6. Design of Water Tanks (For class held on 23 rd, 24 th, 30 th April 7 th and 8 th May 07)

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1 6. Design of Water anks (For class held on 3 rd, 4 th, 30 th April 7 th and 8 th May 07) By Dr. G.S.Suresh, Professor, Civil Engineering Department, NIE, Mysore (Ph: , gss_nie@ yahoo.com) 6.1 Introduction: Storage tanks are built for oring water, liquid petroleum, petroleum products and similar liquids. Analysis and design of such tanks are independent of chemical nature of product. hey are designed as crack free ructures to eliminate any leakage. Adequate cover to reinforcement is necessary to prevent corrosion. In order to avoid leakage and to provide higher rength concrete of grade M0 and above is recommended for liquid retaining ructures. o achieve imperviousness of concrete, higher density of concrete should be achieved. Permeability of concrete is directly proportional to water cement ratio. Proper compaction using vibrators should be done to achieve imperviousness. Cement content ranging from 330 Kg/m 3 to 530 Kg/m 3 is recommended in order to keep shrinkage low. he leakage is more with higher liquid head and it has been observed that water head up to 15 m does not cause leakage problem. Use of high rength deformed bars of grade Fe415 are recommended for the conruction of liquid retaining ructures. However mild eel bars are also used. Correct placing of reinforcement, use of small sized and use of deformed bars lead to a diffused diribution of cracks. A crack width of 0.1mm has been accepted as permissible value in liquid retaining ructures. While designing liquid retaining ructures recommendation of Code of Practice for the orage of Liquids- IS3370 (Part I to IV) should be considered. Fractured rength of concrete is computed using the formula given in clause 6.. of IS ie., f cr =0.7f ck MPa. his code does not specify the permissible resses in concrete for resiance to cracking. However earlier version of this code published in 1964 recommends permissible value as cat = 0.7 f ck for direct tension and cbt = 0.37 f ck for bending tensile rength. Allowable resses in reinforcing eel as per IS 3370 are = 115 MPa for Mild eel (Fe50) and = 150 MPa for HYSD bars(fe415) 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.% iii) For thickness between 100 mm to 450 mm = varies linearly from 0.3% to 0.% For concrete thickness 5 mm, two layers of reinforcement be placed, one near water face and other away from water face. Cover to reinforcement is greater of i) 5 mm, ii) Diameter of main bar. In case of concrete cross section where the tension occurs on fibers away from the water face, then permissible resses for eel to be used are same as in the analysis of other sections, ie., =140 MPa for Mild eel and =30 MPa for HYSD bars. 1

2 6. Introduction to Working Stress method: In this method the concrete and eel are assumed to be elaic. At the wor combination of working loads, the resses in materials are not exceeded beyond permissible resses. he permissible resses are found by using suitable factors of safety to material rengths. Permissible resses for different grades of concrete and eel are given in ables 1 and respectively of IS he modular ratio m of composite material ie., RCC is defined as the ratio of modulus of elaicity of eel to modulus of elaicity of concrete. But the code 80 ipulate the value of m as m, where bc is the permissible ress in concrete 3 bc in bending compression. o develop equation for moment of resiance of singly reinforced beams, the linear rain and ress diagram are shown in Fig. 6.1 x b c E c x/3 C d z Section E s Strain Diagram Stress Diagram Fig. 6.1 Singly Reinforced Section he neutral axis depth is obtained from rain diagram as x / E c m m = solving for x; x d kd d x / E s m m where, k, k is known as neutral axis conant m he lever arm z=d-x/3 = d-(kd/3)= d(1-k/3) = jd, where, j=1-k/3; j is known as lever arm conant C= ½ bx; = A Moment of resiance M= C z = z Consider, M=C z = (½ bx) jd = (½ bkd) jd = (½ kj) bd = Q bal bd Where, Q bal is known as moment of resiance factor for balanced section. Now consider M= z = A jd;

