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DESIGN OF REINFORCED CONCRETE
STRUCTURES
1SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
UNIT-II
WATER TANKS
2SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
Needs (or) uses for water tanks
• Water tank is a container for storing liquids
Needs :
Drinking purpose
Irrigation proc...
• What is design…????
• why is it????
4SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
RCC WATER TANK DESIGN BASIS
• RCC Water tank design is based on IS 3370: 2009
(Parts I – IV).
• The design depends on the ...
• The tanks can be made of RCC or even of
steel.
• The overhead tanks are usually elevated
from the roof top through colum...
TYPES OF WATER TANKS
• Based on the location of the tank in a building s
tanks can be classified into three categories. Th...
 In most cases the underground and on ground tanks are
circular or rectangular is shape.
 The shape of the overhead tank...
 A special type of tank named Intze tank is used for
storing large amount of water for an area.
Intze tank 9SKCET / CIVIL...
Spherical
Steel tanks are also used specially in railway yards.
Basing on the shape the tanks can be circular, rectangula...
PERMISSIBLE STRESSES IN CONCRETE
• To ensure impervious concrete mixture linear than M20
grade is not normally recommended...
Grade of
Concrete
Permissible Stresses Shear stress
N/mm2Direct Tension
N/mm2
Tension due to
bending
N/mm2
M15 1.1 1.5 1.5...
THE PERMISSIBLE STRESS IN STEEL
• The stress in steel must not be allowed to exceed the following values
under different p...
BASE FOR FLOOR SLAB
• The floor slab should be strong enough to transmit the load
from the liquid and the structure itself...
MINIMUM REINFORCEMENT FOR WATER
TANK
• Minimum reinforcement required for 200mm thick sections is
0.3 % of the area of con...
UNDER GROUND WATER TANKS
EXAMPLE : 1
To Design an underground RC rectangular water tank of 10m x 3m x
3m. The soil surroun...
DESIGN PROCEDURE
• Given data:
Size = 10m x 3m x 3m
Angle of repose @ dry soil = 30o
Angle of repose @ wet soil = 6o
Densi...
 Step:1 permissible & design values:
Tension due to bending ( concrete),
σct = 1.7 N/mm2 (Pg.no.:2,table- 1, IS 3370;part...
j =1 –(n/3)
n = (1/( 1+ (σst /m* σcbc)))
= (1/(1+(115/13.33*7)))
=0.447
j = 1-(0.447/3)
= 0.85
Q = (1/2)* σcbc * n * j
= (...
 Step : 2 design of long walls:
Ratio for long wall = L/H = 10 / 3 = 3.33 >2
Ratio for short wall = B/H = 3/3 = 1 < 2
Lon...
Step:3 Design of long wall
By considering 1m run of wall
BM calculation for vertical reinforcement:
Max B.M at top (tensio...
• Step 4 Thickness of wall:
Max BM, M = (σct * bD2) / 6
29.18x 106 = (1.7 * 1000 * D2 ) / 6
29.18x 106 = 283.33 * D2
D2 = ...
• step-5 Area of steel required for long wall:
for outer face:
Ast = (max BM / σst * j *d)
= (29.18x 10 6 / 115* 0.85*280 ...
• Step:6 Area of steel for inner face:
Ast = (max BM / σst * j *d)
= (13.07x 10 6 / 115* 0.85* 280)
= 477.5 mm2
provide 12...
• Step 7: Horizontal reinforcement for long wall:
Area of distribution bar = 0.3% (b*D)
= (0.3/100) (1000*320)
= 960 mm2
p...
• step-8 Design Of Short Wall
intensity of earth pressure, p = 48.64 kN/m2
Max BM = pH2/12
= 48.64 * 3*3/ 12
= 36.48 kNm
E...
Step-9 Area of steel for short wall:
Ast = (max BM / σst * j *d)
= (36.48x 10 6 / 115* 0.85* 280)
= 1332.84 mm2
provide 16...
• Step 10: Horizontal reinforcement for short wall:
Area of distribution bar = 0.3% (b*D)
= (0.3/100) (1000*320)
= 960 mm2...
• Step-11 Design Of Roof Slab :
Let us assume the overall thickness as slab , t = 150mm
Assume cover = 25mm
Eff.depth prov...
Moment calculation:
BM (M) = W* D2/ 8
= 6 * 3.32* 3.32/ 8
= 8.26 kNm
check for depth :
d = √M/ Q.b
= √(8.26x106/1.33*1000)...
• Area of steel:
Ast = (max BM / σst * j *d)
= (8.26x 10 6 / 115* 0.85* 125)
= 676.01 mm2
provide 10mm dia bar
Spacing:
s ...
• Distribution bar :
Area of distribution bar = 0.3% (b*D)
= (0.3/100) (1000*150)
= 450 mm2
provide 8 mm dia bar
Spacing:
...
SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II 33
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Introduction & under ground water tank problem

  1. 1. DESIGN OF REINFORCED CONCRETE STRUCTURES 1SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  2. 2. UNIT-II WATER TANKS 2SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  3. 3. Needs (or) uses for water tanks • Water tank is a container for storing liquids Needs : Drinking purpose Irrigation process Fire suspension Chemical manufacture  preparation for well, etc 3SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  4. 4. • What is design…???? • why is it???? 4SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  5. 5. RCC WATER TANK DESIGN BASIS • RCC Water tank design is based on IS 3370: 2009 (Parts I – IV). • The design depends on the location of tanks, i.e. overhead, on ground or underground water tanks. • The tanks can be made in different shapes usually circular and rectangular shapes are mostly used. 5SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  6. 6. • The tanks can be made of RCC or even of steel. • The overhead tanks are usually elevated from the roof top through column. • In the other hand the underground tanks are rested on the foundation. 6SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  7. 7. TYPES OF WATER TANKS • Based on the location of the tank in a building s tanks can be classified into three categories. Those are:  Underground tanks  Tank resting on grounds  Overhead tanks 7SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  8. 8.  In most cases the underground and on ground tanks are circular or rectangular is shape.  The shape of the overhead tanks are influenced by the aesthetical view of the surroundings and as well as the design of the construction. 8SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  9. 9.  A special type of tank named Intze tank is used for storing large amount of water for an area. Intze tank 9SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  10. 10. Spherical Steel tanks are also used specially in railway yards. Basing on the shape the tanks can be circular, rectangular, square, polygonal, spherical and conical. Conical 10SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  11. 11. PERMISSIBLE STRESSES IN CONCRETE • To ensure impervious concrete mixture linear than M20 grade is not normally recommended to make the walls leak proof the concretes near the water face need to such that no crack occurs. • To ensure this member thicknesses are so designed that stress in the concrete is lesser then the permissible as given in table 1. 11SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  12. 12. Grade of Concrete Permissible Stresses Shear stress N/mm2Direct Tension N/mm2 Tension due to bending N/mm2 M15 1.1 1.5 1.5 M20 1.2 1.7 1.7 M25 1.3 1.8 1.9 M30 1.5 2.0 2.2 M35 1.6 2.2 2.7 M40 1.7 2.4 2.7 Table 1 Permissible Stresses In Concrete (For calculations relating to resistance to concrete) 12SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  13. 13. THE PERMISSIBLE STRESS IN STEEL • The stress in steel must not be allowed to exceed the following values under different positions to prevent cracking of concrete. • When steel is placed near the face of the members in contact with liquid  115 N/mm2 for ms Bars and  150 N/mm2 for HYSD bars. • When steel is placed on the face away from the liquid for members 225 mm or more in thickness:  125 N/mm2 for M.S. bars and  190 N/mm2 for HYSD bars.  HYSD bars– High Yielding Strength Deformed bars. 13SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  14. 14. BASE FOR FLOOR SLAB • The floor slab should be strong enough to transmit the load from the liquid and the structure itself to the ground without subsidence. The floor slab is usually 150 to 200 mm thick. • Before laying the slab the bed has to be rammed and leveled then a 75 mm thick layer of lean concrete of M100 grade should be laid and cured. 14SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  15. 15. MINIMUM REINFORCEMENT FOR WATER TANK • Minimum reinforcement required for 200mm thick sections is 0.3 % of the area of concrete section which reduced linearly to 0.2% for 450 mm thick sections. • In case of floor slab for tank resting on ground the minimum reinforcement from practical consideration should not be less than 0.3% of the gross sectional area of the floor slab. 15SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  16. 16. UNDER GROUND WATER TANKS EXAMPLE : 1 To Design an underground RC rectangular water tank of 10m x 3m x 3m. The soil surrounding the tank is in dry states & sometimes wet state. Angle of repose of soil in dry state is 30° and in wet state 6°. Density of soil is 20kN/m3. Adopt M20 grade of concrete & Fe415 steel. 16SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  17. 17. DESIGN PROCEDURE • Given data: Size = 10m x 3m x 3m Angle of repose @ dry soil = 30o Angle of repose @ wet soil = 6o Density (or) unit wt of soil= 20 kN/m3 17SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  18. 18.  Step:1 permissible & design values: Tension due to bending ( concrete), σct = 1.7 N/mm2 (Pg.no.:2,table- 1, IS 3370;part-II) Compression due to bending ( concrete), σcbc = 7 N/mm2 (Pg.no.:3, table – 2, IS 3370;part-II) or (IS 456, page no.81) Steel reinforcement for strength, σst = 115 N/mm2 (Pg.no.:3, table- 4) Design constant, m = 280 / 3* σcbc = 280 / 3*7 = 13.33 18SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  19. 19. j =1 –(n/3) n = (1/( 1+ (σst /m* σcbc))) = (1/(1+(115/13.33*7))) =0.447 j = 1-(0.447/3) = 0.85 Q = (1/2)* σcbc * n * j = (1/2) * 7 * 0.447 * 0.85 = 1.32 SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II 19
