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Flat Slab Analysis & Design, (In accordance with BS8110:PART 1:1997).

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- 1. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Date Calc. by Dr. C. Sachpazis Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Chk'd by Date 18/01/2014 App'd by Date FLAT SLAB DESIGN TO BS8110:PART 1:1997 Slab geometry Span of slab in x-direction; Spanx = 7200 mm Span of slab in y-direction; Spany = 7200 mm Column dimension in x-direction; lx = 400 mm Column dimension in y-direction; ly = 400 mm External column dimension in x-direction; lx1 = 250 mm External column dimension in y-direction; ly1 = 250 mm Edge dimension in x-direction; ex = lx1 / 2 = 125 mm Edge dimension in y-direction; ey = ly1 / 2 = 125 mm Effective span of internal bay in x direction; Lx = Spanx – lx = 6800 mm Effective span of internal bay in y direction; Ly = Spany – ly = 6800 mm Effective span of end bay in x direction; Lx1 = Spanx – lx / 2 = 7000 mm Effective span of end bay in y direction; Ly1 = Spany – ly / 2 = 7000 mm ex B C Span x Span x lx ey 1 A lx l y1 Sp an y lx1 ly Sp an y 2 ly 3 h Slab details Depth of slab; h = 250 mm 1
- 2. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date 2 Characteristic strength of concrete; fcu = 35 N/mm Characteristic strength of reinforcement; fy = 500 N/mm Characteristic strength of shear reinforcement; fyv = 500 N/mm Material safety factor; γm = 1.15 2 2 Cover to bottom reinforcement; c = 20 mm Cover to top reinforcement; c’ = 20 mm Loading details 2 Characteristic dead load; Gk = 7.000 kN/m Characteristic imposed load; Qk = 5.000 kN/m 2 Dead load factor; γG = 1.4 Imposed load factor; γQ = 1.6 Total ultimate load; Nult = (Gk × γG) + (Qk × γQ) = 17.800 kN/m Moment redistribution ratio; βb = 1.0 Ratio of support moments to span moments; i = 1.0 2 DESIGN SLAB IN THE X-DIRECTION SAGGING MOMENTS End bay A-B Effective span; L = 7000 mm Depth of reinforcement; d = 200 mm Midspan moment; m = (Nult × L ) / (2 × (1 + √(1 + i)) ) = 74.823 kNm/m Support moment; m’ = i × m = 74.823 kNm/m 2 2 Design reinforcement Lever arm; 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m / (d × fcu) = 0.053 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 187.3 mm Area of reinforcement designed; 2 As_des = m / (z × fy / γm) = 919 mm /m 2 Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 919 mm /m 2 Provide 20 dia bars @ 150 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 2094 mm /m PASS - Span reinforcement is OK Check deflection Design service stress; fs = 2 × fy × As_req / (3 × As_prov × βb) = 146 N/mm 2 2 2 2 Modification factor; k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.545 Allowable span to depth ratio; 0.9 × 26 × k1 = 36.151 Actual span to depth ratio; L / d = 35.000 2
- 3. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date PASS - Span to depth ratio is OK Internal bay B-C Effective span; L = 6800 mm Depth of reinforcement; d = 202 mm Midspan moment; m = (Nult × L ) / (2 × (√(1 + i) + √(1 + i)) ) = 51.442 kNm/m Support moment; m’ = i × m = 51.442 kNm/m 2 2 Design reinforcement Lever arm; 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m / (d × fcu) = 0.036 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 191.9 mm 2 Area of reinforcement designed; As_des = m / (z × fy / γm) = 617 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 617 mm /m 2 2 Provide 16 dia bars @ 200 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1005 mm /m PASS - Span reinforcement is OK Check deflection 2 Design service stress; fs = 2 × fy × As_req / (3 × As_prov × βb) = 204 N/mm Modification factor; k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.601 Allowable span to depth ratio; 0.9 × 26 × k1 = 37.469 Actual span to depth ratio; L / d = 33.663 2 2 2 PASS - Span to depth ratio is OK HOGGING MOMENTS – INTERNAL STRIP Penultimate column B3 Consider the reinforcement concentrated in half width strip over the support Depth of reinforcement; d’ = 200 mm Support moment; m’ = 2 × i × m = 149.646 kNm/m Lever arm; K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 2 K = m’ / (d’ × fcu) = 0.107 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 172.5 mm Area of reinforcement required; 2 As_des = m’ / (z × fy / γm) = 1996 mm /m 2 Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1996 mm /m 2 Provide 20 dia bars @ 150 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 2094 mm /m PASS - Support reinforcement is OK 3
- 4. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date Internal column C3 Consider the reinforcement concentrated in half width strip over the support Depth of reinforcement; d’ = 200 mm Support moment; m’ = 2 × i × m = 102.884 kNm/m Lever arm; K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 2 K = m’ / (d’ × fcu) = 0.073 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 182.1 mm 2 Area of reinforcement required; As_des = m’ / (z × fy / γm) = 1300 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1300 mm /m 2 2 Provide 20 dia bars @ 200 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1571 mm /m PASS - Support reinforcement is OK HOGGING MOMENTS – EXTERNAL STRIP Penultimate column B1, B2 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm Depth of reinforcement; d’ = 200 mm Support moment; m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 158.265 kNm/m Lever arm; 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m’ / (d’ × fcu) = 0.113 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 170.5 mm Area of reinforcement required; 2 As_des = m’ / (z × fy / γm) = 2134 mm /m 2 Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 2134 mm /m 2 Provide 20 dia bars @ 125 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 2513 mm /m PASS - Support reinforcement is OK Internal column C1, C2 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm Depth of reinforcement; d’ = 200 mm Support moment; m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 108.810 kNm/m 4
