# What is the value of 5*5?

Anonymous

### Question Description

5*5=25

Design and drawing of RC Structures CV61 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: gss_nie@yahoo.com 1 WATER TANKS 2 Learning out Come • • REVIEW DESIGN OF CIRCULAR WATER TANK RESTING ON GROUND WITH RIGID BASE 3 • • When the joints at base are flexible, hydrostatic pressure induces maximum increase in diameter at base and no increase in diameter at top When the joint at base is rigid, the base does not move 4 Design of Circular Tanks resting on ground with Rigid base 6 • 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 stresses 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 • 7 • • 1. 2. 3. 4. The 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 Reissner’s method Carpenter’s simplified method Approximate method IS code method 8 IS code method • Tables 9,10 and 11 of IS 3370 part IV gives coefficients for computing hoop tension, moment and shear for various values of H2/Dt • Hoop tension, moment and shear is computed as T= coefficient ( wHD/2) M= coefficient (wH3) V= coefficient (wH2) 9 • Thickness of wall required is computed from BM consideration M d= Qb where, Q= ½ cbcjk j=1-(k/3) m  cbc k= m  + cbc st b = 1000mm 10 IS code method • 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 A = M st • stjd Area of hoop steel in the form of rings is computed as Ast1 = T st 11 IS code method • Distribution steel and vertical steel for outer face of wall is computed from minimum steel consideration • Tensile stress computed from the following equation should be less than the permissible stress for safe design T  c= 1000 t+ ( m − 1 ) A st the permissible stress is 0.27 fck 12 IS code method • Base slab thickness generally varies from 150mm to 250 mm and minimum steel is distributed to top and bottom of slab. 13 Design Problem No.1 on Circular Tanks resting on ground with Rigid base 14 A cylindrical tank of capacity 7,00,000 liters is resting on good unyielding ground. The depth of tank is limited to 5m. A free board of 300 mm may be provided. The wall and the base slab are cast integrally. Design the tank using M20 concrete and Fe415 grade steel . Draw the following Plan at base Cross section through centre of tank. 15 Step 1: Dimension of tank H= 5-0.3 = 4.7 and volume V = 700 m3 A=700/4.7 = 148.94 m2 D= (4 x 148.94/) = 13.77 14 m 16 Step 2: Analysis for hoop tension and bending moment One meter width of the wall is considered and the thickness of the wall is estimated as t=30H+50 = 191 mm. The thickness of wall is assumed as 200 mm. Referring to table 9 of IS3370 (part IV), the maximum coefficient for hoop tension = 0.575 17 Step 2: Analysis for hoop tension and bending moment (Contd.) Tmax=0.575 x 10 x 4.7 x 7 =189.175 kN Referring to table 10 of IS3370 (part IV), the maximum coefficient for bending moment = -0.0146 (produces tension on water side) Mmax= 0.0146 x 10 x 4.73=15.15 kN-m 18 Step 3: Design of section: For M20 concrete cbc=7, For Fe415 steel st=150 MPa and m=13.33 for M20 concrete and Fe415 steel The design constants are: m  cbc k = = 0 .39 m  +  j=1-(k/3)=0.87 cbc st Q= ½ cbcjk = 1.19 Effective depth is calculated as 6 M 15 . 15 x 10 d = = = 112 . 94 mm Qb 1 . 19 x 1000 19 Step 3: Design of section: (Contd.) Let over all thickness be 200 mm with effective cover 33 mm dprovided=167 mm 6 M 15 . 15 x 10 2 A = = = 695 . 16 mm st jd 150 x 0 . 87 x 167 Spacing  ofst 16 mm diameter bar = 201 x 1000 = 289 . 23 mmc / c 695 . 16 (Max spacing 3d=501mm) Provide #16@275 c/c as vertical reinforcement on water face 20 Step 3: Design of section: (Contd.) Hoop steel: 3 T 189 . 275 x 10 2 A = = = 1261 mm st 1  150 st x 1000 Spacing of 12 mm diameter bar = 113 = 89 . mmc /c 1261 Provide #12@80 c/c as hoop reinforcement on water face Actual area of steel provided 113 x 10002 A = = 141 . 