3 M A ; Let p t be the percentage of eel expressed as jd 100A M 1 50k p tbal 100 bd jd bd Design conants for balanced section is given in table 6.1 able 6.1 Design conants Concrete Grade Steel Grade k j Q bal p tbal M0 Fe Fe M5 Fe Fe Liquid Retaining Members subjected to axial tension only: When the member of a liquid retaining ructure is subjected to axial tension only, the member is assumed to have sufficient reinforcement to resi all the tensile force and the concrete is assumed to be uncracked. For analysis purpose 1m length of wall and thickness t is considered. he tension in the member is resied only by eel and hence ct A and 1000 t ct +(m-1)a or t 1 (m 1) 1000ct Minimum thickness of the member required is tabulate in table 6. able 6. Minimum thickness of members under direct tension (Uncracked condition) Grade of hickness of members in mm for force in N concrete Mild eel HYSD M0 /1377 /1331 M5 /1465 /143 M30 /168 / Liquid Retaining Members subjected to Bending Moment only: For the members subjected to BM only with the tension face in contact with water or for the members of thickness less than 5 mm, the compressive ress and tensile resses should not exceed the value given in IS For the member of thickness more than 5 mm and for the face away from the liquid, this condition need not be satisfied and higher ress in eel may be allowed. he bending analysis is done for cracked and uncracked condition. Cracked condition: he procedure of designing is same as in working ress method except that the resses in eel are reduced. he design coefficients for these reduced resses in eel is given in able 6.3 3

4 able 6.3 Design conants for members in bending (Cracked condition) Concrete Grade Steel Grade k j Q bal p tbal For members less than 5 mm thickness and tension on liquid face M0 Fe Fe For members more than 5 mm thickness and tension away from liquid face M0 Fe Fe Uncracked condition: In this case, the whole section is assumed to resi the moment. Hence the maximum tensile ress in concrete should not be more than permissible value. he section is designed as a homogenous section. b kd d D Section cbt Stress Diagram Fig. 6.1 Singly Reinforced Section aking moments of transformed areas about NA b kd kd/ = b (D-kD) (D-kD)/ + (m-1) A (d-kd) Subituting A = p t bd /100 and simplifying d 100 p t (m 1) k D 00 p t (m 1) Moment of inertia I xx =bd 3 /1 + bd (kd-d/) + (m-1) A (d-kd) subituting A = p t bd /100 and simplifying I xx =(1/3 k(1-k)+(d/d-k) (m-1) p t /100)bD 3 he moment of resiance may be expressed using Bernouli s equation M I xx cb D kd D(1 k) and cbt I xx M D(1 k) 4

5 6.5 Liquid Retaining Members subjected to Combined axial tension and Bending Moment : For the members subjected to combined axial tension and bending moment, two cases are considered: i) ension on liquid face and ii) ension on remote face ension on liquid face IS 3370 requires that the resses due to combination of direct tension and bending moment shall satisfy the following condition f ct f cbt 1 ct cbt where, f ct = calculated direct tensile ress in concrete ct = permissible direct tensile ress in concrete f cbt =calculated ress in concrete in bending tension cbt = permissible ress in concrete in bending tension. ension on remote face For the sections less than 5 mm thick, the procedure explained above for tension on liquid face should be used. For the sections more than 5 mm thick, concrete rain need not be checked. his has two cases: i) ensile force is large ie., the line of action of resultant force lies within the ii) effective depth ensile force is small ie., the line of action of resultant force lies outside the section i) ensile force is large: Steel is provided on both faces. 1 and are tensile forces in eel on remote and water face face respectively. otal tensile force = 1 +. Referring to Fig. 6. and taking moment about cg of eel on water face b D d Fig. 6. d' e=m/ d' 1 (d-d )=(D/+e-d ) but d =D-d D d e 1 d D =- 1 D d e d D 5

6 ii) ensile force is small: If eel is provided on both faces then the equation derived in case 1 is valid. When eel is provided only on tension face and referring to Fig.6.3, an approximate method may be used as given below b d' C D d e=m/ z=jd Fig. 6.3 d' s Equilibrium of forces give s -C= aking moment about centroid of tensile reinforcement Cjd= (e-d/+d ) Let E= e-d/+d = e-d/+(d-d)=e+d/-d E C Subituting in equilibrium equation jd E A jd E A jd 6.6 WAER ANKS : A water tank is used to ore water to facilitate the daily requirements of habitats. ypes of water tank based on placing and shape is given in Fig Circular tanks have minimum surface area when compared to other shapes for a particular capacity of orage required. Hence the quantity of material required for circular water tank is less than required for other shapes. But the form work for a circular tank is very complex and expensive when compared to other shapes. Square and Rectangular water tanks are generally used under ground or on the ground. Circular tanks are preferred for elevated tanks. 6

