AE due to dead loan on beam CD = 27/2 = 13.5 kips
self weight of girder = 490*32.7/144 = 111.27 plf
Total dead load on girder = (24*111.27/1000) + 13.5 = 16.17 kips