Gulati_Saoner_STP

8/4/2019 Gulati_Saoner_STP

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Subject : Analysis and Design of Pump House and Ground Water Sump,for Dust Supression System at Saoner B.G. Railway Siding of WesternCoalfields Limited.

Client Western Coalfields Limited., Nagpur.

Contractor M/s. Gulati Construction Co., Nagpur.------------------------------------------------------------------------------------------------------------

ANALYSIS & DESIGN OF COLUMN,COLUMN FOOTING AND PLINTH BEAMS

Design Data :

Internal room size of Pump House 36.00 Sqm.

Opening for Ventilation Not less than 20%Rolling Shutter 2.00 m x 2.10 mFlooring 50 mm thick IPS Flooring.

S.B.C. of Soil :100 KN/Sqm (10 T/Sqm) at 2.00 M depth.

Permissible Stresses in concrete & steel :

Design of all structural members is carried out limiting to permissible stresses

that of M-20 Grade of Concrete.

Grade of Concrete = R.C.C. M-20 (Design Mix Concrete).

Direct stress in compression 5 Mpa.Bending stress in compression 7 Mpa.Shear Stress As permitted in I.S. 456 2000 (Table -19 )

Grade of Steel = Tor Steel Fe-415Mild Steel Fe-250

Permissible tensile stress in bending

In axial compression = 190 Mpa.Tensile stress in bending = 230 Mpa.

Contd.2.


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-2-

DESIGN CONSTANTS :

Modular ratio m = 280/3 cbc = 280/3x7 = 13.33 Say 13

Co-efficient of Neutral axis.

N1 = (mx cbc )/ st +m cbc = (13x7)/{230+(13x7)} = 0.28

Lever arm, j = (1-n1/3) = (1-0.28/3) = 0.91

Co-efficient of Moment of Resistance,

q = x cbc x n1 x j = x7x0.28x0.91 = 0.89

ANALYSIS & DESIGN OF COLUMNS :

COLUMN C-1 - ( 230 x 350 )

Load Calculation on Column C-1 :

Thickness of Roof Slab = 125 mm

Dead Load = 0.125x25 = 3.125 Kn/Sqm.Live load = 1.50 KN/Sqm.Weight of finishes/tiles. = 1.00 KN/Sqm.

-------------------------------Total Load 5.625 KN/Sqm.

Influence Area of Critically loaded Column. (3.715x2.615 = 9.7147 Sqm.)Effective Wall Length = 3.715+2.615 = 6.33 M

Therefore,Load Transferred from Roof Slab = 5.625 x 9.147 = 51.45 KNLoad Transferred from Wall = 6.33x0.25x4.70x20 = 148.75 KN

--------------------------Total = 200.20 KN.

Add 10 % Weight of footing = 20.02 KN------------------------------------TOTAL = 220.22 KN

Design of Column for 200.20 KN.

Proposing R.C.C. Column of 230 x 350 with 6 -12 tor main reinforcement and 8mm dia. Rings.


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-3-

Load carrying capacity of column

P = (cc x A c) + (

sc x Asc)

Therefore, cc = 5.00 Mpa for M-20 Grade Concrete.

sc = 190.00 MpaAsc = 6x113 = 678 Sqmm.Ac = 230x350 678 = 79822 Sqmm.

P = (79822x5.00) + (678x190)= 527.93 KN > 200.20 KN O.K.

Design of lateral ties (Ring) Provide 8-Tor four legged lateral ties.Spacing It should be min. of following three.

16 times dia. of main bar = 16x12 = 192 mmLeast lateral dimension = 230 mm.48 x dia of lateral ties =48x8 = 384 mm.

Therefore, Provide 8-Tor four legged lateral ties @ 190 c/c throughout, as shownin figure.

Design of Column Footing, for 220.20 KN

Axial load on column = 200.20 Kn.Add 10 % for weight of footing = 20.02 KN

--------------------------Total = 220.22 KN

Area of footing required (Assuming S.B.C. of soil as 100 Kn/Sqm.)

