design of flat slab column connections as per aci 352-4...

Sheet :

Cont'd :

Design of Shear Studs for interior Column as per ACI 421.1R-8

:

: Building Code Requirements for Structural Concrete (ACI 318-08) and Commentary

: Shear Reinforcement for Slabs - Reported by Joint ACI-ASCE Committee 421

: Recommendations for Design of SlabColumn Connections in Monolithic Reinforced Concrete Structures

Data :

The factored forces transferred from the column to the slab are:

=

=

=

d = - - = =

= =

Prepared by : Date :Job No. : J-1023J-1023J-1023J-1023

1 of 34

A B Quadri9/30/20139/30/20139/30/20139/30/2013

Verified by : Calculation Sheet Revision Notes :

A B Quadri Civil & Structural Design Engineers.

Project : Animal Feed Factory

Subject : Design of Flat Slab Column connections as per Aci 352-1-R89

CALCULATION

Ref codes

ACI 318-08-

ACI 421.1R-99

ACI 352.1 R-89

Strength of concrete fc 27.7 Mpa 4018 psi

Strength of reinforement fy 415 Mpa 60191 psi

thickness of slab 175 mm 7 in

Concrete Cover 19 mm 0.75 in

Dead Load DL 1.00 Kn/m2 21 psf

Self Wt of slab 4.20 Kn/m2 88 psf

Finishes 1.00 Kn/m2 21 psf

Total DL 6.20 Kn/m2 129 psf

Live Load LL 2.00 Kn/m2 42 psf

Column Size Cy 510 mm 20.08 in

Column Size Cx 305 mm 12.01 in

Kn 110.16 kip

Muy = 68 Kn.m 50.184

Vu

The five steps of design outlined in Chapter 3 of ACI 421.1R-8 are followed.

Step 1 : —The effective depth of slab 175 19 16 148

kip.ft

Mux = 0 Kn.m 0 kip.ft

flexural reinforcement nominal diameter

16 mm 5/8 in

mm 5.83 in

= 490

Properties of a critical section at d/2 from column face shown in Fig. 3.1 a

bo = 2222 mm 87.48 in Ac = 328856 mm2 509.73 in2

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www.abqconsultants.com

This program Designs and

Optimises Flat Slab Column

Connections as per ACI 352-4-

R89-3

Written and programmed by

:-

A B Quadri


[email protected]

[email protected]

9959010210

9959010211

= =

= =

= =

fig 3.1afig 3.1afig 3.1afig 3.1a

2 2

3 3

2 2

3 3

= =

= =

* * * *

* * * *

* * *

fig 3.1bfig 3.1bfig 3.1bfig 3.1b

bo = 2222 mm 87.48 in 328856 mm2 509.73 in2

Lx1 = 453 mm 17.835 in Ly1 =

1.00E+10 mm4 24025 in4

Note : Calculate Jx and Jy by using sheet "Jx Jy calculation" and enter appropriate value above.

The fraction of moment transferred by shear [Eq. (3.3)]:

658 mm 25.906 in

Jx1 = 1.00E+10 mm4 24025 in4 Jy1 =

γγγγvy = 1 -1

1 -1

=

1 +√√√√

Ly11

γγγγvx = 1 -1

=

=

1 +√√√√

0.356

√√√√17.83525.91

+√√√√

25.906Lx1 17.83

0.446

Lx11 +

Ly1

= 1 -1

2

at y = . 25.91 12.95 in 329 mm

The maximum shear stress occurs at x = 17.83 8.92 in 226.5 mm

x

Ac Jx Jy

and its value is [Eq. (3.2)]:2

vu =Vu

+γvx Mux y

=110.16 1000

+0.446 0

+γvy Muy

12000 12.95

510 24025

+0.356 50.18 12000 8.92

24025

= 295.73 psi = 2.04 Mpa= 216.12 + 0.00 + 79.61

√√√√fc' ok

0.85

Prep By : A B Quadri- Abq Consultants - 9959010210 - [email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-India.-www.abqconsultants.com

= 5.49√√√√

fc' < 6.00vu / φ =295.73

= 347.91 psi

Visit


This program Designs and

Optimises Flat Slab Column

Connections as per ACI 352-4-

R89-3

Written and programmed by

:-

A B Quadri


[email protected]

[email protected]

9959010210

9959010211

Sheet :

Cont'd :

a)

b).: =

*

c)

d)

The shear stress resisted by concrete in the presence of thes hear reinforcement (Eq. 3.10) at the same critical section:

Prepared by : Date :

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A B Quadri9/30/20139/30/20139/30/20139/30/2013


CALCULATION

The nominal shear stress that can be resisted without shear reinforcement at the critical section considered [Eq. (3.5) to (3.7)]: is the minimum of the following:

vn = ( 2 +4

)√√√√

fc'where βc is the ratio of long side to short side of the column

βc cross section.

