6. Punching Shear Calculation

1.5 PUNCHING SHEAR CHECK

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

SAFE

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LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

Punching Shear check

Pent floor level.(450 mm slab Thickness)

Total maximum load on column V = kN (Subtracting beam shear forces 331KN )( Grid F-5 column load) (Point label : 32) (3721 - 331 = 3390)

Moment transmitted to column = kN-m (Unbalanced bending moment from SAFE)along axis of beam My is 122 kN-m

Moment transmitted to column = kN-m

Depth of Slab provided = mm

clear cover assumed = mm

Effective depth of the Slab = mm

Size of the column = mm (Square)

Grade of Reinforcement = N/mm2

PROJECT: SIDRA MEDICAL & RESEARCH CENTER, DOHADOCUMENT NO DATE

1861B-CS-05-00216 08/12/2009

TITLE: CLINIC BUILDING-PUNCHING SHEAR CHECKDESIGNED CHECKED SHEET

KAMAL JSB/MDS

3390

Mx 0

My 122

450

30

372

800

420

Grade of Concrete of Flat Slab = N/mm2

i) Effective shear force at face of column Veff = kN ( As per clause 3.7.6.2 of BS:8110- Part-1-2005)

maximum shear stress at the column face =

uo = = mm

v max = ---------------------------

= N/mm2 < 5 N/mm2

< √fck = 5 N/mm2

Hence safe (refer 3.7.7.2, BS8110-1)

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) = mm

mm

40

3618.8

Veff / (uod)

4 x d 3200

3618750

1190400

3.04

0.8

7664

1916

1916

1916mm

1.5d

1.5d

1.5d

1.5d

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1861B-CS-05-00216 08/12/2009


KAMAL JSB/MDS

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

= kN

VeffDesign punching shear stress v = ---------

u x d

3373.92*1000= ----------------------------

7664 x 372

= N/mm2

Area of main steel given = mm2

% of steel = %

Corresponding shear stress = N/mm2 Grade = 40

Effective shear force at 1.5d from face of column

Veff 3485.511

Net shear force at 1.5d from face of column Veff 3373.9

1.18

8040

2.16

0 82 N/mm2Corresponding shear stress = N/mm2 Grade = 40( refer Table3.8, BS8110-1)

this value to increased by (fck/25)1/3 Vc = N/mm2

v > Vc (refer 3.7.7.4, BS8110-1)

Shear reinforcement is required

Design of Shear Links : - As Per Clause3.7.7.5 of BS 8110 : 1997

v = N/mm2

Vc = N/mm2

For Case = 1.6 x0.9 = N/mm2

Diameter of Shear Links = mm

Area Of Shear Links = PI()*dia^2/4 = mm2

Grade of Shear Links Reinforcement fyv = N/mm2

Angle between the Shear Reinforcement & Plane of the Slab =

Total Shear Reinforcement Required = (v-Vc) x u x d = mm2(0.87 x fyv )

Min Shear Reinforcement Required ∑Asv Sinα = 0.4 x u x d/(0.87 x fyv) = mm2

∑Asv Req < ∑Asv Min So, Provide ∑Asv = mm2

No Of Shear Links In section Considered n = = 28

Provide 28 Nos of 12 mm dia bars

0.82 N/mm

0.96

1.18

0.96

for case v<1.6vc

1.53

12

asv 113.10

420

90 0

∑Asv 1762.05

3121

3121.0

Asv/asv

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1861B-CS-05-00216 08/12/2009


KAMAL JSB/MDS

iii) Punching shear perimeter at 2.25d from face of column :-

u = 4 x ( Column size + 2 x 2.25 x effective depth of slab) = 4x ( 800 + 2 x 2.25 x372)

= mm

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

9896

2474

2474mm


Veff 3464.0

2.25d

2.25d

2.25d

2.25d.

= kN


u x d

3277.91*1000= ----------------------------

9896 x 372

= N/mm2


% of steel = %

Corresponding shear stress = N/mm2 Grade = 40 N/mm2

( refer Table3.8, BS8110-1)


v < Vc (refer 3.7.7.4, BS8110-1)

Punching Shear is OK, No shear reinforcement is required.

