Column Design

LARSEN & TOUBRO LIMITED ECC Division - EDRC

PROJECT: BSL COKE OVEN PLANT NO. 2 DOCUMENT NO DATE

O9017-10-C-CK-005-001 Rev-0 29/4/2010

TITLE: BATTERY PROPER-1DESIGNED CHECKED SHEET

UNIT: ANALYSIS & DESIGN OF END PLATFORM PJM SUR OF

ANALYSIS OF SUMP WALL:-

Wall is subjected to following horizontal pressure:i) Soil Pressureii) Surcharge Pressureiii) Water Pressure

i) Soil Pressure:Unit Weight of soil = γ= 18Ground Water Table is at EL(-)0.840m

Therefore, 18-10 = 8Co-efficient of earth pressure at rest= 0.5

0.5 × 8 × 8.16 = 32.64[i.e. EL(-)9.0 m ]

ii) Surcharge Pressure:Surcharge load on ground= q = 20 kN/m (Assume)

0.5 × 20 = 10iii) Water Pressure:

10 × 8.16 = 10[i.e. EL(-)9.0 m ] = 81.6

EL(-)0.840m

4

EL(-)4.5 m (T.O.Raft)

m 1.5

8

EL(-)7.8 m1.5

81.6 10 32.64 EL(-)9.0 m

Water Pressure at EL(-)7.8 m = 81.6 × (8.16-1.5)/8.16= 66.6Soil Pressure at EL(-)7.8 m = 32.64 × (8.16-1.5)/8.16= 26.64

Water Pressure at EL(-)4.5 m = 10 × 3.66 = 36.6

Soil Pressure at EL(-)4.5 m = 32.64 × (8.16-4.5)/8.16= 14.64

kN/m3

γsub= kN/m3

k0=

Soil Pressure at base= k0.γsub.h= kN/m2

Surcharge Pressure = k0.q= kN/m2

Water Pressure at Base = γw.h = kN/m2 γw kN/m3

kN/m2

kN/m2 kN/m2 kN/m2

(Water Pressure)

(Surcharge Pressure)

(Submerged Soil Pressure)

kN/m2

kN/m2

kN/m2

kN/m2



O9017-10-C-CK-005-001 Rev-0 29/4/2010



DESIGN OF WALL:-

Y2.2 m

a a Hinge

b

a = 1.1 mb = 4.5 m

X61.2 63 O

(Load 1) (Load 2) (Plate fixed along three edges, hinged along one edge)

(Refer Moody's chart for moments and reactions for rectangular plates.)Load 1, Uniform Load (Page 16, Fig.- 10 )of Moody's Chart

a/b= 1.1/4.5= 0.24444 Use a/b = 3/8 =0.375 Chart

Moment =Reaction=

Mxy/b x/a 0 0.2 0.4 0.6 0.8 1

a/b

= 3

/8

-0.0466 0.0061 0.1237 0.1983 0.2397 0.253 1 -0.0466 0 0 0 0 0 0

0.8 0.3265 0.0301 0.0109 -0.0013 -0.0089 -0.013 -0.01430.6 0.3819 0.0403 0.0161 -0.0007 -0.0116 -0.0177 -0.01970.4 0.3624 0.0374 0.0148 -0.0008 -0.0108 -0.0163 -0.01810.2 0.2164 0.0211 0.0075 -0.001 -0.0059 -0.0085 -0.00920 -0.0013 0 0.001 0.0027 0.0043 0.0054 0.0058

Ry -0.0013 0.031 0.1684 0.2663 0.3238 0.3427My

0 0 0 0 0 00.006 0.0008 -0.0038 -0.0073 -0.0095 -0.01020.0081 0.0023 -0.0028 -0.0068 -0.0094 -0.01030.0075 0.002 -0.0028 -0.0065 -0.0089 -0.0097

kN/m2 kN/m2

(Co-efficient)(P.b2)(Co-efficient)(P.b)

Ry2

Rx Rx

Ry2

RxRy



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0.0042 0.0008 -0.0019 -0.0038 -0.0048 -0.00520 0.0051 0.0135 0.0215 0.0269 0.0288

Moody's Chart for uniform load

y/b x/a

1 -12.842 0 -12.842 -0.3582540.8 89.9769 0.2 1.68104 8.542980.6 105.244 0.4 34.0892 46.4076720.4 99.8702 0.6 54.6475 73.3869540.2 59.6355 0.8 66.0565 89.2328040 -0.3583 1 69.7217 94.441266

YHorizontal B.M. (Mx) due to load 1:Max. (+)ve B.M.= 49.9764 kN-m MxMax. (-)ve B.M.= -24.43 kN-m

Vertical B.M. (My) due to load 1:Max. (+)ve B.M.= 35.7152 kN-mMax. (-)ve B.M.= -12.773 kN-m