A = 220.22/100= 2.20 Sqm.

Proposing rectangular footing having equal projections beyond column face.

Length of footing = 0.350 + 2a

Breadth of footing = 0.230 + 2a

Therefore, (0.350 + 2a) x (0.230 + 2a) = 2.20 Sqm.Solving for a = 0.6355 m , L = 1.8 M and B = 1.50 M

Therefore, Provide footing of size 1800 x 1500 giving A = 2.70 Sqm > 2.20 SqmO.K.


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Upward soil pressure = 200.02/2.70 = 74.08 KN/Sqm.

Maximum B.M. = 74.08 x 0.635x0.635/2 = 14.94 KN-M per Mtr.

Effective depth required, d = (14.94 x 10^6/0.90x1000)^0.5= 128.82 mm

Depth required from two-way shear action is more than that required fromB.M. consideration. Therefore proposing the overall depth of 450 mm.Effective depth 450 50 = 400 mm.

Check for two-way shear action.

The permissible shear stress (Limit state method) = 0.5 x fckfck for M-20 Grade Concrete = 20 N/sqmm.

Therefore c = 0.2520 = 1.1 N/sqmm.Shear Stress at section d from the face of column

2 {(230+400) +(400+350)} x 400 x c actual = 1.50x220x1000 N

on solving c actual = 0.29 N/Sqmm < 1.10 N/Sqmm O.K.

Therefore, Provide footing of size 1800 x 1500. Depth at edge of footing150 mm and face of the column 300 mm as shown in figure.Reinforcement = (14.94 x 10^6/230x0.9x400) = 180.43 Sqm,

Provide 10 TOR @ 175 c/c both ways.

The R.C.C. footing shall be casted over P.C.C. 1:3:6 (M-10) 100 mmthick.

Note Same footing have been provided for other columns.

Analysis & Design of Plinth Beam

Plinth Beam marked PB-2 = Span 5.23 MLoad = 0.25x20x3.10 = 15.50 KN/m say 20 KN/m.

Maximum B.M. (assuming partial fixity at support) = wl2/10= + 20 x (5.23)2/10 = 54.70 KN-M

Provide beam of overall size 230x450 mm.Effective depth considering 40 mm effective cover = 410 mm.

Therefore, Area of steel required = 54.70 x 10^6/230x0.9x410


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= 644.52 mm2

-5-Provide 2-12 tor (str.) + 2-16 tor (ckd) = 2 x(113+ 201) = 628 mm2 > 644.52 mm2

and stirrups 8 mm dia @ 200 C/C.

Check for shear :

Maximum shear force = 0.60 WL = 0.60 x 20 x 5.367 = 64.40 KN.

Nominal Shear Stress, v = (64.40 x 1000)/230 x 410 = 0.68 N/mm2

% age steel = (804x100)/(230x410) = 0.85 %

Therefore, c = 0.34 + (0.39-0.34)x 0.1/0.25 = 0.36 N/ mm2

Balance Shear force = (0.68 0.36) x 230 x 410 = 30.176 KN.

Shear taken by bent up bar = ( sv x A sv x sin )

= 230 x 402 x Sin 45= 65.37 KN

but it is assumed that only 50 % of the balance shear is taken by bent up bars( Ref. IS-456 2000), Therefore 50 % of balance shear is taken up by stirrups.

Therefore required spacing of 8 mm dia reinforcement :

Sv = (sv x A sv x d)/Vs

= (230 x 50.28x2 x 410)/0.50x65370= 290.12 c/c, whereas provided spacing is 200 c/c

Similarly other Plinth Beams have been designed.

Analysis & Design of Roof Slab (125 MM) M-20:

Assuming 125 mm thick slab, D.L. = 3.125 KN/Sqm.Live Load = 1.50 KN/Sqm., Finish = 1.00 KN/Sqm.Hence, Total load on Roof Slab = 5.625 KN/Sqm.

Internal Panel Size = 3.715 M x 5.23 MLy/Lx = 5.23/3.715 = 1.40

Assuming the Simply Supported Slab, Spanning in two directions at right anglesto each other.