5 m st 80 21 Step 4: Check for tensile stress: 3 T 189 . 275 x 10 2  = = = 0 . 87 N / m c 1000 t + ( m − 1 ) A 1000 x 200 + ( 13 . 33 − 1 ) x 141 . 5 st Permissible stress = 0.27fck=1.2 N/mm2 > c Safe 22 Step 5: Distribution Steel: Minimum area of steel is 0.24% of concrete area Ast=(0.24/100) x1000 x 200 = 480 mm2 50 . 24 x 1000 = 104 . 7 . mm / c Spacing of 8 mm diameter bar = 480 Provide #8 @ 100 c/c as vertical and horizontal distribution on the outer face. 23 Step 5: Base slab: The thickness of base slab shall be 150 mm. The base slab rests on firm ground, hence only minimum reinforcement is provided. Ast=(0.24/100) x1000 x 150 = 360 mm2 Reinforcement for each face = 180 mm2 50 . 24 x 1000 = 279 . mm / c Spacing of 8 mm diameter bar = 180 Provide #8 @ 250 c/c as vertical and horizontal distribution on the outer face. 24 25 Design Problem No.2 on Circular Tanks resting on ground with Rigid base 26 Design a circular water tank to hold 5,50,000 liters of water. Assume rigid joints between the wall and base slab. Adopt M20 concrete and Fe 415 steel. Sketch details of reinforcements. 27 Step 1: Dimension of tank Volume of tank V=550 m3 Assume H= 4.5 A=550/4.5 = 122.22 m2 D= (4 x 122.22/) = 12.47 12.5 m 28 Step 2: Analysis for hoop tension and bending moment One meter width of the wall is considered and the thickness of the wall is estimated as t=30H+50 = 185 mm. The thickness of wall is assumed as 200 mm. 2 2 H 4 . 5 = = 8 . 1  8 Dt 12 . 5  0 . 2 Referring to table 9 of IS3370 (part IV), the maximum coefficient for hoop tension = 0.575 29 Step 2: Analysis for hoop tension and bending moment (Contd.) Tmax=0.575 x 10 x 4.5 x 6.25 =161.72 kN Referring to table 10 of IS3370 (part IV), the maximum coefficient for bending moment = -0.0146 (produces tension on water side) Mmax= 0.0146 x 10 x 4.53=13.3 kN-m 30 Step 3: Design of section: For M20 concrete cbc=7, For Fe415 steel st=150 MPa and m=13.33 for M20 concrete and Fe415 steel The design constants are: m  cbc k = = 0 .39 m  +  cbc st j=1-(k/3)=0.87 Q= ½ cbcjk = 1.19 Effective depth is calculated as 6 M 13 . 3 x 10 d = = = 105 . 7 mm Qb 1 . 19 x 1000 31 Step 3: Design of section: (Contd.) Let over all thickness be 200 mm with effective cover 33 mm dprovided=167 mm 6 M 13 . 3 x 10 2 A == = 610 . 27 mm st  jd 150 x 0 . 87 x 167 st Spacing of 16 mm diameter bar = 201 x 1000 = 329 . 36 mmc / c 610 . 27 (Max spacing 3d=501mm) Provide #16@300 c/c as vertical reinforcement on water face 32 Step 3: Design of section: (Contd.) Hoop steel: 3 T 161 . 72 x 10 2 A = = = 1078 . 13 mm st 1  150 st Spacing of 12 mm diameter bar = 113 x 1000 = 104 mm /c 1078 . 13 Provide #12@100 c/c as hoop reinforcement on water face Actual area of steel provided 113 x 1000 2 A = = 1130 mm st 100 33 Step 4: Check for tensile stress: 3 T 161 . 72 x 10 2  = = = 0 . 76 N / m c 1000 t + ( m − 1 ) A 1000 x 200 + ( 13 . 33 − 1 ) x 113 st Permissible stress = 0.27fck=1.2 N/mm2 > c Safe 34 Step 5: Distribution Steel: Minimum area of steel is 0.24% of concrete area Ast=(0.24/100) x1000 x 200 = 480 mm2 50 . 24 x 1000 = 104 . 7 . mm / c Spacing of 8 mm diameter bar = 480 Provide #8 @ 100 c/c as vertical and horizontal distribution on the outer face. 35 Step 5: Base slab: The thickness of base slab shall be 150 mm. The base slab rests on firm ground, hence only minimum reinforcement is provided. Ast=(0.24/100) x1000 x 150 = 360 mm2 Reinforcement for each face = 180 mm2 50 . 24 x 1000 = 279 . mm / c Spacing of 8 mm diameter bar = 180 Provide #8 @ 250 c/c as vertical and horizontal distribution on the outer face. 36 37 A TYPICAL DRAWING 38 39 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: gss_nie@yahoo.com 40

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