Mx= 49.9764kN-m

MyX

My= 35.7152 kN-mLoad 2, Uniformly Varying Load (Page 19, Fig.- 13) of Moody's Chart:

Mxy/b x/a 0 0.2 0.4 0.6 0.8 1

a/b

= 3

/8

-0.0205 -0.0283 0.007 0.0313 0.0455 0.05021 -0.0205 0 0 0 0 0 0

0.8 0.0788 0.0091 0.0039 0.0001 -0.0026 -0.0041 -0.00460.6 0.155 0.017 0.007 -0.0001 -0.0048 -0.0075 -0.00840.4 0.212 0.0207 0.0078 -0.0008 -0.0061 -0.0089 -0.00980.2 0.1696 0.0145 0.0045 -0.0012 -0.0042 -0.0057 -0.00610 0.0102 0 0.0008 0.002 0.003 0.0038 0.004

Ry 0.0102 0.0475 0.149 0.2157 0.2528 0.2647My

0 0 0 0 0 00.0018 0.0007 -0.0003 -0.001 -0.0015 -0.00160.0034 0.0011 -0.0009 -0.0025 -0.0035 -0.0038

RxRy1

(Top)Ry2 (Bott.)

ie Raft

RxRy

Mx= -24.4 kN-m

My=-12.77kN-m



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0.0041 0.0009 -0.0021 -0.0044 -0.0058 -0.00630.0029 0.0001 -0.0023 -0.0039 -0.0049 -0.0052

0 0.0039 0.0099 0.0153 0.0188 0.02

Moody's Chart for Uniformly Varying load

y/b x/a

1 -5.8117 0 -5.8117 2.89170.8 22.3398 0.2 -8.023 13.466250.6 43.9425 0.4 1.9845 42.24150.4 60.102 0.6 8.87355 61.150950.2 48.0816 0.8 12.8992 71.66880 2.8917 1 14.2317 75.04245

YHorizontal B.M. (Mx) due to load 2:Max. (+)ve B.M.= 26.408 kN-mMax. (-)ve B.M.= -12.502 kN-m Mx

Vertical B.M. (My) due to load 2:Max. (+)ve B.M.= 25.515 kN-mMax. (-)ve B.M.= -8.0372 kN-m

Mx= 26.408kN-m

MyX

My= 25.515 kN-m

TOTAL MOMENTS ON WALL:

Horizontal B.M. (Mx) due to load 1 and load 2:Max. (+)ve B.M.= 76.3845 kN-m (Outer wall i.e, Soil face)Max. (-)ve B.M.= -36.933 kN-m (Inner wall)

Vertical B.M. (My) due to load 1 and load 2:Max. (+)ve B.M.= 61.2302 kN-m (Outer wall i.e, Soil face)Max. (-)ve B.M.= -20.81 kN-m (Inner wall)

Max. Reaction on wall due to load 1 and load 2= 169.483716 kN

RxRy1

(Top)Ry2 (Bott.)

ie Raft

Mx= -12.5 kN-m

My=-8.037kN-m



O9017-10-C-CK-005-001 Rev-0 29/4/2010



Provide Wall Thickness= D= 450 mm Dia of each Bar = 16 mmConcrete Grade: M30 Steel Grade: Fe500 D= 450 mm b= 1000 mmCover taken= 50 mm d= 392 mm d'= 58 mm

Max. (+) ve B.M. for outer wall= 76.3845 kN-m

0.74563

0.178 Ast reqd.= 801Providing 16 mmΦ bars, Spacing required= 251 mmHence, Provide 16 mmΦ bars @ 200 mm C/C at outer face of wall

Ast provided= 1005 0.2234

0.3475 136.22 kNShear force at critical section = 79.939567 kNVu = 120 kN < 136.22 kNHence, No Shear Reinforcement Required. Hence the section is safe

Max. (-) ve B.M. for inner wall= 36.9325 kN-m

0.36052

0.12 Ast reqd.= 540Providing 12 mmΦ bars, Spacing required= 209 mmHence, Provide 12 mmΦ bars @ 200 mm C/C at inner face of wall

DESIGN OF BASE SLAB:-Y

2 ma

b

a = 2 mb = 2.2 m

X120 O

(Load 1) (Plate fixed along three four edges,)

(Refer Moody's chart for moments and reactions for rectangular plates.)

