Corbel Excel Sheet Works

December 22, 2016 | Author: Ibrahim A.L | Category: N/A
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Short Description

Design of concrete Corbel according to BS 8110...

Description

Design of Reinforced concrete Corbel Given Data 2 45 N/mm 2 460 N/mm

Fcu

=

Fy

=

D.l L.L

= =

Col.dim =

bar dia =

25 mm

cover

45 mm

=

1165 KN 785 KN 2 750 X 750 mm

Required Flexural reinforcement Shear reinforcement

Solution Maximum bearing stress

= 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

16 KN/m 15 2.7 1.4 1.8 175 100 30

KN/m m m m mm mm degs

Design of Free Standing Staircase

View more...

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