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# Unit i

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### Unit i

1. 1. DESIGN OF CONCRETE STRUCTURES - I Problems 1 Design a reinforced concrete beam spaced at a clear distance of 5 m and supported on two walls 230 mm thick. The beam carries a super-imposed load of 2 kN/m. The width of beam is half the depth of the beam. Use M20 concrete and Fe415 steel. Data Given : L = 5 m ; tw = 230 mm ; w l = 2 kN/m; b = 175 mm; M20 ; Fe415 Required : Design a reinforced concrete beam Solution : Effective span = 5 + 0.23 = 5.23 Effective depth = 5230 =261.5 mm 20 Clear cover is 20 mm and use 16 mm dia bars Overall depth ‘D’ = 261.5 + 20 + 8 = 289.5 mm Provide Overall depth ‘D’ = 300 mm Width of the beam ‘b’ = 150 mm Effective depth d = 300 – 28 = 272 mm Again Effective span i. 5.23 m ii. 5+ 0.272 = 5.272 m Consider Le = 5.23 m Calculation of Loads: Dead Load = 0.15 x 0.30 x 25 kN/m = 1.750 kN/m Total = 2.875 Live load = 1.125 kN/m 2.875x5.232 Bending Moment = M = = 9.83 kN-m 8 1 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur
2. 2. DESIGN OF CONCRETE STRUCTURES - I Moment of Resistance = MR = 0.914 x 150 x 2722 = 10.14 kN-m M < MR, the section is underreinforced Area of steel Ast = No. of bars = 9.83x106 = 173.8 mm2 230x0.904x272 173.8x4 = 0.86 2 x16 Provide 2 bars of 16 mm diameter (Ast = 402 mm2) 2. Design a reinforced concrete beam spaced at a clear distance of 7 m and supported on two walls 230 mm thick. The beam carries a super-imposed load of 9 kN/m. The width of beam is restricted to 230 mm. Use M20 concrete and Fe415 steel. Data Given : L = 7 m ; tw = 230 mm ; w l = 9 kN/m; b = D/2; M20 ; Fe415 Required : Design a reinforced concrete beam Solution : Effective span = 7 + 0.23 = 7.23 Effective depth = 7230 =361.5 mm 20 Clear cover is 20 mm and use 16 mm dia bars Overall depth ‘D’ = 361.5 + 20 + 8 = 389.5 mm Provide Overall depth ‘D’ = 400 mm Effective depth d = 400 – 28 = 372 mm Again Effective span (Le) i. 7.23 m ii. 7 + 0.372 = 7.372 m Consider Le = 7.23 m 2 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur
3. 3. DESIGN OF CONCRETE STRUCTURES - I Calculation of Loads: Dead Load = 0.230 x 0.400 x 25 = 2.30 kN/m = 9.00 kN/m Total = 11.30 kN/m Live load Bending Moment = M = 11.300x7.232 = 73.83 kN-m 8 Moment of Resistance (MR) = 0.914 x 230 x 3722 = 29091084.48N-mm or 29.10 kN-m M > MR, the section is Overreinforced Assume n = nc = 13.33x7 x372= 107.40 mm 13.33x7  230 107.40  28 Stress in concrete at compression steel level  / = x7 = 5.175 N/mm2 cbc 107.40 Equating moment of resistance and find the area of steel in compression 230 107.40 7  107.40  6  372   1.5 13.33  1 A sc  5.175 372 28  73.83 10 2 3   Asc = 1323.80 mm2 No. of bars = 1323.80x4 = 6.58 x162 Provide 7 bars of 16 mm diameter (Asc = 1407 mm2) Total Compression = Total Tension, find the Ast 230 107.40 7  1.5 13.33  1 1407 5.175= A st  230 2 Ast = 977.23 mm2 3 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur
4. 4. DESIGN OF CONCRETE STRUCTURES - I No. of bars = 977.23x4 = 4.86 2 x16 Provide 5 bars of 16 mm diameter (Ast = 1005 mm2) Alternative Solution: Lever arm a = 372 107.40 = 336.2 mm 3 M1 for singly reinforced beam = 230 107.40 7  107.40   372  2 3   = 29066843.4N-mm or 29.10 kN-m For this moment, calculate Ast1 Ast1 = 6 29.10 10 = 376.33 mm2 230 336.2 Balance Bending moment = M2 = (73.83-29.10) x 106 = 44.73 x 106 N-mm Tensile Steel required for this BM Ast1 = 6 44.73 10 = 565.34 mm2 230 372 28 Total Ast = 376.33 + 565.34 = 941.67mm2 No. of bars = 941.67  4 = 4.68 2 x16 Provide 5 bars of 16 mm diameter (Ast = 1005 mm2) Compression Steel Asc = No. of bars = 13.33  372  107.40  565.34 =1322.11 mm2 1.5  13.33  1  107.40  28 1322.11x4 = 6.58 x162 Provide 7 bars of 16 mm diameter (Asc = 1407 mm2) 4 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur
5. 5. DESIGN OF CONCRETE STRUCTURES - I 3. Design a slab for living room of the residential building having a dimension of 2.5m x 7m. The simply supported along the edges. The width of bearing of wall is 300 mm. The live load acting on the slab is 2 kN/m2. Adopt M20 grade of concrete and Fe 415 steel. Data Given : Size = 2.5 m x 7 m ; tw = 300 mm ; w l = 2 kN/m2 ; M20 ; Fe415; Simply supported Required : Design a reinforced slab Solution : Effective span Short span = 2.5 + 0.3 = 2.8 m Long span = 7.0 + 0.3 = 7.3 m Span Ratio = 7 .3 = 2.61 > 2, One way slab 2 .8 Effective depth = 2800 =140 mm 20 Clear cover is 15 mm and use 10 mm dia bars Overall depth ‘D’ = 140 + 15 + 5 = 160 mm Provide Overall depth ‘D’ = 175 mm Effective depth d = 175 – 20 = 155 mm Again find the Effective span i. 2.80 m ii. 2.5+ 0.155 = 2.655 m Le = 2.655 m Calculation of Loads: Consider 1 m wide slab Dead load = = 4.375 kN/m2 = 2.000 kN/m2 Total = Live load 0.175 x 25 6.375 kN/m2 5 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur
6. 6. DESIGN OF CONCRETE STRUCTURES - I Bending Moment = M = 6.375x2.6552 = 5.618 kN-m 8 Moment of Resistance = MR = 0.914 x1000 x 1552 = 21.96 kN-m M < MR , the section is designed as underreinforced section Main Reinforcement: 5.618x106 Area of steel Ast = = 174.3 mm2 230x0.904x155 Ast min = 0.12 x1000x175= 210 mm2 100 Spacing of 10 dia bars = 1000xx102 = 374 mm 4x210 Maximum spacing : Whichever is lesser of the following i. 374 mm ii. 300 mm iii. 3 x 130 = 390 mm Spacing Provided = 300 mm Provide 10 mm diameter @ 300 mm c/c (Ast = 378 mm2) Distribution Reinforcement: Asd = 0.12 x1000x175= 210 mm2 100 Spacing of 8 dia bars = 1000xx82 = 239 mm 4x210 Maximum spacing : Whichever is lesser of the following i. 239 mm ii. 450 mm iii. 5 x 154 = 770 mm Provide 8 mm diameter @ 225 mm c/c (Ast = 224 mm2) 6 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur
7. 7. DESIGN OF CONCRETE STRUCTURES - I 4. A simply supported beam of 300 mm wide and 500 mm effective depth carries a uniformly distributed load of 50 kN/m including its own weight over an effective span of 6m. Design the shear reinforcement by vertical stirrups. The tensile reinforcement consists of 3 bars of 20 mm diameter. Adopt M20 grade concrete and Fe415 steel. Data Given : L = 6 m ; b=300 mm ; d = 500 mm ; w = 50 kN/m; Ast =3-20 mm; M20 ; Fe415 Required : Design a vertical stirrups Solution Ast = : 3xx 20 2 = 943 mm2 4 Shear force = V = 50x6 = 75 kN 4 Nominal shear stress = v = %p= 75 x10 3 = 0.5 N/mm2 300 x500 100 x3xx 20 2 = 0.63 4 x300 x500 Table 23 of IS 456,Permissible shear stress in concrete c = 0.326 N/mm2 Table 23 of IS 456, Maximum shear stress cmax = 1.8 N/mm2 Cl. B5.4 of IS 456 v > c < cmax , Shear reinforcement shall be provided. Net shear force = 75000 – 0.326 x 300 x 500 = 26100 N Use two legged 10 mm diameter stirrups sv = 2 xx10 2 x 230 x500 = 692 mm 4 x 26100 7 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur
8. 8. DESIGN OF CONCRETE STRUCTURES - I Spacing of stirrups shall not exceed the following i. 692 mm ii. 450 mm iii. 0.75 x 500 = 375 mm Provide 2 legged 10mm dia stirrups @ 350 mm c/c 8 V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur