STRUCTURAL DESIGN OF 350KL OVERHEAD WATER TANK AT INDIRA GANDHI NATIONAL OPEN UNIVERSITY.docx

STRUCTURAL DESIGN OF 350KL OVERHEAD WATER TANK AT INDIRA GANDHI NATIONAL OPEN UNIVERSITY, TELIBAGH LUCKNOW

DATA1. Type of Tank: Intze Tank2. Capacity of the tank:350KL3. Type of staging: Column & Brace type4. Depth of foundation:2.5m5. Safe Bearing Capacity of Soil:100KN/m26. Type of foundation:Circular Ring &Raft foundation7. Grade of Concrete:M-258. Grade of Steel:Fe-4159. Height of staging:25m10. Type of soil:Soft Clay11. Height of Building up to Terrace:15.6m12. No. of floors in Building:G+313. Basic Wind Pressure:1500N/m214. Sesmic Zone of Lucknow:Zone 315. No.of student in College:200016. Water consumption rate(Per capita demand in litres per day per head):4517. Design period for tank:30 years18. No.of student in hostels:1600

OBJECTIVE1:- To make a study about the analysis and design of water tank2:- To make a study about the guidelines for the design of liquid retaining structure according to IS Code IS: 3370 part 2-2009 IS: 456:20003:- To know about the design philosophy for the safe and economical design of water tank4:- To estimate the overall cost for making the Intze Tank

WATER QUANTITY ESTIMATION IN COLLEGE CAMPUS

Population or the number of students to be served in 2014 = 2000Let population to be increased at rate of 10% per decade Number of students (2014) =2000 Number of students in 2024 = 2200 Number of students in 2034= 2420 Number of students in 2044= 2662Quantity = per capita demand Population = 45 2662= 1,19,790 litres= 120 KL (assume)

FLUCTUATION IN RATE OF DEMAND

Average daily per capita demand in college campus= 45 lpcdIf this average supplied at all the times it will not be sufficient to meet the fluctuation.HOURLY VARIATION

(1) During the entry of college from 8 to 9 in the morning.(2) During the lunch from 12 to 1 in the afternoon.

WATER CONSUMPTION IN HOSTEL

Average daily per capita demand in hostels = 135 lpcd.

Quantity = 136 1600 = 216 KLTotal quantity = 216 + 130 = 346 KL 350 KL

DESIGN REQUIREMENT OF TANK* Concrete mix weaker than M-20 is not used because of higher grade lesser porosity of concrete.* Minimum quantity of cement in concrete shall be not less than 30 KN/m3.* Use of small size bars.* Coefficient of expansion due to temperature=1110-6/C* Coefficient of shrinkage may be taken = 450 10-6 for initial and 200 10-6 for drying shrinkage.* Minimum cover to all reinforcement should be 20 mm or the diameter of main bar whichever is greater.* An overhead liquid retaining structure is design using working stress method avoiding the cracking in the tank and to prevent the leakage and the component of tank can be design using LIMIT STATE METHOD (example:-column ,foundation ,bracing ,stairs etc.).* Code using IS: 3370-PART 2-2009 IS: 456:2000* The leakage is more with higher liquid head and it has been observed thad water head up to 15m does not cause leakage problem.* In order to minimize cracking due to shrinkage and temperature, minimum reinforcement is recommended as-(i) For thickness 100 mm = 0.3%(ii) For thickness 450 mm = 0.2%For thickness between 100 mm to 450 mm = varies linearly from 0.3% to 0.2%* For concrete thickness 225 mm, two layer of reinforcement be placed one near water face and other away from water face.

FROM IS -3370

(i) For load combination water load treated as dead load.(ii) Cracking The maximum calculated surface width of concrete for direct tension and flexure or restrained temperature and moisture effect shall not exceed 0.2 mm with specified cover.(iii) Shrinkage coefficient may be assumed = 300 10-6.(iv) Bar spacing should generally not exceed than 300 mm or the thickness of the section whichever is less.

DETERMINATION OF FOUNDATION

-From testing of soil sampleFor clay =12 r = density of soil = 1.76 gm/cm3 = 17.6 KN/m3 p = 100 KN/m3

9. DETERMINATION OF HEIGHT OF STAGGING

We know turbulent flow occurs in a pipeSo Re 4000 f L = length of pipe, v = mean velocity in pipe of flow d = diameter of piped The kinematic viscosity of water () = 0.0110-4 m2/sAssume diameter of pipe = 15 cm = Volume (V) = 350 m3Only for one hour maximum velocity occurs in the pipe so the discharge during that period 0.097 0.0176 m/sec.Maximum velocity =5.52 m/sec. (O.K.)