Mu/bd2=

pt reqd = mm2

mm2 pt provided=

τc = N/mm2 τcbd =

Mu/bd2=

Min. pt= mm2

kN/m2

Ry2

Rx Rx

Ry2



O9017-10-C-CK-005-001 Rev-0 29/4/2010



Load 1, Uniform Load (Page 40, Fig.- 34 )of Moody's Chart

a/b= 2/2.2= 0.90909 Use a/b = 1 Chart

Moment =Reaction= (Co-efficient)(P.a)

Mxy/b x/a 0 0.05 0.1 0.2 0.3 0.4 0.5

a/b

= 1

-0.0377 -0.0391 0.0503 0.2341 0.3608 0.4319 0.4546 0.5 0.4389 0.0500 0.0306 0.0156 -0.004 -0.0147 -0.0198 -0.02130.4 0.4189 0.047 0.0286 0.0144 -0.0039 -0.0137 -0.0184 -0.01970.3 0.3551 0.0382 0.0229 0.0112 -0.0033 -0.0109 -0.0143 -0.01530.2 0.2373 0.0244 0.0143 0.0062 -0.002 -0.0061 -0.0078 -0.00820.1 0.0585 0.0082 0.052 0.0028 0.0007 0.0003 0.0006 0.0007

0.05 -0.0316 0.0024 0.0021 0.0018 0.0026 0.004 0.005 0.00540 -0.0351 0 0.0005 0.0016 0.0049 0.0076 0.0094 0.01

Ry -0.0351 -0.0316 0.0585 0.2373 0.3551 0.4189 0.4389My

0.01 0.0054 0.0007 -0.0082 -0.0153 -0.0197 -0.02130.0094 0.005 0.0006 -0.0078 -0.0143 -0.0184 -0.01980.0076 0.004 0.0003 -0.0061 -0.0109 -0.0137 -0.01470.0049 0.0026 0.0007 -0.002 -0.0033 -0.0039 -0.0040.0016 0.0018 0.0028 0.0068 0.0112 0.0144 0.01560.0005 0.0021 0.0052 0.0143 0.0229 0.0286 0.0306

0 0.0024 0.0082 0.0244 0.0382 0.047 0.05Moody's Chart for uniform load

y/b x/a

0.5 105.336 0 -9.048 -8.4240.4 100.536 0.05 -9.384 -7.5840.3 85.224 0.1 12.072 14.040.2 62.6472 0.2 56.184 56.9520.1 14.04 0.3 86.592 85.224

0.05 -7.584 0.4 103.656 100.5360 -8.424 0.5 109.104 105.336

YHorizontal B.M. (Mx) due to load 1:Max. (+)ve B.M.= 24.96 kN-m MxMax. (-)ve B.M.= -10.224 kN-m

(Co-efficient)(P.a2)

Rx Ry1 Ry2

RxRy

Mx= -10.2 kN-m

My=-10.2kN-m



O9017-10-C-CK-005-001 Rev-0 29/4/2010



Vertical B.M. (My) due to load 1:Max. (+)ve B.M.= 24 kN-m Mx= 24.96Max. (-)ve B.M.= -10.224 kN-m kN-m

My= 24 kN-m My X

Max. Reaction on wall due to load 1 = 109.104 kNProvide Slab Thickness= D= 450 mm Dia of each Bar = 16 mmConcrete Grade: M30 Steel Grade: Fe500 D= 450 mm b= 1000 mmCover taken= 50 mm d= 392 mm d'= 58 mm

Max. (+) ve B.M. for outer face= 24.96 kN-m

0.24365

0.22 Ast reqd.= 990Providing 16 mmΦ bars, Spacing required= 203 mmHence, Provide 16 mmΦ bars @ 200 mm C/C at bottom of slab bothways


0.3475 136.22 kNShear force = 109.104 kNShear force at d dist from face = 62.064Vu = 93.1 kN < 136.22 kNHence, No Shear Reinforcement Required. Hence the section is safe

Max. (-) ve B.M. for inner face= 10.224 kN-m

0.0998

0.12 Ast reqd.= 540Providing 12 mmΦ bars, Spacing required= 209 mmHence, Provide 12 mmΦ bars @ 200 mm C/C at top of slab bothways

Mu/bd2=

Provide pt = mm2

mm2 pt provided=

τc = N/mm2 τcbd =

Mu/bd2=

Min. pt= mm2

Mx= -10.2 kN-m

My=-10.2kN-m



O9017-10-C-CK-005-001 Rev-0 29/4/2010





O9017-10-C-CK-005-001 Rev-0 29/4/2010



DESIGN OF STAIR CASE:-

Size of each Tread= 300 mm Size of each Riser= 150 mm

Dead Load for inclined portion:

