Question

Please solve following Concrete Reinforcement question ( ACI318-14 ) to be used.

3) fe - 3000 Psi fue 6o kei FIND 0 Mn CHECK E+ O R 2-#7 24" 3-#11 14"

Answer 1

Answer 2

ANSWER :
V D med '= 2.5" 287 d = 26.5 3 #11 2.5" b=14" wildth (b) = 4 =1911 Elective depth (d ) = (24+2,5) =26.5' 11 = 2.5" d' As '= 2#7 = 220.68= 1.2 in2 As - 3411 = 3x1-56= 4.68 102 fc' - 3000 psi = 3 ksi fy.e. Goksi Write equ li brium equations assuming ofs'=fy): we know that: Asfy = 0.85 fe ' b B, C + Asify 4.68x60 0,85 X3x14x 0185C +1,2x60 (26188 in a = (B,C) = 0.805 X 6-88 = 5-84 in

To find out strains in compression steel to verity the assumption that it is yielding: we know that: 6,88 -2.5 3(ed')= ( )x0.003 X0.003 = 0.001909. G = 0.003 6.88 ES 0:00206 by = fy Goooopsi 29000000 psi Ey> so the assumption is not valid, we have to use the equilibrium equation that is based on Since >Eg 1 (fs) not yielding الم۔ Asly = 0,85 fe B,ebt to! ( cd ) 0:00 3XES. 4-68 x 60 = 0.85% 3x 0-85CX 14+ 1-2 (C-257x0.00 3 x 29000 ksi 280.8 = 30 34c t 104C-261 C. 280,8 = 30-34c? +104-40-261 280.8C = 30,34c? 7104-40-26) 30-34c2_176.40-261-0 ) equation, on solving quadratic we get C = 7.03 . and a= (B,C) =0,8587-03 -5.9710

→ Calculation of strains, stresses and steel arecco 7.03-25 know that & = cod! -205) Ox0-00 c C-d' x0·003 - 7.03 C 0.00 1933 <Ey. fs = Es es Z 0.001933 X 29000 KSI = 56.06 kgi. 1.2X56060 psi 60000 psi As = As'f's ( 1.28 J-1 lo (2 in 2 fy As, = As - As 14.68–4:12) = 3.55182 4 = (d-c) x0.003 26.5-7.03 X 0.003 (26.5-7. Jx I 7-03 Et = 0.00830 > 0.005 then $ 209 = p mn = $[Ast fy (da) + As'f' (d-d') 50.9( 3.55X60 (26-5-5.97 ) +1-2x56-06 (26.5-2.5)) 50,9C 5008.69 +1614-52]. = 0.9x 66 23-21 5960.8910-k! "ØMn = 496.74 k-ft Thus, weget i. Design moment strength of mn) = 496 74 k-ft &= 0.00830