Net shear force at 2.25d from face of column

Veff 3278

0.89

8040

2.16

0.82

0.96

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Total maximum load on column V = kN (Subtracting beam shear forces 305KN ( Grid E-4 column load) (Point label : 56) from safe output,2750 -102 = 2648)

Moment transmitted to column = kN-m (Unbalanced bending moment from SAFE)Max. of Mx and My is 578 kN-m





Size of the column d = mm (circuar)




1861B-CS-05-00216 08/12/2009


KAMAL JSB/MDS

2648

Mx 578

My

450

30

372

700

420

40

i) Effective shear force at face of column = Vt * {1+ ( As per clause 3.7.6.2 of BS:8110- Part-1-2005)Vt x Where,Mt- Design moment about x/y axis

Effective shear force at face of column Veff = kN


uo = = mm

v max = ---------------------------

= N/mm2 < 5 N/mm2

< √fck = 5 N/mm2




= 4x ( 700 + 2 x 1.5 x372) = mm

mm

1.5Mt }

x - length of the side of the perimeter considered parallel to axis of bending= 2648X{1+ (1.5X578) + 0}(2648X700/1000)

3886.6

Veff / (uod)

∏ x d 2199.11

3886571

(2199.12x372)

4.75

0.8

7264

1816mm

1816

1816mm

1.5d

1.5d

1.5d

1.5d

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KAMAL JSB/MDS

= Vt * {1+Vt x

= kN

=

= -{((0.45*24+4)*1.4+2.5*1.6)*1816*1816/1000000}

= kN

Vnet

Design punching shear stress v = ---------u x d

3043.9*1000= ----------------------------

7264 x 372

= N/mm2


Veff 1.5Mt } ( As per clause 3.7.6.2 of BS:8110- Part-1-2005)

Veff = 2648X{1+ (1.5X578) + 0}(2648X1816/1000)


Veff 3125.42


Veff - {((SW+DL)*1.4+LL*1.6)*Lx*Ly/1000000}

3125.4


Vnet 3043.9

1.13


% of steel = %

Corresponding shear stress = N/mm2 Grade = 40( refer Table3.8, BS8110-1)


v > Vc (refer 3.7.7.4, BS8110-1)



v = N/mm2

Vc = N/mm2











8040

2.16

0.83 N/mm2

0.97

1.13

0.97

for case v<1.6vc

1.550807

12

asv 113.10

420

90 0

∑Asv 1162.49

2958

2958.1

Asv/asv

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1861B-CS-05-00216 08/12/2009


KAMAL JSB/MDS



= mm

= Vt * {1+ ( As per clause 3.7.6.2 of BS:8110- Part-1-2005)Vt x

= kN

9496

2374mm

2374mmEffective shear force at 2.25d from face

of columnVeff 1.5Mt }

Veff = 2648X{1+ (1.5X578) + 0}(2648X2374/1000)


Veff 3013.2

2.25d

2.25d

2.25d

2.25d.

=

= -{((0.45*24+4)*1.4+7.5*1.6)*2374*2374/1000000}

= kN

Vnet

Design punching shear stress v = ---------u x d

2873.89*1000= ----------------------------

9496 x 372

= N/mm2


% of steel = %




v < Vc (refer 3.7.7.4, BS8110-1)


of columnNet shear force at 2.25d from face of

columnVeff - {((SW+DL)*1.4+LL*1.6)*Lx*Ly/1000000}

3013.2


Vnet 2873.89

0.81

8040

2.16

0.83

0.97

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Total maximum load on column V = kN (Subtracting beam shear forces 283KN & 273KN ( Grid C-9 column load) (Point label : 14) from safe output,1819 - 283 - 273 = 1263)

Moment transmitted to column = 0 kN-m (Unbalanced bending moment from SAFE)Along axis of the beam My is 1162kN-m





Size of the column = mm (Circular)




1861B-CS-05-00216 08/12/2009


KAMAL JSB/MDS

1263

Mx

My 1162

450

30

372

700

420

40Grade of Concrete of Flat Slab = N/mm

i) Effective shear force at face of column Veff = 1.25 X V ( As per clause 3.7.6.3 of BS:8110- Part-1-2005)

i) Effective shear force at face of column Veff = kN


uo = = mm

v max = ---------------------------

= N/mm2 < 5 N/mm2

< √fck = 5 N/mm2



u = ( Column size + 2 x 1.5 x effective depth of slab)+2 ( Column size + 1.5 x effective depth of slab)

= ( 700 + 2 x 1.5 x372)+2(700+1.5x372) = mm

mm

40

1578.8

Veff / (uod)

∏ x d 2199.11

1578750

(2199.12x372)

1.93

0.8

4332

1258mm

1258

1816mm

1.5d

1.5d

1.5d

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1861B-CS-05-00216 08/12/2009


KAMAL JSB/MDS

Veff = 1.25 X V ( As per clause 3.7.6.3 of BS:8110- Part-1-2005)