7 Fig. 6.4 WAER ANK BASED ON PLACEMEN OF ANK BASED ON SHAPE OF ANK 1. RESING ON GROUND. UNDER GROUND 3. ELEVAED 1. CIRCULAR. RECANGULAR 3. SPHERICAL 4. INZ 5. CONICAL BOOM Reing on ground Under Ground 7

8 Elevated Circular Rectangular Spherical Intz Conical Bottom 8

9 6.6.1 Circular anks reing on ground : Due to hydroatic pressure, the tank has tendency to increase in diameter. his increase in diameter all along the height of the tank depends on the nature of joint at the junction of slab and wall as shown in Fig 6.5 ank with flexible base ank with rigid base Fig

10 When the joints at base are flexible, hydroatic pressure induces maximum increase in diameter at base and no increase in diameter at top. his is due to fact that hydroatic pressure varies linearly from zero at top and maximum at base. Deflected shape of the tank is shown in Fig When the joint at base is rigid, the base does not move. he vertical wall deflects as shown in Fig Design of Circular anks reing on ground with flexible base: Maximum hoop tension in the wall is developed at the base. his tensile force is computed by considering the tank as thin cylinder D H ; Quantity of reinforcement required in form of hoop eel is computed HD / as A or 0.3 % (minimum) When the thickness of the wall is less than 5 mm, the eel placed at centre. When the thickness exceeds 5mm, at each face A / of eel as hoop reinforcement is provided In order to provide tensile ress in concrete to be less to be less than permissible ress, the ress in concrete is computed using equation HD / c If c cat, where cat =0.7f ck, then the A (m 1)A 1000t (m 1)A c section is from cracking, otherwise the thickness has to be increased so that c is less than cat. While designing, the thickness of concrete wall can be eimated as t=30h+50 mm, where H is in meters. Diribution eel in the form of vertical bars are provided such that minimum eel area requirement is satisfied. As base slab is reing on ground and no bending resses are induced hence minimum eel diributed at bottom and the top are provided Design Problem: Design a circular water tank with flexible connection at base for a capacity of 4,00,000 liters. he tank res on a firm level ground. he height of tank including a free board of 00 mm should not exceed 3.5m. he tank is open at top. Use M 0 concrete and Fe 415 eel. Draw to a suitable scale: i) Plan at base ii) Cross section through centre of tank. Solution: Step 1: Dimension of tank Depth of water H= = 3.3 m Volume V = 4,00,000/1000 = 400 m 3 Area of tank A = 400/3.3 = 11. m Diameter of tank D 4A 1.4m 13 m 10

11 he thickness is assumed as t = 30H+50= mm Step : Design of Vertical wall D Max hoop tension at bottom H 14.5kN Area of eel A 1430 mm 150 Minimum eel to be provided A min =0.4%of area of concrete = 0.4x 1000x160/100 = 384 mm he eel required is more than the minimum required Let the diameter of the bar to be used be 16 mm, area of each bar =01 mm Spacing of 16 mm diameter bar=1430x 1000/01= mm c/c Provide 140 c/c as hoop tension eel Step 3: Check for tensile ress Area of eel provided A provided =01x1000/140 = mm Modular ratio m= Stress in concrete c 1. N/mm 1000t (m 1)A ( )1436 Permissible ress cat =0.7f ck = 1. N/mm Actual ress is equal to permissible ress, hence safe. Step 4: Curtailment of hoop eel: Quantity of eel required at 1m, m, and at top are tabulated. In this table the maximum spacing is taken an 3 x 160 = 480 mm Height from top Hoop tension =HD/ (kn) A = / Spacing of #16 mm c/c.3 m m op 0 Min eel (384 mm ) 400 Step 5: Vertical reinforcement: For temperature and shrinkage diribution eel in the form of vertical reinforcement is 0.4 % ie., A =384 mm. Spacing of 10 mm diameter bar = 78.54x1000/384=04 mm c/c 00 mm c/c Step 6: ank floor: As the slab res on firm ground, minimum 0.3 % is provided. hickness of slab is assumed as 150 mm. 8 mm diameter bars at 00 c/c is provided in both directions at bottom and top of the slab. 11

