Question

I designed and detailed the beam in problem 3 as you see. Redesign the reinforced concrete beam of problem 3, If the value of factored bending moment increased to 1,600 k.ft where the section needs to be designed as a doubly reinforced beam.

3) Design and detail the reinforced concrete beam shown below if it is subjected to a factored bending moment of 1,100 k.ft f'=4,800psi fx=60 ksi. Include skin steel. 36" 15" 2#6 36 24 kino f's = 4800 I 514 in 6#11 fy = 60000 1,5 clear cover 15" Asa 9.36 in? $Mn=1182 K.ft 4) Redesign the reinforced concrete beam of problem 3 if the value of factored bending moment increased to 1,600 k.ft where the section needs to be designed as a doubly reinforced beam.

Answer 1

Solution:

Given! •fc = 4800 Psi. fy = 60,000 Psi D=36" . b= 15". deff. = D-cover = 36 – 13 = 33 Inches. Mu=1600k-ft. solution! @ Schematic Representation for doubly Plf beams, is" K 0.003 Cu As' 33" 36" T- doc As Asi TELY . 0.0075 4-4-44 strain Block stress Block. For Mr. 0.003 ad I I ES fiskes As' C. (d-d') + Asa fy. TS2 IZTUZ Por Mnz strain Block stress Block.

1 Singly Reinforced design (Mr. Let, & + = 0.0075; Hese, 8 = 0.0138.... For fc=4000psi . P = 0.0162 for fc = 5000psi.& fy = 60 ks). But, for, at fc=4800 psi; P=0.01572. As, = 80.003s x bxd= 0.01572x15x38= 7.78 in? Asgfy 7.78 x 60 = 7.6874 inch 0.88 fc.b. 0.85* 4.8 X 15 Now, Mng = Ts, x(d-%). Asix fyx (d-0/2). = 7.78X 60 x ( 33-7.6274). = 13624.16 k-inch. = 13624.16 K-ft = 1735.35k-ft. 12lft ..Omni (0.9)x1185-35= 1021.81 lip-ft Here, o Mon, < Mu=1600H-k. a = - Mni © Additional moment cap needed, Mn2. 0 (Mat Mna) > Mu. Mn2> My - Mai . Mnza 1600 1021-81 0.9 = 755.97 k-ft. - Mar

Addi Honal tension Steel needed (ASe):7. Mn2=Tse (d-d'). 1.755.97 K-H - Asz. fy.(d-d'). .. 755.97X12"/F+ = As2X60X( 339-3) 755.97X 12 AS2 = 60X30 As2 = 5.0898 inch? So, As = As,+Asa=7.7845-0398 - 12.82 inch © compression Steel: (As'):). Strain in compression Steel (es): Due to similar traingles in strain block for Mn29 ES (c=917 7.6274 y = 8.97") c-d' 0.003 0.85 c-d' X 0.003 = 8-97-3 -X0-003 8-97 ES' = 0.001996. <&= 0.00207 ----(Not Yeilded). fy (Hooke's Low) for steel ६५ Young's Modulus = 29000 = 60 ६५ Ey = 0.00207

fs=&s & Es. fs' = 0.001996X29000 ks). fs'=57.884 Ksi- Now, consider stress block for Mna. Tse=Cs As2.fy- Asifs ... 5.0398x60 = AS'X 57.884. As'=5.2240 inch². 6 Rebar calculation.. O tension As-12.82 in? , ii) compression ! As'= 5.2240 inch? No. of bars= 12.82 No. of bars= 5.2240 (All bar) TIX (1/3)2 #lo bons) TGX (10/? = 8.63 = 425 > I Nos. 5 Nos. 15" K #10-5 NOS. 36" TB'. Ta". #ll- 9 Nos. K