= kN

Net shear force at 1.5d from face of column =

= -{((0.45*24+5.5)*1.4+4*1.6)**/1000000}

= kN


u x d

1509.31*1000= ----------------------------

4332 x 372

= N/mm2




Veff 1578.75

Veff - {((SW+DL)*1.4+LL*1.6)*Lx*Ly/1000000}

1579Net shear force at 1.5d from face of

column Veff 1509.3

0.94

4020

% of steel = %



v > Vc (refer 3.7.7.4, BS8110-1)



v = N/mm2

Vc = N/mm2

For Case = 1.6 x0.77 = N/mm2










1.08

0.66 N/mm2

0.77

0.94

0.77

for case v<1.6vc

1.23

12

asv 113.10

420

90 0

∑Asv 731.42

1764

1764.1

Asv/asv


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KAMAL JSB/MDS


u = ( Column size + 2 x 2.25 x effective depth of slab)+2 ( Column size + 2.25 x effective depth of slab)

= ( 700 + 2 x 2.25 x372)+2(700+2.25x372)

= mm

Veff = 1.25 X V ( As per clause 3.7.6.3 of BS:8110- Part-1-2005)

5448

1537mm

2374mm


2.25d

2.25d

2.25d

= kN

=

= -{((0.45*24+5.5)*1.4+4*1.6)**/1000000}

= kN


u x d

1467.83*1000= ----------------------------

5448 x 372

= N/mm2


% of steel = %




v < Vc (refer 3.7.7.4, BS8110-1)



Veff 1578.75

Net shear force at2.25d from face of column Veff - {((SW+DL)*1.4+LL*1.6)*Lx*Ly/1000000}

1579


Veff 1467.83

0.72

4020

1.08

0.66

0.77

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Total maximum load on column V = kN( Grid H-8 column load) (Point label : 41)

Moment transmitted to column = kN-m (Unbalanced bending moment from SAFE)Max. of Mx and My is 447 kN-m





Size of the column = mm (Square)



i) Effective shear force at face of column Veff = Vt* {1+ ( As per clause 3.7.6.2 of BS:8110- Part-1-2005)

30

372

800

420

40

1 5Mt }

2830

Mx 0

My 447

450


KAMAL JSB/MDS


1861B-CS-05-00216 08/12/2009

) eff Vt {1+ ( p )Vt x Where, Mt- Design moment about x or y axis

Effective shear force at face of column Veff = kN ( As per clause 3.7.6.2 of BS:8110- Part-1-2005)


uo = = mm

v max = ---------------------------

= N/mm2 < 5 N/mm2

< √fck = 5 N/mm2




= 4x ( 800 + 2 x 1.5 x372) = mm

mm

x - length of the side of the perimeter considered parallel to axis of bendingVeff = 2830X{1+ + 0}

(2830X800/1000)(1.5X447)

0.8

7664

1916

1916

1916mm

3668.1

Veff / (uod)

4 x d

3.08

3200

3668125

1190400

1.5Mt }

1.5d

1.5d

1.5d

1.5d

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KAMAL JSB/MDS


1861B-CS-05-00216 08/12/2009

Vt* {1+Vt x

= kN

=

= -{((0.45*24+5.5)*1.4+4*1.6)**/1000000}

= kN


u x d

3068.35*1000= ----------------------------

7664 x 372

= N/mm2


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

Veff = 2830X{1+ + 0}(2830X1916/1000(1.5X447)

Veff 3068.3

1.08

8040

3180



Veff - {((SW+DL)*1.4+LL*1.6)*Lx*Ly/1000000}

3179.95


Veff = 1.5Mt }


Veff

% of steel = %



v > Vc (refer 3.7.7.4, BS8110-1)



v = N/mm2

Vc = N/mm2











3121

3121.0

Asv/asv

asv 113.10

420

90 0

∑Asv 834.69

0.97

1.08

0.97

for case v<1.6vc

1.550807

12

2.16

0.83 N/mm2

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KAMAL JSB/MDS


1861B-CS-05-00216 08/12/2009



= mm

= Vt * {1+ ( As per clause 3.7.6.2 of BS:8110- Part-1-2005)Vt x

(1 5X447)

2474mm


Veff 1.5Mt }

9896

24742.25d

2.25d

2.25d

2.25d.