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13 Design of Circular anks reing on ground with rigid base: Due to fixity at base of wall, the upper part of the wall will have hoop tension and lower part bend like cantilever. For shallow tanks with large diameter, hoop resses are very small and the wall act more like cantilever. For deep tanks of small diameter the cantilever action due to fixity at the base is small and the hoop action is predominant. he exact analysis of the tank to determine the portion of wall in which hoop tension is predominant and the other portion in which cantilever action is predominant, is difficult. Simplified methods of analysis are i) Reissner s method ii) Carpenter s simplified method iii) Approximate method iv) IS code method Use of IS code method for analysis and design of circular water tank with rigid base is udied in this course. IS code method ables 9,10 and 11 of IS 3370 part IV gives coefficients for computing hoop tension, moment and shear for various values of H /Dt Hoop tension, moment and shear is computed as = coefficient ( w HD/) M= coefficient ( w H 3 ) V= coefficient ( w H ) hickness of wall required is computed from BM consideration ie., M d Qb where, Q= ½ jk m k m j=1-(k/3) b = 1000mm Providing suitable cover, the over all thickness is then computed as t = d+cover. Area of reinforcement in the form of vertical bars on water face is computed as M A. Area of hoop eel in the form of rings is computed as A 1 jd Diribution eel and vertical eel for outer face of wall is computed from minimum eel consideration. ensile ress computed from the following equation should be less than the permissible ress for safe design c and the permissible ress is 0.7 f ck 1000t (m 1)A Base slab thickness generally varies from 150mm to 50 mm and minimum eel is diributed to top and bottom of slab. 13

14 Design Problem No.1: A cylindrical tank of capacity 7,00,000 liters is reing on good unyielding ground. he depth of tank is limited to 5m. A free board of 300 mm may be provided. he wall and the base slab are ca integrally. Design the tank using M0 concrete and Fe415 grade eel. Draw the following i) Plan at base ii) Cross section through centre of tank. Solution: Step 1: Dimension of tank H= = 4.7 and volume V = 700 m 3 A=700/4.7 = m D= (4 x /) = m Step : Analysis for hoop tension and bending moment One meter width of the wall is considered and the thickness of the wall is eimated as t=30h+50 = 191 mm. he thickness of wall is assumed as 00 mm. H Dt Referring to table 9 of IS3370 (part IV), the maximu m coefficient for hoop tension = max =0.575 x 10 x 4.7 x 7 = kn Referring to table 10 of IS3370 (part IV), the maximum coefficient for bending moment = (produces tension on water side) M max = x 10 x =15.15 kn-m Step 3: Design of section: For M0 concrete =7, For Fe415 eel =150 MPa and m=13.33 for M0 concrete and Fe415 eel he design conants are: m k 0.39 m j=1-(k/3)=0.87 Q= ½ jk = 1.19 Effective depth is calculated as Step 3: Design of section: For M0 concrete =7, For Fe415 eel =150 MPa and m=13.33 for M0 concrete and Fe415 eel he design conants are: m k 0.39 m j=1-(k/3)=0.87 Q= ½ jk = 1.19 Effective depth is calculated as d M Qb x mm 1.19x