- 5. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Lever arm; Date Chk'd by Date 18/01/2014 App'd by Date 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m’ / (d’ × fcu) = 0.078 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 180.9 mm 2 Area of reinforcement required; As_des = m’ / (z × fy / γm) = 1383 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1383 mm /m 2 2 Provide 20 dia bars @ 200 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1571 mm /m PASS - Support reinforcement is OK Corner column A1 Depth of reinforcement; d’ = 206 mm Total load on column; S = ((Spanx / 2) + ex) × ((Spany / 2) + ey) × Nult = 247 kN Area of column head; A = lx × ly1 = 0.100 m Support moment; m’ = S × (1 – (Nult × A / S) ) / 2 = 99.639 kNm/m Lever arm; K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 1/3 2 2 K = m’ / (d’ × fcu) = 0.067 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 189.3 mm 2 Area of reinforcement required; As_des = m’ / (z × fy / γm) = 1211 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1211 mm /m 2 2 Provide 16 dia bars @ 150 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1340 mm /m PASS - Support reinforcement is OK Edge column A2, A3 Depth of reinforcement; d’ = 202 mm Total load on column; S = Spanx × (Spany / 2 + ey) × Nult = 477 kN Area of column head; A = lx1 × ly = 0.100 m Support moment; m’ = S × (1 – (Nult × A / S) ) / 5.14 = 78.476 kNm/m Lever arm; K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 1/3 2 2 K = m’ / (d’ × fcu) = 0.055 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 188.8 mm Area of reinforcement required; 2 As_des = m’ / (z × fy / γm) = 956 mm /m 2 Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 956 mm /m 2 Provide 16 dia bars @ 175 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1149 mm /m 5
- 6. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date PASS - Support reinforcement is OK Between columns 1-2, 2-3 Around the perimeter between the column heads provide a minimum of 50% of the required end span bottom reinforcement. Area of reinforcement required; 2 As_req = Asx1 / 2 = 1047 mm /m Provide 16 dia bars @ 150 centres - 'U' bars with 1600 mm long legs Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1340 mm /m PASS - Edge reinforcement is OK Distribution reinforcement Provide 12 dia bars @ 300 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 377 mm /m DESIGN SLAB IN THE Y-DIRECTION SAGGING MOMENTS End bay 1-2 Effective span; L = 7000 mm Depth of reinforcement; d = 220 mm Midspan moment; m = (Nult × L ) / (2 × (1 + √(1 + i)) ) = 74.823 kNm/m Support moment; m’ = i × m = 74.823 kNm/m 2 2 Design reinforcement Lever arm; 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m / (d × fcu) = 0.044 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 208.6 mm 2 Area of reinforcement designed; As_des = m / (z × fy / γm) = 825 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 825 mm /m 2 2 Provide 20 dia bars @ 200 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1571 mm /m PASS - Span reinforcement is OK Check deflection 2 Design service stress; fs = 2 × fy × As_req / (3 × As_prov × βb) = 175 N/mm Modification factor; k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.579 Allowable span to depth ratio; 0.9 × 26 × k1 = 36.942 Actual span to depth ratio; L / d = 31.818 2 2 2 PASS - Span to depth ratio is OK Internal bay 2-3 Effective span; L = 6800 mm 6
- 7. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date Depth of reinforcement; d = 222 mm Midspan moment; m = (Nult × L ) / (2 × (√(1 + i) + √(1 + i)) ) = 51.442 kNm/m Support moment; m’ = i × m = 51.442 kNm/m 2 2 Design reinforcement Lever arm; 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m / (d × fcu) = 0.030 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 210.9 mm 2 Area of reinforcement designed; As_des = m / (z × fy / γm) = 561 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 561 mm /m 2 2 Provide 16 dia bars @ 200 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1005 mm /m PASS - Span reinforcement is OK Check deflection 2 Design service stress; fs = 2 × fy × As_req / (3 × As_prov × βb) = 186 N/mm Modification factor; k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.798 Allowable span to depth ratio; 0.9 × 26 × k1 = 42.062 Actual span to depth ratio; L / d = 30.631 2 2 2 PASS - Span to depth ratio is OK HOGGING MOMENTS – INTERNAL STRIP Penultimate column C2 Consider the reinforcement concentrated in half width strip over the support Depth of reinforcement; d’ = 220 mm Support moment; m’ = 2 × i × m = 149.646 kNm/m Lever arm; K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 2 K = m’ / (d’ × fcu) = 0.088 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 195.7 mm Area of reinforcement required; 2 As_des = m’ / (z × fy / γm) = 1758 mm /m 2 Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1758 mm /m 2 Provide 20 dia bars @ 150 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 2094 mm /m PASS - Support reinforcement is OK Internal column C3 Consider the reinforcement concentrated in half width strip over the support Depth of reinforcement; d’ = 220 mm 7
