= 0.8Fcu = 18 Assumming effective length = 350 mm factored applied load = 1.4D.L + 1.6L.L = 2887 KN Fu = 2887 KN from applied shear =
Fu
= Vu
=
bc
=
bc
=
Fu/A
459 mm taking bc to equal the width of the column which is 750 mm 750 mm taking depth at the face of the column such that shear developed at column face is less then 0.8√Fcu 2 0.8√Fcu = 5.366563 maximum premitted value is 5N/mm v d Hieght at col face =
let v = V/bcd = 1078
2 3 N/mm 3.773856
= 1284 mm
let d
=
1020 0.75
av
=
200
av/d
=
0.196078
z/d z
= =
0.71 724.2 mm
v/Fcu
=
0.083863
x
=
658 mm
from Strain compatibility we have Es
=
0.001926
Fy
=
386
Ft
=
798 KN
Fc
=
2996 KN
Fh
= 0.2 V = 577.4
Asmain
=
2068
Using T25 for reinforcement no of bars
= =
spacing of bars =
Asmin/ Asbar 5 bars 115 mm
Shear reinforcement approvide at 2/3 of d using 10 mm dia no of bars = spacing =
Amount of concrete = amount of steel in metters =
=
1034 6 bars
152 mm
0.271453875
Length of one main bar loop = Total length of main steel bars = Steel amount in kg =
4.2 m 16.8 m 64.75 Kg
Shear Reinforcement loop length = Total length of steel shear bars = Amount of steel in Kg =
3.7 m 18.5 29.2115
Total amout of steel used in single corbel =
93.9615 Kg
poundage
=
2 346.1417 Kg/m
v
=
765
if d=
v 1020.3
649
14.4
Design of Free S
Given data Ultimate load on flight
=
nf
=
nl = Ultimate load on landing = length of flight =a a = width of flight and landing = b = distance b/w centre lines of flights = b1 = height of landing = hf = height of flight = hf = slope of flight measured from horizontal = Φ = the supports are simple supports. Required A) Flexural reinforcement B) shear reinforcement if required Solution H = {nl(bl+b)(1+(b/2a)secΦ+nfacos Φ}/2tan Φ
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