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The B.M. Coefficients as per TABLE No. 27 of I.S.-456 : 2000 (P-91)

x = 0.099,

y = 0.051

M x = x x w x L2

M y = y x w x L2

L = Span = 3.715W = 5.625 KN/Sqm.

Therefore Max. B.M.

M x = 0.099 x 5.625 x 3.7152

= 7.68 KN-M

M y = 0.051 x 5.625 x 3.7152

= 3.96 KN-M


Depth required from stiffness criteria = span/35 = 3715/35 = 106.14 mm.Whereas overall provided depth is 125 mm O.K.

Calculation of Steel,

Astxx = (7.68 x 10^6/230x0.9x 105) = 353.35 Sqm

Provide 10 TOR @ 175 c/c, Alternate bars are cranked near support, extra top10 Tor @ 350 c/c.Astxx = (3.96 x 10^6/230x0.9x 105) = 182.19 Sqm

Provide 8 TOR @ 200 c/c, Alternate bars are cranked near support, extra top 8 Tor @ 400 c/c.

2) DESIGN OF GROUND WATER SUMP :Data : Capacity = 50,000 Ltrs.

Internal Diameter = 4600 mmHeight of tank = 3.05 M + 0.3 meter free Board.


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Capacity Calculation = 0.7854x4.60x4.60x3.05 = 50.688 Cum > 50 Cum. O.K.-7-

The total height of tank = 3.35 M (1.00 M above G.L., and 2.35 M below G.L.)Assume Wall thickness = 150 mm.The ratio H2/Dt = (3.05x3.05)/(4.60x0.15) = 13.50

The Coefficients of Hoop Tension, B.M. and S.F are calculated from Table-9,10& 11 of I.S. 3370 (part-IV)

Sr.no. Hoop Tension Max. B.M. Max. S.F.

Coeiff. 0.65742 0.00936 0.13760

Value Tank Empty

21.32000 KN 0.729 KN-M 4.56 KN

Tank Full

46.11 KN-M 2.6568 Kn 12.80 KNAst. 307.40 mm2 145.77 mm2 ------ mm2

Astmin 450.00 mm2 450.00 mm2 ------ mm2

Astprovided

10 Tor @ 200 c/chorizontal bars,

staggered on eachface.

8 Tor @ 200 c/chorizontal bars,

staggered on eachface.

Design of Raft :

The raft is resting on firm ground. Provide 250 mm thick raft.

Critical Condition Tank is Empty & raft is subjected to upward soil pressure.The Max. B.M. = 0.032 x (50-5.125) x 4.6 x 4.6 = 30.47 KN-MDepth required = (30.47 x 10^6/0.65x1000)^0.5

= 216.51Therefore, Provide overall depth of 250 mm.

Ast = (30.47 x 10^6/230x0.9x 215) = 684.64 Sqm

Provide 10 TOR @ 200 c/c in both directions and staggered on each face.

DESIGN OF TOP SLAB :

Total Load on Slab = 3.125+1.50+1.00 = 5.625 KN/Sqm.The Max. B.M. = 0.062 x (5.625) x 4.6 x 4.6 = 7.3625 KN-MDepth required = (7.3625 x 10^6/0.65x1000)^0.5

= 106.42Therefore, Provide overall depth of 125 mm.Ast = (7.3625 x 10^6/230x0.9x 105) = 338.74 Sqm


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Provide 8 TOR @ 200 c/c in both directions and staggered on each face.

Analysis & Design of Roof Slab (100 MM) M-20:

Assuming 100 mm thick slab, D.L. = 2.50 KN/Sqm.

Live Load = 0.75 KN/Sqm., Finish = 0.25 KN/Sqm.Hence, Total load on Roof Slab = 3.50 KN/Sqm.

Internal Panel Size = 3.27 M x 3.27 MLy/Lx = 3.27/3.27 = 1.00

Assuming the Simply Supported Slab, Spanning in two directions at right anglesto each other.

