Question

15 points 9. A rectangular concrete beam of width b=24 in. is limited by architectural considerations to a maximum total depth h=16 in. It must carry a total factored load moment Mu 400 k-ft. Design the flexural reinforcement for this member, using compression steel if necessary. Allow 3-in, to the center of the bars from the compression or tension face of the beam. Material strengths are fy=60 ksi and fc=4 ksi. Select reinforcement to provide the needed areas, complete necessary checks, and show a sketch of your final design, including provisions for No. 4 stirrups.

Answer 1

ANSWER:

flexursal ceinforcement It is a mechanical answer to the problem of when a struckere needs cither an added semforcement late in the building stage due to overlooked Capocity essors or intermittent extreme forces due to local such as fault linee os tornado alley. Given that 622นโท ho16in Mu = 400 koft. Material strength: e fy - 60000 psi - 60 kg fe- Hooopsis 4 ksi Go for a singly reinforced Section P- Poos : 0.0181 (0-0.4) Calculate the area of Steel As = pbd ..20.01814 24x13 = 5.65in2 Astfy 085feb. a - 5-65x60 08544x24 • 339 816

0 4.15 in calaulate the maximum moment Mu - Asfy (d-4) - 565x60(13 - 4:15) -565x6 (13-2015) - 339x10925 = 3405 in kips Mul = 309 fl-kips Mul <Mu from, fs<fy a 4.15 24.88 in fo: Estax cod 4.88-3 - 29000X0:00 3x 488 227000*00054.88 - 2fooc X0 003x38524 : € -33599 Calculate the moment MN = My Mul 2 4444-309 MN - 135 kips-ft

calculate the reinforcement Ast= MN , ہے۔ ہی) 135812 335(3-3) 1620 33.5x10 : 1620 335 24.8 Use 4#10 Dose as the tension reinforcement calculate the positive reinforcement A5 -564749673360 -564+49620:55 .834n2 Use 7#lo bate 22 Pustive reinforcement As Ao > As for P-0.018) check for ! (889-50893760 0.854424 (889-5-08X0583|x60 085x4x 24 -8.89-2.96860 0.85X4x24. 73558 8106

- 433in C= 4.83 0.85 - 5:09 in check for fs fs'- 29000 20:03 xborot :29000X0:03x 200 - 290007 00 3X 0-4106 - 357222 354. Calculate the moment Mn= 5089255(13:3)+(889-509 356 ) 760 (13-433) Mn +5-08X47215+(8-89-509X0.5916)x60 (13-2165) Mn = 2398.522 +5878 X 60x10 835 22398.522 +5.878x650.1. -5629 kips-in 469 Kips-ft: check for beam for shead Et = 0.03da 2003 13-509 5.09 2003x7.91 5.09 20-0371554 -0.00466

As it is more than 0.005 6:0.9 $M1 = 4693081 +382 kips-ft Revise the design by increasing the compression Sieinforcement Try 4#11 bars as compression reinforcement increases. Use trial and emot a: (889-508 x 355/60 0-8584x24 Q = 4.33 in E 4:33 085 25.09 in. check for ' f' 290004 0.003 x 417-3 - 29000 YO 003X 0 3710 calculate the moment Mnr 624x31:5(13-3)+(889-624x32\x60(13-4.03) 25656 kips-in 2471 kips-ft check for bearn for shear {1:003x de

no 8 -0.03% 13-477 4.77 +0.00518 As it is more than 0.005, 6=0.9. & Mn = 471x09 424 kips.ft Hence ok. - Use 4#10 al steel in compress and 2#10 a Steel in tension - Drawing the cross section of the beam- 4# 24 7#lo