= kN

=

= -{((0.45*24+5.5)*1.4+4*1.6)**/1000000}

= kN


u x d

2914.96*1000= ----------------------------

9896 x 372

= N/mm2


% of steel = %




v < Vc (refer 3.7.7.4, BS8110-1)


+ 0}(2830X2474/1000(1.5X447)

Veff - {((SW+DL)*1.4+LL*1.6)*Lx*Ly/1000000}

Veff =

3101

2914.95Net shear force at2.25d from face of column

Net shear force at2.25d from face of column

0.79

8040

2.16

0.83

0.97


Veff 3101.0

Veff

2830X{1+

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column reactions SAFE v8.1.0 File: PENT 100% IFC-R1-A KN-m Units PAGE 1 December 8,2009 14:10 C O L U M N S P R I N G R E A C T I O N S COLUMN GRID I GRID J LOAD FZ MX MY 27 146 360 201DLLL 3125.88 21.929 -516.283 28 187 360 201DLLL 3341.12 103.121 44.318 29 242 360 201DLLL 3482.27 242.866 -90.420 30 334 360 201DLLL 3213.45 -4.574 546.858 32 187 319 201DLLL 3721.77 -684.568 122.078 33 242 319 201DLLL 3432.84 -261.686 254.082 37 146 381 201DLLL 3108.99 17.364 -380.802 38 187 381 201DLLL 3509.99 -100.274 -183.408 39 242 381 201DLLL 1853.03 -2214.176 190.887 40 334 381 201DLLL 2888.75 -391.765 186.729 44 146 319 201DLLL 3263.71 -259.402 -594.068 45 334 319 201DLLL 2823.30 697.513 -201.976 12703 65 403 201DLLL 1048.77 -1037.733 -544.526 2 103 403 201DLLL 1750.15 -1219.550 -6.037 3 121 403 201DLLL 1768.52 -1190.928 -4.717 4 146 403 201DLLL 1731.65 -1163.862 22.717 5 187 403 201DLLL 1912.87 -1155.474 -177.265 6 242 403 201DLLL 2563.29 -1183.205 240.105 7 334 403 201DLLL 1632.16 -869.736 -84.454 8 368 403 201DLLL 1908.98 -1227.246 -165.732 9 422 403 201DLLL 1078.68 -881.495 896.770 10 422 381 201DLLL 1897.27 114.903 1194.250 11 422 360 201DLLL 1439.22 363.720 754.878 12 422 216 201DLLL 1940.88 208.823 1098.559 13 422 154 201DLLL 1736.81 -28.074 1178.978 14 422 123 201DLLL 1819.29 -46.446 1162.077 15 422 83 201DLLL 1443.77 337.470 281.116 16 368 83 201DLLL 2250.72 139.229 -50.821

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column reactions 21 242 154 201DLLL 2891.83 -518.330 -222.424 12808 10 360 201DLLL 829.05 220.878 -147.953 12704 28 381 201DLLL 1929.01 -52.584 -1361.145 12705 38 319 201DLLL 666.42 24.627 -139.568 25 103 360 201DLLL 2087.56 190.407 -487.001 31 368 360 201DLLL 2422.07 353.664 -228.866 34 334 123 201DLLL 2506.54 351.444 396.406 35 103 381 201DLLL 2891.90 391.224 499.701 36 121 381 201DLLL 2554.04 422.071 -3.969 41 368 381 201DLLL 2830.88 362.621 -447.198 42 103 319 201DLLL 1909.99 198.657 -620.210 43 121 319 201DLLL 2456.00 -76.756 155.574 46 368 216 201DLLL 3028.41 517.718 -70.985 47 368 154 201DLLL 2561.80 -19.266 -426.608 48 334 154 201DLLL 2704.81 266.563 277.454 19 368 123 201DLLL 2626.03 -76.232 -418.615 12809 48 293 201DLLL 557.16 117.564 -305.160 60 98 274 201DLLL 1406.59 77.261 -286.853 61 126 234 201DLLL 1863.01 744.813 3.197 64 153 181 201DLLL 1131.90 5.182 -533.954 56 146 216 201DLLL 2749.58 577.628 -446.383 66 195 142 201DLLL 2109.29 205.390 -565.669 12812 241 113 201DLLL 2055.77 1604.747 -308.640 994 368 19 201DLLL 1151.05 434.521 -153.288 2044 121 360 201DLLL 2363.97 -57.572 105.784 15074 356 19 201DLLL 763.68 24.868 109.875 14772 292 96 201DLLL 1982.77 -280.430 97.892 14771 347 77 201DLLL 1371.57 161.908 57.799 14769 340 49 201DLLL 2333.68 -232.485 521.574

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6. Punching Shear Calculation