i) Tread and Riser= 1/2×0.3×0.15 = 0.0225

ii) Waist Slab(200 thk)= 0.20×0.335 = 0.067

iii) Finish (40 thk.)= (0.3+0.15)×0.04 = 0.018Total = 0.1075

Dead Load on Plan Area= 0.1075×25/0.3 = 8.95833≈ 10

Live Load on Plan Area= = 5

Dead Load for landing portion:i) s/w of slab (200 thk)= 0.2×25 = 5

ii) Finish (40mm thk)= 0.04×24 = 0.96iii) Plaster = 0.006×24 = 0.144

Total = 6.104

Live Load = = 5

DL= 10DL= 6.104 LL= 5 DL= 6.104LL= 5 LL= 5

1.2 m A 3 m 1.25 m B

DL= 45.4 kN DL= 25.3 kNLL= 17.5 kN LL= 9.8 kN

m2

m2

m2

m2

kN/m2

kN/m2

kN/m2

kN/m2

kN/m2

kN/m2

kN/m2

kN/m2

kN/m2

kN/m2 kN/m2 kN/m2

kN/m2 kN/m2



O9017-10-C-CK-005-001 Rev-0 29/4/2010



Maximum Span B.M= 37.3 kN-m (L/C:- DL+LL)Maximum Support B.M= 11.4 kN-m (L/C:- DL+LL)Maximum Shear= 43.85 kN-m (L/C:- DL+LL)

Overall Depth, D= 200 mm, Clear Cover= 20 mm, Dia. of Bar = 16 mmEff. Depth, d= 172 mm, Grade of Concrete= M 30 Grade of Steel= Fe 500Considering Width, b= 1000 mmMain Reinforcement :Design for Bending Moment = 37.3 kN-m

1.89

0.475 Ast reqd.= 950

Providing 16 mmΦ bars, Spacing required= 212 mmHence, Provide 16 mmΦ bars @ 200 mm C/C at top and bottom as main reinforcement


0.5 86 kNVu = 65.8 > VuHence, No Shear Reinforcement Required.Distribution Reinforcement :

0.12 Ast reqd.= 240

Providing 10 mmΦ bars, Spacing required= 327 mmHence, Provide 10 mmΦ bars @ 200 mm C/C at top and bottom as Distribution reinforcement

Check for Deflection:Span (l) = 4.25 m Eff. Depth(d) = 172 mmHence, Span (l) and Effective Depth (d) ratio(l/d) = 24.7093As per IS 456:2000, Fig. 4

274.045Modification Factor= 1.1As per clause 23.2.1a) of IS 456:2000 allowable ratio for Span to effective depth of continuous slab = 1.1×26 = 28.6 > 24.7093 Hence, the section is Safe

Landing Beam:

At A, Landing Beam mkd. LB-2:Reaction= 45.4+17.5 = 62.9 kN/m

62.9 kN/m

3.1 m97.495 kN 97.495 kN

(Beam size 300mm×400mm)

Mu/bd2=

pt reqd.= mm2

mm2 Pt provided=

τc = N/mm2 τcbd =τcbd

pt reqd.= mm2

fs= 0.58 fy (Ast reqd./Ast provided)=



O9017-10-C-CK-005-001 Rev-0 29/4/2010



Concrete Grade: M30 Steel Grade: Fe500 D= 400 mm b= 300 mmCover taken= 25 mm d= 365 mm d'= 35 mm

d'/d ≈ 0.1

Maximum Bending Moment= 62.9×3.1×3.1/ 8 kN-m = 75.5586 kN-m2.83575

0.750 Ast reqd. = 821

Hence, Provide 3 nos 28 mmΦ at bottomand Provide 2 nos 25 mmΦ and 1 nos 20 at topAst provided= 1296 1.18348

0.7 76.65 kNVus/d= 1.90664 kN/cmHence, Provide 2 legged 10 mmΦ@ 200 C/C as Shear Reinforcement

35.1 kN/m

At B, Landing Beam mkd. LB-1: 1.91 1.19Load W= 25.3 + 9.8 = 35.1 kN/m(Beam size 300mm×400mm)Concrete Grade: M30 Steel Grade: Fe500 D= 400 mm b= 300 mmCover taken= 25 mm d= 365 mm d'= 35 mm

d'/d ≈ 0.1

Maximum Bending Moment = 35.1×1.91×1.91/ 2 kN-m = 64.0242 kN-m

Maximum Shear = 35.1×1.91/ 2 kN = 33.5205 kN

2.40286

0.620 Ast reqd. = 679

Hence, Provide 3 nos 25 mmΦ at topand Provide 3 nos 20 mmΦ at bottom


0.73 79.935 kNVus/d= No shear reinforcement reqdHence, Provide 2-L-8mmΦ@200 C/C as Shear Reinforcement

Mu/bd2=

Pt required= mm2

mm2 Pt provided=

τc = N/mm2 τcbd =

Mu/bd2=

Pt required= mm2

mm2 Pt provided=

τc = N/mm2 τcbd =



O9017-10-C-CK-005-001 Rev-0 29/4/2010



LARSEN & TOUBRO LIMITED ECC Division - EDRC, Kolkata

PROJECT: BSL Coke Oven Plant No 2DOCUMENT NO. Rev Date 08.06.10

O9017-10-C-CK-005-001 A SheetUNIT: Battery Proper 1 DESIGNED CHECKED APPROVED

OF TITLE: Analysis & Design of End Platform PJM SUR KCH

NET BASE PRESSURE:

From STAAD Analysis results, the Net Base Pressure Summary for the Critical load Conditions are as follows.

for Base Pressure Diagram refer Annexure-A

Horizontal Vertical Horizontal

Node L/C Fx MTon/m2 Fy MTon/m2 Fz MTon/m2

Max Px 5 101 DL + LL + SOIL PR. LOAD 0 0 0

Min Px 5 101 DL + LL + SOIL PR. LOAD 0 0 0

Max Py 6423 157 DL + LL + SL+X 0 11.529 0

Min Py 6267 115 DL + SOIL PR. LOAD + WIND LOAD IN (+)Z-DIR 0 -0.375 0

Max Pz 5 101 DL + LL + SOIL PR. LOAD 0 0 0

Min Pz 5 101 DL + LL + SOIL PR. LOAD 0 0 0

< 30 T/m2

The Base Raft has been designed separately for strip below grid mkd P1/1 and P1/4 for which maximum Moments at

face of column and Max. shear at d dist from the Face of column are as follows (from STAAD Analysis Results)

Plate L/C

Max Mx 12178 265 89.167 -21.039 209.159 134.178 -17.489Min Mx 12178 31 -33.202 9.561 -71.498 -35.519 14.148Max My 10487 265 89.44 -17.535 158.684 150.459 11.013Min My 12124 31 -32.75 -9.277 -70.795 -37.291 10.158

Max abs(SQx) 12210 271 -48.101 -9.174 81.137 99.436 -5.065Max abs(SQy) 10317 271 -0.896 37.02 122.874 37.43 -12.757

Maximum Moments at face of column and Max. shear at d dist from the Face of column at areas other than strip

below P1/1 & P1/4 are as follows (from STAAD Analysis Results)

Plate L/C

Max Mx 10902 208 -23.057 70.732 164.806 100.773 0.586Min Mx 10052 207 -34.437 -22.409 -46.835 -11.585 -32.691Max My 10509 207 -56.432 -7.125 118.958 122.381 -16.245Min My 11135 302 -19.982 -39.846 -15.136 -135.799 -38.657

Max abs(SQx) 12191 278 -34.515 -7.194 58.512 60.924 8.454Max abs(SQy) 10820 285 0.436 -36.285 0.442 26.382 -17.202

SQX MTon/m2

SQY MTon/m2

MX MTon-m/m

MY MTon-m/m

MXY MTon-m/m

SQX MTon/m2

SQY MTon/m2

MX MTon-m/m

MY MTon-m/m

MXY MTon-m/m





Design of base raft for strip below grid mkd P1/1 & P1/4Factored Moments & Shear Forces

Grade of Concrete = 30 N/mm2 Max (+VE) Mx = 209.159 t-m/m

Grade of Steel = 500 N/mm2 Max (-VE) Mx = -71.498 t-m/m

Max (+VE) My = 150.459 t-m/m

Overall depth of section = 1500 mm Max (-VE) My = -37.291 t-m/m

Clear cover = 50 mm Max Qx = 48.101 t/m2Direction of main reinforcement = Local X mm Max Qy = 37.02 t/m2

Reinforcement along local XOn +Ve Local Z face Bar dia = 32 mm Spacing = 100 mmOn -Ve Local Z face Bar dia = 28 mm Spacing = 200 mm

Reinforcement along local YOn +Ve Local Z face Bar dia = 32 mm Spacing = 200 mmOn -Ve Local Z face Bar dia = 28 mm Spacing = 200 mm

Calculation of reinforcement along local XOn +Ve Local Z faceMaximum +ve moment (factored) = 209.159 t-m/m

Effective depth = 1434 mm

Mu/bd2 = 209.159 x 10000 / ( 1434 x 1434 ) = 1.017

for which pt = 0.244

Area of steel required = 3499

Minimum steel required = 1800

Area of steel provided = 8042 > Ast required, Hence O.K.

On -Ve Local Z faceMaximum -ve moment (factored) = 71.498 t-m/m


Mu/bd2 = 71.498 x 10000 / ( 1436 x 1436 ) = 0.347for which pt = 0.081




Check for Shear :Maximum factored shear force = 48.101 x 1.5 = 72.1515 t/m

touv = 72.1515 x 10000 / ( 1000 x 1434 ) = 0.503 N/mm2100 Ast/bd = ( 100 x 8042 ) / ( 1000 x 1434 ) = 0.561For which touc = 0.521

mm2/m width

mm2/m width

mm2/m width

mm2/m width

mm2/m width

mm2/m width





For overall depth of slab of 1500 mm value of k = 1Design shear strength = 1 x 0.521 = 0.521 N/mm2 > touv, Hence O.K.