- 8. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Support moment; m’ = 2 × i × m = 102.884 kNm/m Lever arm; Date K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 2 K = m’ / (d’ × fcu) = 0.061 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 204.0 mm Area of reinforcement required; 2 As_des = m’ / (z × fy / γm) = 1160 mm /m 2 Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1160 mm /m 2 Provide 20 dia bars @ 200 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1571 mm /m PASS - Support reinforcement is OK HOGGING MOMENTS – EXTERNAL STRIP Penultimate column A2, B2 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm Depth of reinforcement; d’ = 220 mm Support moment; m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 158.265 kNm/m Lever arm; 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m’ / (d’ × fcu) = 0.093 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 194.1 mm Area of reinforcement required; 2 As_des = m’ / (z × fy / γm) = 1875 mm /m 2 Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1875 mm /m 2 Provide 20 dia bars @ 150 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 2094 mm /m PASS - Support reinforcement is OK Internal column A3, B3 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm Depth of reinforcement; d’ = 220 mm Support moment; m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 108.810 kNm/m Lever arm; 2 K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 K = m’ / (d’ × fcu) = 0.064 8
- 9. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 203.0 mm 2 Area of reinforcement required; As_des = m’ / (z × fy / γm) = 1233 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 1233 mm /m 2 2 Provide 20 dia bars @ 200 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1571 mm /m PASS - Support reinforcement is OK Edge column B1, C1 Depth of reinforcement; d’ = 222 mm Total load on column; S = (Spanx / 2 + ex) × Spany × Nult = 477 kN Area of column head; A = ly1 × lx = 0.100 m Support moment; m’ = S × (1 – (Nult × A / S) ) / 5.14 = 78.476 kNm/m Lever arm; K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176 2 1/3 2 2 K = m’ / (d’ × fcu) = 0.045 Compression reinforcement is not required z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 210.1 mm 2 Area of reinforcement required; As_des = m’ / (z × fy / γm) = 859 mm /m Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm /m Area of reinforcement required; As_req = max(As_des, As_min) = 859 mm /m 2 2 Provide 16 dia bars @ 175 centres Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1149 mm /m PASS - Support reinforcement is OK Between columns A-B, B-C Around the perimeter between the column heads provide a minimum of 50% of the required end span bottom reinforcement. Area of reinforcement required; 2 As_req = Asy1 / 2 = 785 mm /m Provide 16 dia bars @ 200 centres - 'U' bars with 1600 mm long legs Area of reinforcement provided; 2 2 As_prov = π × D / (4 × s) = 1005 mm /m PASS - Edge reinforcement is OK PUNCHING SHEAR Corner column A1 Design shear transferred to column; Vt = ((0.45 × Spanx) + ex) × ((0.45 × Spany) + ey) × Nult = 202 kN Design effective shear transferred to column; Veff = 1.25 × Vt = 252 kN Area of tension steel in x-direction; Asx_ten = Ascorner = 1340 mm /m Area of tension steel in y-direction; Asy_ten = Ascorner = 1340 mm /m Column perimeter; uc = lx1 + ly = 650 mm Average effective depth of reinforcement; d = h – c - φp = 214 mm 2 2 9
- 10. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Date Calc. by Dr. C. Sachpazis Chk'd by Date 18/01/2014 Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 1.811 N/mm App'd by Date 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 1292 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 1731 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.707 N/mm 2 Nominal design shear stress at perimeter; v = Veff / (u × d) = 0.911 N/mm Shear reinforcement required at perimeter; Asv_req = (v - vc) × u × d / (0.95 × fyv) = 119 mm vc < v <= 1.6 × vc 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 1613 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 2161 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.707 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.730 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 16 mm Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 1934 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 2592 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.707 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.609 N/mm v < vc no shear reinforcement required Penultimate edge column A2 Design shear transferred to column; Vt = ((0.45 × Spanx) + ex) × (1.05 × Spany) × Nult = 453 kN Design effective shear transferred to column; Veff = 1.4 × Vt = 634 kN Area of tension steel in x-direction; Asx_ten = Asx_edge = 1148 mm /m 2 10