  20. 20.  Step : 2 design of long walls: Ratio for long wall = L/H = 10 / 3 = 3.33 >2 Ratio for short wall = B/H = 3/3 = 1 < 2 Long walls are designed as vertical cantilevers and Short walls are designed as continuous slab. pressure intensity @ wet soil, P = w*h* (1-sinφ/1+sin φ) = 20 *3*(1-sin 6o /1+sin6o ) = 48.64 kN/m2 20SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  21. 21. Step:3 Design of long wall By considering 1m run of wall BM calculation for vertical reinforcement: Max B.M at top (tension near water face ) Max BM @ top = pH2/33.5 = (48.65 * 3 * 3 ) / 33.5 = 13.06kNm Max B.M @ bottom = pH2/15 = 48.64 * 3 * 3 / 15 = 29.18kNm Walls of underground water tank must also be designed cracking stress consideration. 21SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  22. 22. • Step 4 Thickness of wall: Max BM, M = (σct * bD2) / 6 29.18x 106 = (1.7 * 1000 * D2 ) / 6 29.18x 106 = 283.33 * D2 D2 = 29.18x 106 / 283.33 D = 323mm provide overall depth, D = 320mm cover, d’ = 40mm Effective depth, d eff = 320-40 = 280 mm 22SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  23. 23. • step-5 Area of steel required for long wall: for outer face: Ast = (max BM / σst * j *d) = (29.18x 10 6 / 115* 0.85*280 ) = 1066.13 mm2 provide 16mm dia bar Spacing: s = ( ast / Ast )*1000 = ( (π/4*(162)/1066.13)*1000 = 189.15 mm = 180mm Provide 16mmΦ at 180 mm c/c along vertical direction 23SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  24. 24. • Step:6 Area of steel for inner face: Ast = (max BM / σst * j *d) = (13.07x 10 6 / 115* 0.85* 280) = 477.5 mm2 provide 12mm dia bar Spacing: s = ( ast / Ast )*1000 = ( (π/4*(122)/477.5)*1000 = 236.85mm = 230mm Provide 12mmΦ at 230 mm c/c along vertical direction 24SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  25. 25. • Step 7: Horizontal reinforcement for long wall: Area of distribution bar = 0.3% (b*D) = (0.3/100) (1000*320) = 960 mm2 provide 10 mm dia bar Spacing: s = ( ast / Ast )*1000 = ( (π/4*(102)/960)*1000 = 81.87mm = 100mm Provide 10mmΦ at 100 mm c/c along horizontal direction on both faces. 25SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  26. 26. • step-8 Design Of Short Wall intensity of earth pressure, p = 48.64 kN/m2 Max BM = pH2/12 = 48.64 * 3*3/ 12 = 36.48 kNm Effective span = clear span +( thickness/2) = 3 + (0.32/2)+(0.32/2) = 3.32 m Effective depth, d = √M/ Q.b = √(36.48x106/1.33*1000) = 165.61 < 280 mm adopt, d = 280 mm 26SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  27. 27. Step-9 Area of steel for short wall: Ast = (max BM / σst * j *d) = (36.48x 10 6 / 115* 0.85* 280) = 1332.84 mm2 provide 16mm dia bar Spacing: s = ( ast / Ast )*1000 = ( (π/4*(162)/1332.84)*1000 = 150mm Provide 16mmΦ at 150 mm c/c on both faces. 27SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  28. 28. • Step 10: Horizontal reinforcement for short wall: Area of distribution bar = 0.3% (b*D) = (0.3/100) (1000*320) = 960 mm2 provide 10 mm dia bar Spacing: s = ( ast / Ast )*1000 = ( (π/4*(102)/960)*1000 = 81.87mm = 100mm Provide 10mmΦ at 100 mm c/c. 28SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  29. 29. • Step-11 Design Of Roof Slab : Let us assume the overall thickness as slab , t = 150mm Assume cover = 25mm Eff.depth provided = 150 -((25/2)+(25/2)) = 125mm • Load calculation self wt of slab/ m2 = (1 * 0.15* 1 * 20) = 3 kN/m2 Live load/ m2 = 2.5 kN/m2 Floor finish = 0.5 kN/m2 Total load = 6 kN/m2 Ratio of Ly/Lx = 10.35 / 3.35 = 3.09 > 2 Hence one way slab Max. 29SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  30. 30. Moment calculation: BM (M) = W* D2/ 8 = 6 * 3.32* 3.32/ 8 = 8.26 kNm check for depth : d = √M/ Q.b = √(8.26x106/1.33*1000) = 78.80 < 100 mm adopt, d= 100 mm 30SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  31. 31. • Area of steel: Ast = (max BM / σst * j *d) = (8.26x 10 6 / 115* 0.85* 125) = 676.01 mm2 provide 10mm dia bar Spacing: s = ( ast / Ast )*1000 = ( (π/4*(162)/761.12)*1000 = 103.198mm = 100mm Provide 10mmΦ at 100 mm c/c. 31SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  32. 32. • Distribution bar : Area of distribution bar = 0.3% (b*D) = (0.3/100) (1000*150) = 450 mm2 provide 8 mm dia bar Spacing: s = ( ast / Ast )*1000 = ( (π/4*(82)/450)*1000 = 111.70mm = 110mm Provide 8mmΦ at 110 mm c/c spacing. 32SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II
  33. 33. SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II 33
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Clear introduction and a problem of under ground water tank.

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