15 Let over all thickness be 00 mm with effective cover 33 mm d provided =167 mm 6 M 15.15x10 A mm jd 150x0.87x167 01x1000 Spacing of 16 mm diameter bar = 89.3mmc / c Provide #16@75 c/c as vertical reinforcement on water face x10 Hoop eel: A 1 161mm x1000 Spacing of 1 mm diameter bar = 89.mmc / c 161 Provide #1@80 c/c as hoop reinforcement on water face 113x1000 Actual area of eel provided A 141.5mm 80 (Max spacing 3d=501mm) Step 4: Check for tensile ress: x10 c 0.87N / mm 1000t (m 1)A 1000x00 ( )x141.5 Permissible ress = 0.7f ck =1. N/mm > c Safe Step 5: Diribution Steel: Minimum area of eel is 0.4% of concrete area A =(0.4/100) x1000 x 00 = 480 mm Spacing of 8 mm diameter bar = 50.4x mmc / c Provide 100 c/c as vertical and horizontal diribution on the outer face. Step 5: Base slab: he thickness of base slab shall be 150 mm. he base slab res on firm ground, hence only minimum reinforcement is provided. A =(0.4/100) x1000 x 150 = 360 mm Reinforcement for each face = 180 mm 50.4x1000 Spacing of 8 mm diameter bar = 79.mmc / c 180 Provide 50 c/c as vertical and horizontal diribution on the outer face. 15

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17 Design Problem No.: Design a circular water tank to hold 5,50,000 liters of water. Assume rigid joints between the wall and base slab. Adopt M0 concrete and Fe 415 eel. Sketch details of reinforcements. Solution: Step 1: Dimension of tank Volume of tank V=550 m 3 Assume H= 4.5 A=550/4.5 = 1. m D= (4 x 1./) = m Step : Analysis for hoop tension and bending moment One meter width of the wall is considered and the thickness of the wall is eimated as t=30h+50 = 185 mm. he thickness of wall is assumed as 00 mm. H Dt Referring to table 9 of IS3370 (part IV), the maximum coefficient for hoop tension = max =0.575 x 10 x 4.5 x 6.5 =161.7 kn Referring to table 10 of IS3370 (part IV), the maximum coefficient for bending moment = (produces tension on water side) M max = x 10 x =13.3 kn-m Step 3: Design of section: For M0 concrete =7, For Fe415 eel =150 MPa and m=13.33 for M0 concrete and Fe415 eel he design conants are: m k 0.39 m j=1-(k/3)=0.87 Q= ½ jk = M 13.3x10 Effective depth is calculated as d 105.7mm Qb 1.19x1000 Let over all thickness be 00 mm with effective cover 33 mm d provided =167 mm 6 M 13.3x10 A 610.7mm jd 150x0.87x167 01x1000 Spacing of 16 mm diameter bar = 39.36mmc / c Provide #16@300 c/c as vertical reinforcement on water face x10 Hoop eel: A mm 150 (Max spacing 3d=501mm) 17

18 113x1000 Spacing of 1 mm diameter bar = 104mmc / c Provide #1@100 c/c as hoop reinforcement on water face 113x1000 Actual area of eel provided A 1130mm 100 Step 4: Check for tensile ress: x10 c 1000t (m 1)A 1000x00 ( )x1130 Permissible ress = 0.7f ck =1. N/mm > c Safe 0.76N / mm Step 5: Diribution Steel: Minimum area of eel is 0.4% of concrete area A =(0.4/100) x1000 x 00 = 480 mm Spacing of 8 mm diameter bar = 50.4x mmc / c Provide 100 c/c as vertical and horizontal diribution on the outer face. Step 5: Base slab: he thickness of base slab shall be 150 mm. he base slab res on firm ground, hence only minimum reinforcement is provided. A =(0.4/100) x1000 x 150 = 360 mm Reinforcement for each face = 180 mm 50.4x1000 Spacing of 8 mm diameter bar = 79.mmc / c 180 Provide 50 c/c as vertical and horizontal diribution on the outer face. 18

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20 6.6. Rectangular tank with fixed base reing on ground : Rectangular tanks are used when the orage capacity is small and circular tanks prove uneconomical for small capacity. Rectangular tanks should be preferably square in plan from point of view of economy. It is also desirable that longer side should not be greater than twice the smaller side. Moments are caused in two directions of the wall ie., both in horizontal as well as in vertical direction. Exact analysis is difficult and such tanks are designed by approximate methods. When the length of the wall is more in comparison to its height, the moments will be mainly in the vertical direction, ie., the panel bends as vertical cantilever. When the height is large in comparison to its length, the moments will be in the horizontal direction and panel bends as a thin slab supported on edges. For intermediate condition bending takes place both in horizontal and vertical direction. In addition to the moments, the walls are also subjected to direct pull exerted by water pressure on some portion of walls. he walls are designed both for direct tension and bending moment. ` B C p=h B A D L BASE FBD OF AB IN PLAN FBD OF AD IN PLAN + 0.5b 0.5b y - a Bending moment diagram x IS3370 (Part -IV) gives tables for moments and shear forces in walls for certain edge condition. able 3 of IS3370 provides coefficient for max Bending moments in horizontal and vertical direction. Maximum vertical moment = M x w a 3 ( for x/a = 1, y=0) Maximum horizontal moment = M y w a 3 (for x/a = 0, y=b/) 0