Calculation of reinforcement along local YOn +Ve Local Z faceMaximum +ve moment (factored) = 150.459 t-m/m


Mu/bd2 = 150.459 x 10000 / ( 1402 x 1402 ) = 0.765for which pt = 0.181




On -Ve Local Z face

Maximum -ve moment (factored) = 37.291 t-m/m


Mu/bd2 = 37.291 x 10000 / ( 1408 x 1408 ) = 0.188

for which pt = 0.044




Check for Shear :

Maximum factored shear force = 37.02 x 1.5 = 55.53 t/m

touv = 55.53 x 10000 / ( 1000 x 1402 ) = 0.39

100 Ast/bd = ( 100 x 4021 ) / ( 1000 x 1402 ) = 0.287

For which touc = 0.392

For overall depth of slab of 1500 mm value of k = 1

Design shear strength = 1 x 0.392 = 0.392 N/mm2 > touv, Hence O.K.

mm2/m width

mm2/m width

mm2/m width

mm2/m width

mm2/m width

mm2/m width

N/mm2

Design of Column mkd. C1 for Maximum Axial Load Case : ( STAAD Member No's: 6238,6514,6562,6613,6661)

Column Size = 1200 mm × 650 mm Slenderness Ratio= L/D OR L/b

b = 1200 mm ( Local Y Dir.) = 3.85 < 12

D = 650 mm ( Local Z Dir.) Hence, this is a short column

L = 2500 mm

Eccentricity(Y) = L/500 + b/30 = 45 mm Eccentricity(Z) = L/500 + D/30 = 26.67 mm

Grade of Concrete = M 25 & Grade of Steel = Fe 500 Clear Cover = 40 mmDiameter of Bar= 20 mm Diameter of Tie = 8 mm

-19.61 kN (Member No: 6562 ,L/C : 107) p = 0.8

16.67 kN-m 25

-0.52 kN-m (Due to Eccentricity) d'= 50 mm

253.02 kN-m

-0.88 kN-m (Due to Eccentricity) Z

Additional Moment due to Slenderness :

0.00 kN-m 1200

0.00 kN-m

Hence, Final Design Parameter

-19.61 kN Y 650 Y

16.67 kN-m

253.02 kN-m

ZMoment Carrying Capacity in ZZ direction

0.032

d'/b= 0.042

Refer Chart 48 of SP-16

-0.001

0.05 (value from chart)

0.05 × 25 × 650 × 1200 × 1200 = 1170.00 > 253.02 kN-m

Hence Safe

Moment Carrying Capacity in YY direction

0.032

d'/D= 0.077


-0.001


0.05 × 25 × 1200 × 650 × 650 = 633.75 > 16.67 kN-m

Hence Safe

= {0.45 × 25 × ( 1 - 0.008 ) × 1200 × 650 } + { 0.75 × 500 × 0.008 × 650 × 1200 } ≤ 0.2 1

= 11044.8 kN ≥ 0.8 2

-19.61 / 11044.8 = -0.002 1.00

0.24 < 1 Hence, O.K

= ( 0.8 × 650 × 1200 ) / 100 Ast prov. = 6283.185 Hence, O.K

= 6240

Calculation of number of bars and spacing of bars

Diametre of bar used = 20 Ф

Area of single bar = ( 0.785 × 20 × 20 )

= 314.0

Number of Bars required = ( 6240 / 314 )

= 20 #

Provide 20 Nos. 20 Ф

PU =

MUY = fck= N/mm2

MUY=

MUZ =

MUZ =

MUy =

MUz =

MZ

PU =

MUY = MY

MUZ =

p/fck =

PU / ( fck.b.D) =

MUZ1 / (fck.D.b2) =

MUZ1 =

p/fck =

PU / ( fck.b.D) =

MUY1 / (fck.b.D2) =

MUY1 =

PUZ = 0.45 × fck × Ac + 0.75 × fy × Asc PU/PUZ αn

PU / PUZ = αn =

(MUY/ MUY1)αn + (MUZ/ MUZ1)αn =

Ast reqd. mm2

mm2

mm2

Calculation of diameter of Lateral Tie and Pitch

(i) Pitch of lateral tie : Min of Below Three conditions (ii) Diameter of lateral tie

* Least lateral dimension = 650 mm Dia of lateral tie should not be less than 1 / 4 × 20

* 16 times smallest dia. of RFT = ( 16 × 20 ) mm = 5 mm

= 320 mm

* Should not be more than = 300 mm

Hence Provide 20 nos of 20 Ф as Main Reinf. for column and Lateral Ties 8 mm Ф at Spacing 250 mm C/C

