Steel framework is used to support a reinforced concrete floor slab in a classro
ID: 1711546 • Letter: S
Question
Steel framework is used to support a reinforced concrete floor slab in a classroom (see idealized framing plan in Fig. 1)
The floor slab is 6 inches thick and has a density of 150 pcf. Neglect other dead loads. Use Table 1.4 for the live load. Complete your analysis for the 1.2D + 1.6L load combination.
The slab is supported directly by both the beams and girders (like the picture on the right side on Page 43).
Beams AB and GH and Girders AG and BH are connected directly to the columns as shown on Fig. 2-12(a) on Page 43.
a) For S1=10’, S2=20’, and S3=30’: show the loadings and reactions (with magnitudes) for members AB, CD, and AG. Then show calculation of load in Column A from the end reactions on the beams and girders.
b) For S1=10’, S2=10’,S3=10’: show the loadings and reactions (with magnitudes) for members AB, CD, and AG.
AI IH Fig. 1 Idealized Framing PlarnExplanation / Answer
Slab thickness = 6 in=0.5 ft
Self weight of slab=150*0.5=75 PSF
Live load = 40 PSF
Total factored area load=(1.2*75)+(1.6*40)=154 PSF
a)tributary width of AB =(1/2)S1 =5 feet
Uniform load on AB = 154*5=770 plf
Span of AB = S3 = 30 ft
Reaction at supports of AB=770*30/2=11.55 kips
Tributary width of CD = (1/2)(S1 + S2)=15 ft
Uniform.load on CD =154*15=2310 plf
Span of CD = 30 ft
Reaction at supports of CD =2310*30/2=34.65 kips
Girder AG is symmetrically loaded with the reactions from beams CD and EF. Reaction at supports of AG =34.65 kips
Load on column A = reaction of beam AB+reaction of girder AG =11.55+34.65 =46.2 kips
b)uniform load on AB =154*(10/2)=770 plf
Reaction at supports of AB = 770*10/2 =3.85 kips
Uniform load on CD =154*10=1540plf
Reaction at supports of CD =1540*10/2=7.7 kips
Girder AG is symmetrically loaded with the reactions from beams CD and EF. Reaction at supports of AG =7.7 kips
Load on column A = reaction of beam AB+reaction of girder AG =3.85+7.7 = 11.55 kips
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