Minimum length of pipe requirement= 2 height of building up to 3 storeys from the level + lateral distance up to the centre of tank= 2 15.6 + 18 = 49.2 m 50 m

Head loss = 5.40 mHEIGHT OF STAGGING

Total hydrostatic pressure on tank P = ghTotal head = Minor loss (assume) = 1 m. Using total head = 29.5Height of stagging = 29.5 4.5 = 25 m

DESIGN OF TOP DOME

Assume thickness of top dome = 100 mm.Meridional thrust at edges Dead load of top dome = 0.100 25 = 2.5 KN/m2Live load on top dome = 0.75 KN/m2 (assume)Total load P = 3.25 KN/m2

N/mMeridional stress = MPa < 5 MPa (OK)

Maximum hoop stress occurs at the centre and its magnitude 0.30 N/mm2 0.3 MPa < 5MPa (OK)Provide nominal reinforcement of 0.24%. 2Use 8 mm bars. 2Spacing 205 mm c/c.Provide 8 mm bars @ 205 mm c/c radially and circumtentially as shown in figure.The 205 mm c/c for radial bar is provided at the springing of the dome.At the crown the spacing reduces to zero.Hence the curtailment of radial bars may be carried out at the appropriate distance.

DIMENSION OF TANK

Inner diameter of cylindrical portion D = 12 mRise of top dome h1 = 1 mRise of bottom dome h2 = D/8 = 1.5 m (centre)Free board = 0.15 mDiameter of ring beam Do = 5/8 D = 7.5 = 8 mRise of bottom dome (side) ho = 3/16 D = 2.25 m = 2.5 mCapacity of tank:- Radius of bottom circular dome:- 1.5 (2R2 1.5) = 42 2R2 1.5 = 10.67 R2 =6 mSin 2 = 2 = 41.8o

Radius of top circular dome:-1 (2R1-1) = 6 6R1 = 18.5 mSin 1 = 6/18.51 = 18.92oDesign of top ring beam:-A ring beam is provided at the junction of top dome and the vertical wall to resist hoop tension induced by the top dome.Horizontal component of meridional thrust P1 = T1 cos 1 = 30897.15 cos 18.92o = 29227.8 N/m.Total hoop tension tending to rupture of beam = = Permissible stress in HYSD bars = 150 N/m2Ash = 175366.8/150 = 1170 mm2Provide 20 mm bars (314.15) as hoop.Number of 12 mm bars = 1170 / 314.15 = 3.72 = 4Actual Ash = 4 /4 202 = 1256.63 mm2 = 1257 mm2Provide 4-20 mm hoop and 8 mm bars tie @ 205 mm c/c.Hence the cross sectional area of concrete1.3Ac = 124841.53Provide ring beam of 320 mm 400 mm.

Design of cylindrical wall:-In the membrane analysis the tank wall is assumed to be free at top and bottom maximum hoop tension occurs at the base of the wall and its magnitude:-=Hoop tension at any depth x from the top X (m) Hoop tension (N/m)

0 0

1 58800

2 117600

Minimum thickness of cylindrical wall = 3 H + 5 = 3 2 + 5 = 11 cm.Provide 20 cm at the bottom and taper it to 12 cm at top.At x = 1 m.Area of steel Ash = 58800/150 = 392 mm2Provide 8 mm bars.A = 50.26 mm2Spacing = (1000 50.26) / 392 = 130 mm c/c.At x = 2 m.Area of steel Ash = 117600/150 = 784 mm2Provide 10 mm bars.A = 78.53 mm2Spacing = (1000 78.53) / 784 = 100 mm c/c.The hoop steel may be curtailed according to hoop tension at different height along the wall provided 0.24% of minimum vertical reinforcement.Average thickness of wall = (120+200) / 2 = 160 mm. Ash = mm2Provide 8 mm . A = 50.26 mm2Spacing = mm c/c.

Design of ring beam B3:-Thickness =100 mmRise = 1.5 m (centre)Base dia. = 8 mRaidus of curvature = 6 mCos 41.8o = 0.745The ring beam connect the tank wall within conical dome. The vertical load at the junction of the wall with conical dome is transferred to the ring beam B3 by horizontal thrust. In the conical dome the horizontal component of thrust causes hoop tension at the junction.

