6. Punching Shear Calculation

1.5 PUNCHING SHEAR CHECK

66 of 188

SAFE

SAFE v8.1.0 - File: PENT 100 IFC-R1-A - December 5,2009 11:40 - Scale: Fit to Page Structural Layer Plan View - KN-m Units

67 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

PROJECT:

SIDRA MEDICAL & RESEARCH CENTER, DOHA

TITLE:

CLINIC BUILDING-PUNCHING SHEAR CHECK

DOCUMENT NO

DATE

1861B-CS-05-00216

08/12/2009

DESIGNED

CHECKED

KAMAL

JSB/MDS

SHEET

Punching Shear check Pent floor level.(450 mm slab Thickness) Total maximum load on column ( Grid F-5 column load) (Point label : 32)

V =

3390

kN

(Subtracting beam shear forces 331KN ) (3721 - 331 = 3390)

Moment transmitted to column

Mx =

0

kN-m

Moment transmitted to column

My =

122

kN-m

Depth of Slab provided

=

450

mm

clear cover assumed

=

30

mm

Effective depth of the Slab

=

372

mm

Size of the column

=

800

mm

Grade of Reinforcement

=

420

N/mm2

Grade of Concrete of Flat Slab

=

40

N/mm2

Veff

i) Effective shear force at face of column maximum shear stress at the column face

=

3618.8

=

Veff / (uod)

uo =

v max

4xd

kN

=

(Unbalanced bending moment from SAFE) along axis of beam My is 122 kN-m

(Square)

( As per clause 3.7.6.2 of BS:8110- Part-1-2005)

3200

mm

3618750 = --------------------------1190400 =

3.04

N/mm2

<

2 5 N/mm

N/mm2 < 5 0.8 √fck = (refer 3.7.7.2, BS8110-1)

Hence safe ii) Punching shear perimeter at 1.5d from face of column : u = =

4 x ( Column size + 2 x 1.5 x effective depth of slab) 4x ( 800 + 2 x 1.5 x372)

=

7664

mm

1.5d

1916 1.5d

1.5d 1916

mm 1.5d

1916mm

68 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

PROJECT:

SIDRA MEDICAL & RESEARCH CENTER, DOHA

TITLE:

CLINIC BUILDING-PUNCHING SHEAR CHECK

Effective shear force at 1.5d from face of column Net shear force at 1.5d from face of column

Veff =

Veff =

Design punching shear stress

v

3485.511 kN

3373.9

DOCUMENT NO

DATE

1861B-CS-05-00216

08/12/2009

DESIGNED

CHECKED

KAMAL

JSB/MDS

SHEET

( As per clause 3.7.6.2 of BS:8110- Part-1-2005)

kN

Veff = --------uxd 3373.92*1000 = ---------------------------7664 x 372 =

1.18

N/mm2

Area of main steel given

=

8040

mm2

% of steel

=

2.16 %

Corresponding shear stress ( refer Table3.8, BS8110-1)

=

0 82 N/mm2 0.82

Vc =

0.96 N/mm2

this value to increased by (fck/25)

1/3

v

> Vc

Grade

=

40

N/mm2

(refer 3.7.7.4, BS8110-1)

Shear reinforcement is required As Per Clause3.7.7.5 of BS 8110 : 1997

Design of Shear Links : v

=

1.18 N/mm2

Vc

=

0.96 N/mm2 for case v Vc

Grade

=

40

N/mm2

(refer 3.7.7.4, BS8110-1)

Shear reinforcement is required As Per Clause3.7.7.5 of BS 8110 : 1997

Design of Shear Links : v

=

1.13 N/mm2

Vc

=

0.97 N/mm2 for case v Vc

Grade

=

2 40 N/mm

(refer 3.7.7.4, BS8110-1)

Shear reinforcement is required As Per Clause3.7.7.5 of BS 8110 : 1997

Design of Shear Links : v

=

0.94 N/mm2

Vc =

0.77 N/mm2

For Case

for case v Vc

Grade

=

2 40 N/mm

(refer 3.7.7.4, BS8110-1)

Shear reinforcement is required As Per Clause3.7.7.5 of BS 8110 : 1997

Design of Shear Links : v

=

1.08 N/mm2

Vc

=

0.97 N/mm2 for case v