Design of Column mkd. C1 for Maximum Muy Case


b = 600 mm ( Local Y Dir.) = 5.56 < 12


L = 2500 mm

Eccentricity(Y) = L/500 + b/30 = 25 mm Eccentricity(Z) = L/500 + D/30 = 20.00 mm

Grade of Concrete = M 30 & Grade of Steel = Fe 500 Clear Cover = 40 mm

Diameter of Bar= 25 mm Diameter of Tie = 8 mm

350.72 kN (Member No: 6613 ,L/C : 110) p = 2

204.54 kN-m 30

7.01 kN-m (Due to Eccentricity) d'= 52.5 mm

21.77 kN-m

8.77 kN-m (Due to Eccentricity) Z


0.00 kN-m 600

0.00 kN-m


PU =

MUY = fck= N/mm2

MUY=

MUZ =

MUZ =

MUy =

MUz =

MZ

350.72 kN Y 450 Y

204.54 kN-m

21.77 kN-m

Z

Moment Carrying Capacity in ZZ direction

0.067

d'/b= 0.088


0.043


0.125 × 30 × 450 × 600 × 600 = 607.50 > 21.77 kN-m

Hence Safe


0.032

d'/D= 0.117


0.043


0.11 × 30 × 600 × 450 × 450 = 400.95 > 204.54 kN-m

Hence Safe

= {0.45 × 30 × ( 1 - 0.02 ) × 600 × 450 } + { 0.75 × 500 × 0.02 × 450 × 600 } ≤ 0.2 1

= 5597.1 kN ≥ 0.8 2

350.72 / 5597.1 = 0.063 1.00

0.55 < 1 Hence, O.K

= ( 2 × 450 × 600 ) / 100 Ast prov. = 5890.486 Hence, O.K

= 5400




= 490.6

Number of Bars required = ( 5400 / 490.625 )

= 11 #





* 16 times smallest dia. of RFT = ( 16 × 25 ) mm = 6.25 mm

= 400 mm



PU =

MUY = MY

MUZ =

p/fck =

PU / ( fck.b.D) =

MUZ1 / (fck.D.b2) =

MUZ1 =

p/fck =

PU / ( fck.b.D) =

MUY1 / (fck.b.D2) =

MUY1 =


PU / PUZ = αn =


Ast reqd. mm2

mm2

mm2

Design of Column mkd. C1 for Maximum Muz Case


b = 600 mm ( Local Y Dir.) = 5.56 < 12


L = 2500 mm

Eccentricity(Y) = L/500 + b/30 = 25 mm Eccentricity(Z) = L/500 + D/30 = 20 mm



612.65 kN (Member No: 6562 ,L/C : 108) p = 2

6.49 kN-m 30


138.57 kN-m



0.00 kN-m 600

0.00 kN-m


612.65 kN Y 450 Y

12.26 kN-m

138.57 kN-m

Z


0.067

d'/b= 0.088


0.076


0.12 × 30 × 450 × 600 × 600 = 583.20 > 138.57 kN-m

Hence Safe


0.067

d'/D= 0.117


0.076

PU =

MUY = fck= N/mm2

MUY=

MUZ =

MUZ =

MUy =

MUz =

MZ

PU =

MUY = MY

MUZ =

p/fck =

PU / ( fck.b.D) =

MUZ1 / (fck.D.b2) =

MUZ1 =

p/fck =

PU / ( fck.b.D) =


0.11 × 30 × 600 × 450 × 450 = 400.95 > 12.26 kN-m

Hence Safe

= {0.45 × 30 × ( 1 - 0.02 ) × 600 × 450 } + { 0.75 × 500 × 0.02 × 450 × 600 } ≤ 0.2 1

= 5597.1 kN ≥ 0.8 2

612.65 / 5597.1 = 0.109 1.00

0.27 < 1 Hence, O.K

= ( 2 × 450 × 600 ) / 100 Ast prov. = 5890.486 Hence, O.K

= 5400



Area of single bar = ( 0.785 × 2.5 × 2.5 )

= 490.6


= 11 #






= 400 mm



MUY1 / (fck.b.D2) =

MUY1 =


PU / PUZ = αn =


Ast reqd. mm2

mm2

cm2

Design of Column mkd. C2 for Maximum Axial Load Case ( STAAD Member No's: 6229, 6513, 6561, 6612, 6660)


b = 600 mm ( Local Y Dir.) = 8.14 < 12


L = 3665 mm

Eccentricity(Y) = L/500 + b/30 = 27.33 mm Eccentricity(Z) = L/500 + D/30 = 22.33 mm

Grade of Concrete = M 30 & Grade of Steel = Fe 500 Clear Cover = 40 mmDiameter of Bar= 25 mm Diameter of Tie = 8 mm

1051.04 kN (Member : 6612, L/C : 106) p = 2

19.26 kN-m 30


87.55 kN-m



0.00 kN-m 600

0.00 kN-m


1051.04 kN Y 450 Y

23.47 kN-m

87.55 kN-m

ZMoment Carrying Capacity in ZZ direction

0.067

d'/b= 0.088


0.130


0.12 × 30 × 450 × 600 × 600 = 583.20 > 87.55 kN-m

Hence Safe


0.067

d'/D= 0.117


0.130


0.11 × 30 × 600 × 450 × 450 = 400.95 > 23.47 kN-m

Hence Safe

= {0.45 × 30 × ( 1 - 0.02 ) × 600 × 450 } + { 0.75 × 500 × 0.02 × 450 × 600 } ≤ 0.2 1

