Question

Q2 (50 pts): Design, verify and detail a reinforced concrete beam with fixed dimensions of b= 16 in and h = 24 in with f. = 5,000 psi and fy = 60,000 psi subjected to the bending moments: (a) Mu = 460 ft.k, and (b) Mu=- 590 ft.k

Answer 1

Design of section (a) as singly reinforced and section (b) as doubly reinforced section with complete solution is given below.

fc :) 0.003 0.003 +0.005 - 14 Find etc and associated moment loc 0.85 B, C 05) fy 0.05 ( fel-) 20.85 -0.05 (5-4) (BI 0.85*0.8 x) () B,0.85 = 0.8 ao lic ет 0.02125 As etc od = 0.02)25 x 16 x el Asta Tolh in? a= Aspfy 8.85 feb Tol4 x 60 0.85X5X16 6.3 in Ma = Asify [d- a = 414x60 * (21 – 6:3] Mn = 7.646.94 k-in 637.24 k-ft Ma =

27 Design of section @ Mr 460 k-ft Mn, neg 460 0.9 - 511011 K-pt = 0133. 33 k-in Mr > Marea . Singly neinforced section & Mn = As fy [d- & J x 0.9 (460x12) As x 60x (21 – 6:30) 6.73) *0.9 | As = 5073 in ² Assume #11 baus 5.73 Number of bans = 1.56 11 3.67 4 provide 4 # 1 4 # 1 baus (As = 6.24 in X Check As fy 6.24 x 60 0.85 X0.8 X 5X 16 11 6.88 0.85 fe fib $mn = 0.9 x As Fryx ( - fic (21-0.8*6.68] = 0.9% 6.24 x 60 x

= 6148.85 k-in фмп (6Mn = 512.4 K-ft > 460 Safe Ma- 590 kott 590 0.9 655.56 k-ft= 7866.67 k-in Mn,neq Ma,req > Mn 2. Doubly reinforced section Additional capacity required * Mn Mn' Mn, reg 7866.67 - 7646.94 219.73 k-in Tma! Additional steel * 6.3 7.875 Bi fs' 0.003 Es -0.003 X29000 x 7.875-3) 7.875 (se) ksi) fs! 53.86 ksi

As! : 11 Mn! 219.13 fs (d-d) 53.86 x (21-3) As' 1=0.221 in ² * Total tension steel required t's! Aś (to le As2 = :) ( s 53.86 60 LAs a = 0.204 inzy . As = Asit Asz Tola + O. 204 As to 344 in² * provide 2# 4 bars for compression stec As'= Oode in 2 provide 5# 11 bain for tension steel As 5x 1.56 As = Tosin²

Check moment capacity Mn = {As' x ky * (d-d'))+((13- Ad) fry (6-2)] (0.4x60x (21-3)) + ((7.8 -0.4) 60 (21-6 8357.4 k-in & Mn 7521.66 k-in фMv 826.81 k-ft 590 Safe K16 6 I-3 244 21 21 4#11 5#11 3 3 Section for section for case ca) Case (6)