= ( 2 +4

)(Eqn 3.5 )

1.672

vn = (αs * d

+ 2 )√√√√

√√√√fc' = 4.39

√√√√fc'

fc'where αs is 40 for interior columns, 30 for edge columns, 20 for corner columns.

bo αs 40

= (40 5.83

+√√√√

fc'(Eqn 3.6 )

87.48= 4.66

vn = 4.00√√√√

fc'(Eqn 3.7 )

2 )√√√√

fc'

psi = 1.75 Mpa

Step 2

The quantity vn/φ is greater than vn, indicating that shear reinforcement is required; the same quantity is less than the upper limit vn = 6 √fc', which means that the slab thickness is adequate.

At a critical section outside the shear-reinforced zone,

use the smallest value of eqn 3.5, 3.6 and 3.7 :

.: vn = 4.00√√√√

fc' = 253.54

vn = 2.00√√√√

fc'(Eqn 3.8 )

126.77 psi = 0.87 Mpa (Eqn 3.10 )vc = 2.00√√√√

fc' =

Use of Eq. (3.1), (3.9), and (3.11) gives:

vu <= φ vn vu

vn = vc + vs φ

Av vs * bo *

s

Av * fyv

bo * s

so <= * d = * = =

s <= * d = * = =

3 //// 8 in = diameter studs welded to a bottom anchor strip

3 //// 16 x 1 in = x Taking cover of 3 //// 4 in

at top and bottom the length of stud ls should not exceed =

cross-section area of studs = in2 =

lsmax = - 2 * 3 - 3 = in =

4 16

in =

√√√√

= 1.52 Mpa(Eqn 3.9 )

vc = 2.00√√√√

fc'(Eqn 3.10 )

347.91 - 126.77 = 221.14 psi(Eqn 3.1 )

vs >= - vc =

Step 3

0.4 0.4 5.83 2 1/3 in

in 8.16 mmfyv 60191

vs =(Eqn 3.11 )

= =221.14 87.48

= 0.32

This example has been provided for one specific type of shear stud reinforcement, but the approach can be adapted and used also for other types mentioned in Appendix A.

Try 9.525 mm

4.7625 25.4 mm

59.20 mm

0.5 0.5 5.83 3 in 74.00 mm

19.05 mm

0.11 71 mm2

7 5 1/5 132 mm

or the overall height of the stud, including the two anchors, should not exceed 5 1/2 140 mm


Sheet :

Cont'd :

lsmin = lsmax - one bar diameter of flexural reinforcement

lsmin = - = in =

in =

studs per peripheral line

= in =

= in =

*

Allowable

This value is , indicating that the choice of studs and their spacing is

Properties of critical section at 4d from column face fig 3.1b

= = * = =

= in = mm = in = mm

= in = mm = in = mm

= in = mm = in = mm

= = = =

The maximum shear stress in the critical section occurs on line AB at:


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A B Quadri9/30/20139/30/20139/30/20139/30/2013


Choose ls = 5 1/4 133 mm

with 10

CALCULATION

Also, ls should not be smaller than:

5 1/2 5/8 4 7/8 124 mm

and the spacing between column face and first peripheral line so 2 1/4 57 mm ok

choose the spacing between peripheral lines s 2 3/4 70 mm ok

in = 9.63 mms 3

Av=

10 0.11= 0.38

Step 4

α 4.0 αd 4.0 5.83

8.16 mm oks

greater than 0.32 adequate

>Av

= 0.32 in =

23 1/3 in 592.00 mm

lx1 17.8 453 ly1 25.9 658

ℓ 30.2 767 bo 202.8 5152

lx2 58.6 1489 ly2 66.7 1694

x =58.6

= 29.31 in

Ac 1182 in2 762533

= 7452

4.495E+05 in4 1.871E+11 mm4mm2 Jy

in = 724 mmfill your value of x

here if differentmm 28.50adopt x =

Eq. (3.2) gives:* * * *

* * * *

* * *

The value = = is than vn = 2√fc = psi =

which indicates that shear stress should be checked at α = n = peripheral lines of studs

distance between column face and outermost peripheral line of studs:

= so + (n-1) s = + ( - 1 ) * = in =

Check shear stress at a critical section at a distance from column face:

= + = + = in =

Muy x

Jx Jy

2

vu =Vu

+γvx

Ac

=110.16 1000

+0.446 0

Mux y+

12000 12.95

1182 449500

γvy

+0.356 50.18 12000 28.50

449500

= 0.74 Mpa= 106.80 psi= 93.21 + 0.00 + 13.60

vu / φ =106.80

= 125.65 psi

126.77 0.87 mpa

4 try 8

√√√√fc'

0.85

vu / φ 125.65 psi 0.87 mpa <=

= 1.98√√√√

fc' < 2.00

2

α =25.64

= 4.4d

mm

αd 22.72 d/2 22.72 5.83 25.64 651.20 mm

αd 2.33 8 2.91 22.72 577.20

Mpa

vu / φ = 124.04 psi = 0.86 Mpa

12.23 = 105.44 psi = 0.73vu = 93.21 + 0.00 +


Sheet :

Cont'd :


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CALCULATION






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