21 ension in short wall is computed as s =pl/ ension in long wall L =pb/ Horizontal eel is provided for net bending moment and direct tensile force M' A =A 1 +A ; A 1 ; M =Maximum horizontal bending moment x; x= d-d/ jd A =/ d D/ x Design problem No.1 Design a rectangular water tank 5m x 4m with depth of orage 3m, reing on ground and whose walls are rigidly joined at vertical and horizontal edges. Assume M0 concrete and Fe415 grade eel. Sketch the details of reinforcement in the tank Solution: Step1: Analysis for moment and tensile force E C A Free a=h=3m Fixed D L=5m B F b=4m i) Long wall: L/a= ; at y=0, x/a=1, M x =-0.074; at y=b/, x/a=1/4, M y =-0.05 Max vertical moment = M x w a 3 = Max horizontal moment = M y w a 3 = ; long = w ab/=60 kn ii) Short wall: B/a= ; at y=0, x/a=1, M x =-0.06; at y=b/, x/a=1/4, M y = Max vertical moment = M x w a 3 = -16. Max horizontal moment = M y w a 3 = ; short = w al/=75 kn 1

22 Step: Design conants =7 MPa, =150 MPa, m=13.33 m k 0.38 m j=1-(k/3)=0.87 Q= ½ jk = 1.15 Step3: Design for vertical moment For vertical moment, the maximum bending moment from long and short wall (M max ) x = kn-m d M Qb x mm 1.15x1000 Assuming effective cover as 33mm, the thickness of wall is t= =164.8 mm170 mm d provided =170-33=137mm 6 M 19.98x10 A mm jd 150x0.87x137 Spacing of 1 mm diameter bar = 113x Provide 100 mm c/c Diribution eel Minimum area of eel is 0.4% of concrete area A =(0.4/100) x1000 x 170 = 408 mm Spacing of 8 mm diameter bar = 50.4x mmc / c (Max spacing 3d=411mm) 13.19mmc / c Provide 10 c/c as diribution eel. Provide 10 c/c as vertical and horizontal diribution on the outer face. Step4: Design for Horizontal moment Horizontal moments at the corner in long and short wall produce unbalanced moment at the joint. his unbalanced moment has to be diributed to get balanced moment using moment diribution method. A B 14.4 C K AC DF DF AC AB 1 1 ; K AC ; 5 5 1/ / 0 1/ / 0 K 9 0

23 Moment diribution able Joint A Member AC AB DF FEM Diribution Final Moment he tension in the wall is computed by considering the section at height H 1 from the base. Where, H 1 is greater of i) H/4, ii) 1m, ie., i) 3/4=0.75, ii) 1m; H 1 = 1m Depth of water h=h-h 1 =3-1-m; p= w h=10 x = 0 kn/m ension in short wall s =pl/=50 kn ension in long wall L =pb/= 40 kn Net bending moment M =M-x, where, x= d-d/=137-(170/)=5mm M = x 0.05= kn-m x10 A mm 150x0.87x x10 A mm 150 A =A 1 +A = mm 113x1000 Spacing of 1 mm diameter bar = 13 mmc / c (Max spacing 3d=411mm) Provide #1@10 mm c/c at corners Step5: Base Slab: he slab is reing on firm ground. Hence nominal thickness and reinforcement is provided. he thickness of slab is assumed to be 00 mm and 0.4% reinforcement is provided in the form of 00 c/c. at top and bottom A haunch of 150 x 150 x 150 mm size is provided at all corners 3

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