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1.5 PUNCHING SHEAR CHECK 66 of 188

# Punching Shear Calculation

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1.5 PUNCHING SHEAR CHECK

66 of 188

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

67 of 188

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)

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

68 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

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

= 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

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

69 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

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

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

Effective shear force at 2.25d from face of column

Veff 3464.0

2.25d

2.25d

2.25d

2.25d.

= kN

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

u x d

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

9896 x 372

= N/mm2

Area of main steel given = mm2

% of steel = %

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

( 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)

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

70 of 188

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 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

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 d = mm (circuar)

Grade of Concrete of Flat Slab = 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

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

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 ( 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

71 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

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

= 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

Effective shear force at 1.5d from face of column

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

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

Effective shear force at 1.5d from face of column

Veff 3125.42

Net shear force at 1.5d from face of column

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

3125.4

Net shear force at 1.5d from face of column

Vnet 3043.9

1.13

Area of main steel given = mm2

% of steel = %

Corresponding 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

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 = = 26

Provide 26 Nos of 12 mm dia bars

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

72 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

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

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 ( 700 + 2 x 2.25 x372)

= 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)

Effective shear force at 2.25d from face of column

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

Area of main steel given = mm2

% of steel = %

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

( 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)

Punching Shear is OK, No shear reinforcement is required.

of columnNet shear force at 2.25d from face of

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

3013.2

Net shear force at 2.25d from face of column

Vnet 2873.89

0.81

8040

2.16

0.83

0.97

73 of 188

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 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

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 (Circular)

Grade of Concrete of Flat Slab = 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

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

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 = ( 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

74 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

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

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

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

u x d

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

4332 x 372

= N/mm2

Area of main steel given = mm2

Effective shear force at 1.5d from face of column

Effective shear force at 1.5d from face of column

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 = %

Corresponding 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 x0.77 = N/mm2

Diameter of Shear Links = mm

Area Of Shear Links = PI()*dia^2/4 = 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 = = 16

Provide 16 Nos of 12 mm dia bars

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

Provide 16 Nos of 12 mm dia bars

75 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

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

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

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

Effective shear force at 2.25d from face of column

2.25d

2.25d

2.25d

= kN

=

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

= kN

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

u x d

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

5448 x 372

= N/mm2

Area of main steel given = mm2

% of steel = %

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

( 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)

Punching Shear is OK, No shear reinforcement is required.

Effective shear force at 2.25d from face of column

Veff 1578.75

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

1579

Net shear force at 2.25d from face of column

Veff 1467.83

0.72

4020

1.08

0.66

0.77

76 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

Punching Shear check

Pent floor level.(450 mm slab Thickness)

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

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 Concrete of Flat Slab = N/mm2

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

TITLE: CLINIC BUILDING-PUNCHING SHEAR CHECKDESIGNED CHECKED SHEET

KAMAL JSB/MDS

PROJECT: SIDRA MEDICAL & RESEARCH CENTER, DOHADOCUMENT NO DATE

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)

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

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

77 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

TITLE: CLINIC BUILDING-PUNCHING SHEAR CHECKDESIGNED CHECKED SHEET

KAMAL JSB/MDS

PROJECT: SIDRA MEDICAL & RESEARCH CENTER, DOHADOCUMENT NO DATE

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

Vt* {1+Vt x

= kN

=

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

= kN

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

u x d

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

7664 x 372

= N/mm2

Area of main steel given = 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

Net shear force at 1.5d from face of column

Net shear force at 1.5d from face of column

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

3179.95

Effective shear force at 1.5d from face of column

Veff = 1.5Mt }

Effective shear force at 1.5d from face of column

Veff

% of steel = %

Corresponding 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

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

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

78 of 188

LARSEN & TOUBRO LIMITED GLOBAL ENGINERING SERVICES

TITLE: CLINIC BUILDING-PUNCHING SHEAR CHECKDESIGNED CHECKED SHEET

KAMAL JSB/MDS

PROJECT: SIDRA MEDICAL & RESEARCH CENTER, DOHADOCUMENT NO DATE

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

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

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

(1 5X447)

2474mm

Effective shear force at 2.25d from face of column

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

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

u x d

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

9896 x 372

= N/mm2

Area of main steel given = mm2

% of steel = %

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

( 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)

Punching Shear is OK, No shear reinforcement is required.

+ 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

Effective shear force at 2.25d from face of column

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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