W = Load transferred through the tank wall at the top of conical dome / unit length.o = Inclination of conical dome.T = Meridional thrust in conical dome at the junction.tan o = 2/2.5o = 38.65

= 30897.15 sin 18.92 = 10018.32 N/m(ii) Load due to ring beam B1 = 320 mm depth = 400 mm width = 0.32(0.4-0.1)125000 = 2400 N/m(iii) Load due to tank wall = = 8000 N/m(iv) Seif load of beam B3 (1m 0.6m say) =(1-0.3) 0.625000 = 10500 N/mTotal W = 10018.32 + 2400 + 8000 + 10500 = 30918.32 N/mSin o = sin 38.65 = 0.62 , cos 38.65 = 0.78Force Pw due to load Pw1 = W tan o = 30918.32 tan 38.65 = 24725.97 N/mForce Pw caused due to water pressure at top of conical dome Pw2 = rw hd3h = depth of water upto centre of ring beamd3 = depth of ring beam Pw2 = 9800 2 0.6 = 11760 N/mHence hoop tension in the ring beam is given by :-P = =This is to be resisted by steel hoops the area of which isAsh = = 1460 mm2Use 20 mm bars = 314.15Number of 20 mm bars = = 5 barsHence provide 5 ring of 20 mm dia bars.Actual area As = /4 20 5 = 1570 mm2Stress in equivalent section = = 0.35 N/mm2 < 1.2 N/mm2 (SAFE) (OK)The 10 mm diameter distribution bara (vertical bars) provided in the wall@ 100 mm c/c should be taken round the above ring to act as stirrups.

Design of conical dome :-

(a)Meridional thrust :-Ww = Total weight of water on the conical domeW = Weight of top dom, cylindrical wall etc.Ws = Self weight of conical domeWw = = = =3920098 NLet the thickness of conical slab = 400 mmWw = l = = 3.2 mWs = = 1005309.649 NWeight W at B3 = 30918.32 N/m.Hence vertical load W2 per metre run is given byW2 = = =242353.22 N/mMeridional thrust To in the conical domeTo = W2 / cos o = 242353.22 / cos 38.65 = 310321.06 N/m.Meridional stress = 310321.06 / (1000400) = 0.775 N/mm2 < 5 N/mm2 (Safe).

(b) Hoop stress

Diameter of conical dome at any height h above base isD = 8 + (12-8)/2 h = 8 + 2hIntensity of water pressure P = [(4+2)-h]9800 = [6-h]9800 N/mm2Self weight q = 0.4 25000 = 10000 N/mm2Hoop tension Po= 333150.48 + 12548.4 h2 + 12548.4h2 h Hoop Tension

0 333150.48 N

1 353695.41 N

2 349144.86 N

2.5 337457.95 N

For maximum 33093.99 25096.8 h= 0 h= 1.31 m.Maximum Po = 354969.2977 N

(c) Design of walls:-Meridional stress = 0.775 N/mm2Maximum hoop stress = 354969.29 NWhich is to be resisted by steelAs = 354969.29/150 = 2366.46 mm2Area of each face = 1183.23 mm2Spacing of 16 mm dia bars = (1000 201)/1183.23 = 170 mm c/cHence provide 16 mm dia hoops @170 mm c/c on each face.Actual As = (1000 201)/170 = 1182 mm2Maximum tensile stress in composite section = =0.83 N/mm2 < 1.2 N/mm2 (Safe) (OK)In the meridional direction the reinforcement = = 840 mm2Or 420 mm2 on each face Use 10mm diameter bars A = 78.53 mm2 Spacing = = 180 mm c/cHence provide 10 mm bars @ 180 mm c/c on each face with a clear cover 20 mm DESIGN OF BOTTOM DOMER2= 6 m2= 41.8Weight of water on w0 on the dome is given by W0=w D0=8,H0 =2.5 ,R2=6,h2=1.5, W0=850471.35NLet the thickness of bottom dome =250mmSelf Weight Ws =2R2h2t225000R2=6, h2=1.5, t2=0.25 Ws=353429.17NTotal Weight=1203900.52NMeridional thurst=T2= D0=8T2=71866.98N/mMeridional Stress= =0.287N/mm2