The B.M. Coefficients as per TABLE No. 27 of I.S.-456 : 2000 (P-91)

x = 0.062,

y = 0.062

M x = x x w x L2

M y = y x w x L2

L = Span = 3.27 + Wall thickness (0.15) = 3.42 MW = D.L. + L.L. + Finishes = 2.50 + 0.75 + 0.25 = 3.50 Kn/Sqm.

Therefore Max. B.M.

M x = 0.062 x 3.50 x 3.422

= 2.54 KN-M

M y = 0.062 x 3.50 x 3.422

= 2.54 KN-M



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Depth required from stiffness criteria = span/35 = 3420/35 = 97.71 mm.Whereas overall provided depth is 100 mm O.K.

Calculation of Steel,

Ast = (2.54 x 10^6/150x0.9x 80) = 235.18 Sqm

Provide 8 TOR @ 200 c/c both ways, Alternate bars are cranked near support,extra top 8 Tor @ 400 c/c.

Analysis & Design of Gallary Slab (100 MM) M-20 :

Maximum B.M. (assuming partial fixity at support) = wl2/10

= + 17.88 x (4.23)2/10 = 31.99 KN-M

Therefore, Area of steel required = 31.99 x 10^6/230x0.9x315

= 448.00 mm2

Provide 2-12 tor + 2-12 tor = 4 x 113 = 452 mm2 > 448 mm2

And stirrups 6 mm dia @ 200 C/C.

Check for shear :

Maximum shear force = 0.60 WL = 0.60 x 17.88 x 4.23 = 45.37 KN.

Nominal Shear Stress, v = (45.37 x 1000)/230 x 315 = 0.62 N/mm2

% age steel = (452x100)/(230x315) = 0.62 %

Therefore, c = 0.29 + (0.34-0.29)x 0.12/0.25 = 0.30 N/ mm2

Balance Shear force = (0.62 0.31) x 230 x 315 = 22.46 KN.

Shear taken by bent up bar = ( sv x A sv x sin )

= 230 x 226 x Sin 45= 36.755 KN


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but it is assumed that only 50 % of the balance shear is taken by bent up bars( Ref. IS-456 2000), Therefore 50 % of balance shear is taken up by stirrups.

Therefore required spacing of 6 mm dia reinforcement :

Sv = (sv x A sv x d)/Vs

= (140 x 28.50x2 x 315)/0.50x22460= 223.83 c/c,

whereas provided spacing is 200 c/c O.K.

DESIGN OF R.C.C. COLUMNS & FOOTINGS :

Maximum load on central column

= Reaction transferred from roof beam + wall loads.

= 69.06 + (5.23+4.23)3.3x0.23x20 = 212.66 KN.

Proposing R.C.C. Column of 230 x 350 with 6-12 tor main reinforcement and 6mm dia. Rings.

Load carrying capacity of column

P = ( cc x A c) + ( sc x Asc)

Therefore, cc = 4.00 Mpa for M-15 Grade Concrete.

sc = 190.00 Mpa

Asc = 6x113 = 678 Sqmm.Ac = 230x350 452 = 79822 Sqmm.

P = (79822x4.00) + (678x190)= 448.12 KN > 212.66 KN

Design of Column Footing,Area of footing required

A = 1.10 x 212.66/100

(Assuming S.B.C. of soil as 100 Kn/Sqm.)

Area of footin2.33 SQM. Provide square footing of size 1.60 M x 1.80 Mgiving area = 2.88 Sqm. O.K. Provide depth at edge = 150 mm and at columnface = 450 mm.

Upward soil pressure = 212.66/2.88 = 73.84 KN/Sqm.


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Maximum B.M. = 73.84 x 0.57x0.57/2 = 11.99 KN-M per Mtr.

Effective depth required, d = (11.99 x 10^6/0.65x1000)^0.5

= 135.84 mm

Whereas overall provided depth is 450 mm O.K.

Reinforcement = (11.99 x 10^6/230x0.9x410) = 141.27 Sqm

Provide 10 TOR @ 175 c/c both ways.

