Question

1. For the given framing plan, first identify all unique beams and columns. After that, find out the final load transferred on each beam and girder. Finally draw load diagrams for all beams and girders, and determine the support reactions. 50 ft 50 ft 50 ft 25 ft I(2 B1 25 ft 3 1 25 ft 4 25 ft A B D Use: Live load 110 PSF ead load excluding weight of the slab Thickness of concrete slab = 5 inch 15 PSF (Includes all beam self weight) The beams and girders are flushed on the top. Total Load: tR PSF 87.5

Answer 1

Given: 50 ft 50 ft 50 ft 25 ft I(2 25 ft 3 25 ft T 4 25 ft I 5 B Solution: Step-1: calculate the area of the given section: know, the length is 50ft which is same for every section As we The breadth is 25ft, which is same for each section. Length 50 breadth 25 2 1 2 - Therefore, it is a one-way slab. HA

Step-2: Calculate tributary width in beams: 50ft = 10ft B t A-25ftx10ft -250ft Step-3: Calculate girder loads G, G2, G,,G,,G 25 G (125) lb -1563 ft 25 25 + 2 (125 G2 2 lb 3125 ft So far the loads will be san ame: 25 25 _ (125) G 2 2 lb =3125 25 25 + (125 G4 2 2 lb =3125 ft 25 25 (125) +_ 2 2 lb = 3125 ft

Slabwidth-5 inch. 5 -x150 12 bxdxUnit wt.=10x lb -625.05 ft Therefore, the self-weight of the slab is 625.05 ft Step-4: Column loads: Column-1 Column-2 Column-3 A=12.5x25 A,-12.5x50 A,=12.5x25 -375ft2 self weight of column: -312.5ft2 -625ft self weight of column: self weight of column: 0.042x 625 x150=3937.5lb0.042x 375 x 150-2343.8lb 0.042x 312.5 x150-1968.75lb Loads of the column: Loads of the column: Loads of the column: (115 15x375 = 48750/b (115 15x 312.5 = 40625lb (115 15x625 = 81250/b columnreactions columnreactions columnreactions (1968.75x40625lb = 42.593kij (3937.5)x81250/b 85.19/ (2343.75)+48750/b = 51.094k Column-4 A-25x50 =1250ft self weight of column 0.042 x 1250 x 150=7875lb Loads of the column: (115 15 x1250 162500/b columnreactions: (7875)x162500/b =170.38kips