= 5597.1 kN ≥ 0.8 2

1051.04 / 5597.1 = 0.188 1.00

0.21 < 1 Hence, O.K

= ( 2 × 450 × 600 ) / 100 Ast prov. = 5890.486 Hence, O.K

= 5400

PU =

MUY = fck= N/mm2

MUY=

MUZ =

MUZ =

MUy =

MUz =

MZ

PU =

MUY = MY

MUZ =

p/fck =

PU / ( fck.b.D) =

MUZ1 / (fck.D.b2) =

MUZ1 =

p/fck =

PU / ( fck.b.D) =

MUY1 / (fck.b.D2) =

MUY1 =


PU / PUZ = αn =


Ast reqd. mm2

mm2




= 490.6


= 11 #






= 400 mm



Design of Column mkd. C2 for Maximum Muy Case


b = 600 mm ( Local Y Dir.) = 8.14 < 12


L = 3665 mm

Eccentricity(Y) = L/500 + b/30 = 27.33 mm Eccentricity(Z) = L/500 + D/30 = 22.33 mm


mm2


591.27 kN (Member : 6612, L/C : 110) p = 2

222.19 kN-m 30


27.29 kN-m



0.00 kN-m 600

0.00 kN-m


591.27 kN Y 450 Y

222.19 kN-m

27.29 kN-m

Z


0.067

d'/b= 0.088


0.073


0.125 × 30 × 450 × 600 × 600 = 607.50 > 27.29 kN-m

Hence Safe


0.067

d'/D= 0.117


0.073


0.115 × 30 × 600 × 450 × 450 = 419.175 > 222.19 kN-m

Hence Safe

= {0.45 × 30 × ( 1 - 0.02 ) × 600 × 450 } + { 0.75 × 500 × 0.02 × 450 × 600 } ≤ 0.2 1

= 5597.1 kN ≥ 0.8 2

591.27 / 5597.1 = 0.106 1.00

0.57 < 1 Hence, O.K

= ( 2 × 450 × 600 ) / 100 Ast prov. = 5890.486 Hence, O.K

= 5400




= 490.6


= 11 #






= 400 mm


PU =

MUY = fck= N/mm2

MUY=

MUZ =

MUZ =

MUy =

MUz =

MZ

PU =

MUY = MY

MUZ =

p/fck =

PU / ( fck.b.D) =

MUZ1 / (fck.D.b2) =

MUZ1 =

p/fck =

PU / ( fck.b.D) =

MUY1 / (fck.b.D2) =

MUY1 =


PU / PUZ = αn =


Ast reqd. mm2

mm2

mm2


Design of Column mkd. C2 for Maximum Muz Case


b = 600 mm ( Local Y Dir.) = 12.39 > 12

D = 450 mm ( Local Z Dir.) Hence, this is a long column

L = 5576 mm

Eccentricity(Y) = L/500 + b/30 = 31.152 mm Eccentricity(Z) = L/500 + D/30 = 26 mm



420.66 kN (Member : 6561, L/C : 117) p = 1.9

7.11 kN-m 30


309.53 kN-m



14.53 kN-m 600

19.38 kN-m


420.66 kN Y 450 Y

25.54 kN-m

328.91 kN-m

Z


PU =

MUY = fck= N/mm2

MUY=

MUZ =

MUZ =

MUy =

MUz =

MZ

PU =

MUY = MY

MUZ =

0.063

d'/b= 0.088


0.052


0.11 × 30 × 450 × 600 × 600 = 534.60 > 328.91 kN-m

Hence Safe


0.063

d'/D= 0.117


0.052


0.105 × 30 × 600 × 450 × 450 = 382.725 > 25.54 kN-m

Hence Safe

= {0.45 × 30 × ( 1 - 0.019 ) × 600 × 450 } + { 0.75 × 500 × 0.019 × 450 × 600 } ≤ 0.2 1

= 5499.495 kN ≥ 0.8 2

420.66 / 5499.495 = 0.076 1.00

0.68 < 1 Hence, O.K

= ( 1.9 × 450 × 600 ) / 100 Ast prov. = 5890.486 Hence, O.K

= 5130



Area of single bar = ( 0.785 × 2.5 × 2.5 )

= 490.6


= 10 #






= 400 mm



p/fck =

PU / ( fck.b.D) =

MUZ1 / (fck.D.b2) =

MUZ1 =

p/fck =

PU / ( fck.b.D) =

MUY1 / (fck.b.D2) =

MUY1 =


PU / PUZ = αn =


Ast reqd. mm2

mm2

cm2

Documents

Column Design