Clacul Ginda
1
M Ed = 20 kNm f ctm = 26 daN/cmp f cd = 166.7 daN/cmp η= 1 λ= 0.8 f yd = 3000 daN/cmp f yk = 3550 daN/cmp ξ b = 0.55 0.8 · ξ b = 0.55, pentru betoane <C28/35; h b = 165 cm 0.8 · ξb =0.5, pentru betoane clasa ≥C28/35 b b = 60 cm b b = 60 cm a s = 5 cm d = h - a s = 160 cm λx=d · (1 - √(1 - (2·M Ed / b · d 2 ·ηf cd ))) = 0.13 cm ξ= x/d = 0.001 ξ= x/d > ξ b → se mareste sectiunea sau se opteaza pt o sectiune dublu armata ξ= x/d ≤ ξ b → A s rqd = b · λx · ηf cd / f yd A s rqd = 0.417 cmp A s min ≥ 0.20· f ctm · b t · d / f yk fundatii A s min = 14.062 20 5 A s eff = 15.708 cmp 5Φ20 15.708 cmp p%ef. = 0.16 ξ= x/d ≤ ξ b → M rd = A s · f yd · (d - λx/2)= 753.69 kNm ξ= x/d > ξ b → M rd = b ·d 2 ·ηf cd · ξ b (1-λξ b /2) M rd = 753.69 kNm
description
Clacul Ginda
Transcript of Clacul Ginda
-
MEd = 20 kNmfctm = 26 daN/cmpfcd = 166.7 daN/cmp
= 1= 0.8
fyd = 3000 daN/cmp fyk= 3550 daN/cmpb = 0.55 0.8 b = 0.55, pentru betoane b se mareste sectiunea sau se opteaza pt o sectiune dublu armata= x/d b As
rqd = b x fcd / fydAs
rqd = 0.417 cmpAs
min 0.20 fctm bt d / fyk fundatiiAs
min= 14.062 20 5As
eff = 15.708 cmp 520 15.708 cmpp%ef. = 0.16
= x/d b Mrd = As fyd (d - x/2)= 753.69 kNm
= x/d > b Mrd = b d2fcd b(1-b/2)
Mrd = 753.69 kNm