COMPARATIVE STUDY OF QUANTITIES AS PER OLD AND NEW DESIGN.

Sr.No. Description andcalculations

Qnty as perOld design

Qnty as per newdesign.

Difference

1 Earth Work inExcavation

1x5.40x5.40x4.00= 116.64 Cum.

4x1.50x1.50x2.00 =18.00 Cum.

98.64Cum.

2 P.C.C. 1:4:8 4x4.20x1.00x0.1 =1.68 Cum.

4x1.50x1.50x0.1 =0.90 Cum.

0.78 Cum.

3 R.C.C. (M-20) grade of concrete in

(a) Bottom slab BS-1. 4.50x4.50x0.175 =3.54 Cum.

4.50x4.50x0.10 =2.02 Cum.

(b) Beam WB-1 4x4.35x0.30x0.425= 2.22 Cum.

4x4.35x0.23x0.40 =1.60 Cum. +2x4.20x0.23x0.40 =0.77 Cum.

(c) Column 4x14x0.30x0.30 =5.04 Cum.

4x12x0.23x0.23 =2.54 Cum.

(d) Foundation 4x4.20x1.0x0.15 =2.52 Cum. + 4 x4.20x(1+0.3)/2x0.3=3.276 Cum, total= 5.796 Cum.

4x1.2x1.2x0.15 =0.864 Cum. + 4 x0.25/3x(1.44+0.0529+1.2x.23) =0.589Cum, total = 1.453Cum.

(e) Beam at foundationFB-1

4x4.20x0.30x0.15= 0.756 Cum.

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Total Concrete 17.352 Cum. 8.383 Cum.4 Steel Reinforcement :

(a) Roof Slab 10-Tor @ 200 c/c2x23x4.40x0.62 =125.488 Kg.

8-Tor @ 200 c/c2x23x4.40x0.39 =78.936 Kg.

(b) Wall : Vertical bars oninner face 12 Tor@ 100 c/c4x43x4.20x0.89 =642.936 Kg.Vertical bars on

Vertical bars oninner face 10 Tor@ 100 c/c upto 1.00M4x43x2.20x0.62 =234.60 Kg.


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outer face 12 Tor@ 200 c/c4x22x3.95x0.62 =215.512 Kg.Horizontal bars 10Tor @ 200 c/c.

2x4x16x4.35x0.62= 345.216 Kg.8-Tor @ 250 c/ccorner bars4x13x1.50x0.39 =30.42 Kg.

For remaining height10 Tor @ 200 c/c4x43x2.0x0.62 =109.12 Kg.Vertical bars onouter face 8 Tor @

200 c/c4x22x3.95x0.39 =135.56 Kg.Horizontal bars 10Tor @ 200 c/c.2x4x16x4.35x0.62 =345.216 Kg.

(c) Bottom Slab 12-Tor @ 100 c/c2x43x4.20x0.89 =321.46 Kg.Extra top 12 Tor

@ 200 c/c4x21x1.00x0.89 =74.76 Kg.

8-Tor @ 100 c/c2x43x4.20x0.39 =140.86 Kg.Extra top 8 Tor @

200 c/c16x21x0.50x0.39 =65.52 Kg.

(d) Beam WB-1 4x6x4.35x1.58 =164.95 Kg.

6x6x4.35x1.58 =247.42 Kg.

(e) Column 8x14x4x1.58 =707.84 Kg.

4x12x4x0.89 =170.88 Kg.

(f) Foundation 10 Tor @ 150 c/c4x29x0.90x0.62 =62.50 Kg.Dist. 8-Tor4x5x4.20x0.39 =

32.76 Kg.Beam FB-14x8x4.20x2.47 =331.96 Kg.Stps.4x33x1.55x0.39 =78.62 Kg.Total = 3134.422Kg.

10 Tor @ 200 c/cbothways.4x2x7x1.1x0.62 =38.19 Kg.

Total = 1566.29 Kg. Total =1568.132Kg. + 5 %overlap =1650 Kg.

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