Problem 2: A concrete floor system consists of parallel T beams spaced 10 ft on
ID: 1712869 • Letter: P
Question
Problem 2: A concrete floor system consists of parallel T beams spaced 10 ft on centers and spanning 32 ft between simple supports. The 6 in. thick slab is cast monolithically with T beam webs having width bu=14 in, and total depth, measured from the top of the slab, of h=28 in. The effective depth, d will be taken 3 in. less than the total depth. In addition to its own weight, each T beam must carry a superimposed dead load of 50 psf and service live load of 225 psf. Material strengths are fy-60,oo0 psi and fe-4,00o psi. Determine the required tensile steel area and select the reinforcement needed for a typical interior T- beam member. Make sure to include a sketch of the midspan cross section, including the reinforcement. Problem 3: Repeat Problem 2 above assuming that beff is limited by a maximum of 25 in. and the slab is 4 in. thick.Explanation / Answer
2) center to center spacing of beam=10 ft
Span = 32 ft
Slab thickness=6 in
width of web = 14 in
Total depth of beam=28 in
Effective depth of beam=28-3=25 in
Super dead load=50 psf
Live load=225 psf
grade of concrete=4000 psi
grade of steel=4000 psi
Let us determine the effective flange width of T-beam
i)span /4=32/4=8ft=96 in
ii) (8*6)+14+(8*6)=110 in
iii)53+14+53=120in
Therefore, effective flange width = 96 in
Dead load on beam due to slab self weight=(150*6/12)=75 psf
Self weight of beam web = 150*(14/12)*(22/12)=320.83 plf
Uniform load on beam due to slab weight=75*10=750 plf
Uniform load on beam due to superdead load = 50*10=500 plf
Uniform load on beam due to live load = 225*10=2250 plf
Total dead load = 320.83+750+500=1570.83 plf=1.57klf
Total live load = 2250plf=2.25klf
Total factored load on beam per 1.2D+1.6L combination=(1.2*1.57)+(1.6*2.25)=5.485 klf
Maximum factored moment in beam=5.485*322/8=702.08 kip-ft=8424960 lb-in
design of reinforcement:
Let the area of steel required be Ast, depth of compression zone be a.Also,let us assume the depth of compression zone is within flange
0.85*96*4000*a = 60000*Ast
Or, a=0.184Ast
Lever arm length = 25-a/2=25-0.092Ast
0.75*(25-0.092Ast)*60000*Ast = 8424960
Or, 1125000Ast-4140Ast2 = 8424960
Ast = 7.7 in2
a=0.184*7.7=1.42in(<6 in) .Therefore, the assumption that compression zone is within flange is correct.
Provide 2 layers of 3-#11 rabars such that the centroid of the rebars is at 3" from bottom of beam.
Total steel provided = 6*1.56=9.36 in2
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.