Bracket - 4 Design

BRIGADE MALL

PROJECT NAME: DESIGNED:

SRIDHAR 4TH FLOOR

Fctored load

BRACKET NUMBER:

BRTK- 4

=

927

b

=

800

mm

f y

=

500

Materials

D

f ck N/mm2 =

25

kN

=

1200

mm

Distance b/w col. face to load av Ʈcmax Maximum shear stress

= =

250 3.125

mm N/mm

Effective depth

=

1150

mm

m = P/(0.889*fck*b*d) P/(0.889*fck*b*d)

=

0.0453

n = av/d

=

0.2174

=

0

d

13-05-2013 HNP

Pu

Width Overall Depth

DATE: CHECKED:

2

N/mm2

av 

COLUMN

Here av/d =

0.217

P 

BRACKET  

<1

Hence design as BRACKET

Calculate Z & Xu:

2

(Z/d) - (n/(m+n))(z/d)+(m/(m+n) (n/(m+n))(z/d)+(m/(m+n))n )n

2

Percentage of steel required Pt

Lever arm Z

=

940

mm

But Here, Take

Depth of N.A. Xu = 2.22(d-Z)

=

466

mm

Provide Ast

Limiting value Xumax = 0.46d

=

529

mm

Here

=

0.116

%

Here Pt < 0.4%; Hence Provide Minimum 0.4%.

Provide Dia

Xu < Xumax Under reinforced section; Provide Minimum Compression Steel.

Pt min 20

mm; No's

=

0.2

%

=

1840

mm

2

=

5.855

Say 6 No's

Hence provided Ast

=

1886

mm

Percentage of steel Pt

=

0.205

%

=

943

=

6.000

=

125

2

Computation Computatio n of Horizontal steel:

Computation Computat ion of Tensile force T:

 Ash = 0.5 Ast

2

mm

Tactual = P(av/Z)

=

247

kN

Provide Dia

Tmin = 0.5 P

=

464

kN

Spacing S = (2/3)d/No's

=

464

kN

Provide 6 No's 2-Legged links of 10mm dia in upper 2/3 depth or 765mm

Here Tact

<

Tmin;

T

Computation of Tensile steel Ast:

10

mm; No's

Check for vertical shear:

Strain in steel corresponding to maximum strain of 0.0035 in concrete

From SP:16, Cl.4.1, beta

=

14.162

es = 0.0035(d-Xu)/Xu

Ʈc Design shear stress Enhanced shear  Ʈce = Ʈc(2d/av)

= =

0.334 3.076

=

0.004109

Maximun strain in tension reinforcement @ failure esmax = (fy/1.15Es)+0.002 Here es

<

Say 6 No's mm

=

esmax

0.004174

Ʈcmax Here Ʈce < Shear capacity Vu = Ʈce bd

Hence Ok

Stress in steel fs = 0.87fy

=

435

 Area of steel required Ast = T/fs

=

1066

2

N/mm 2

mm

 Actual shear stress Ʈv = P/bd Here Ʈv

<

Ʈce

Hence take Ʈce = =

=

2830

2

N/mm

2

N/mm

3.076 N/mm2 kN

2 1.008 N/mm Hence SAFE in shear.

D