- 11. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Date Calc. by Dr. C. Sachpazis Chk'd by Date 18/01/2014 App'd by Date 2 Area of tension steel in y-direction; Asy_ten = Asy1e = 2094 mm /m Column perimeter; uc = (2 × lx1)+ ly = 900 mm Average effective depth of reinforcement; d = h – c - φp = 214 mm Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.292 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2184 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 3588 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.757 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.356 N/mm 1.6 × vc < v <= 2 × vc Shear reinforcement required at perimeter; Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 947 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2826 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 4628 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.756 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.048 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 372 mm Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 3468 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 5669 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.756 N/mm 2 Nominal design shear stress at perimeter; v = Veff / (u × d) = 0.854 N/mm Shear reinforcement required at perimeter; Asv_req = (v - vc) × u × d / (0.95 × fyv) = 154 mm vc < v <= 1.6 × vc 2 11
- 12. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Date Calc. by Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Dr. C. Sachpazis Chk'd by Date 18/01/2014 App'd by Date Shear reinforcement at a perimeter of 3.75d - (803 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4110 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 6710 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.755 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.721 N/mm v < vc no shear reinforcement required Internal edge column A3 Design shear transferred to column; Vt = ((0.45 × Spanx) + ex) × Spany × Nult = 431 kN Design effective shear transferred to column; Veff = 1.4 × Vt = 604 kN Area of tension steel in x-direction; Asx_ten = Asx_edge = 1148 mm /m 2 2 Area of tension steel in y-direction; Asy_ten = Asye = 1570 mm /m Column perimeter; uc = (2 × lx1)+ ly = 900 mm Average effective depth of reinforcement; d = h – c - φp = 214 mm Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.135 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2184 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 2989 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = Veff / (u × d) = 1.292 N/mm Shear reinforcement required at perimeter; Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 945 mm 1.6 × vc < v <= 2 × vc 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2826 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 3862 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.712 N/mm 12
- 13. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 Nominal design shear stress at perimeter; App'd by Date 2 v = Veff / (u × d) = 0.998 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 365 mm Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 3468 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 4734 mm2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.712 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.814 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 159 mm Shear reinforcement at a perimeter of 3.75d - (803 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4110 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 5607 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.711 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.686 N/mm v < vc no shear reinforcement required Penultimate edge column B1 Design shear transferred to column; Vt = (1.05 × Spanx) × ((0.45 × Spany) + ey) × Nult = 453 kN Design effective shear transferred to column; Veff = 1.4 × Vt = 634 kN Area of tension steel in x-direction; Asx_ten = Asx1e = 2513 mm /m Area of tension steel in y-direction; Asy_ten = Asy_edge = 1148 mm /m 2 2 Column perimeter; uc = lx + (2 × ly1) = 900 mm Average effective depth of reinforcement; d = h – c - φp = 214 mm Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.292 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2184 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 4066 mm 2 13
- 14. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.789 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.356 N/mm 1.6 × vc < v <= 2 × vc Shear reinforcement required at perimeter; Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 789 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2826 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 5241 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.788 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.048 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 331 mm Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 3468 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 6416 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.788 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.854 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 104 mm Shear reinforcement at a perimeter of 3.75d - (803 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4110 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 7592 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.787 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.721 N/mm v < vc no shear reinforcement required 14
- 15. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date Penultimate central column B2 Design shear transferred to column; Vt = (1.05 × Spanx) × (1.05 × Spany) × Nult = 1017 kN Design effective shear transferred to column; Veff = 1.15 × Vt = 1170 kN Area of tension steel in x-direction; Asx_ten = Asx1e = 2513 mm /m Area of tension steel in y-direction; Asy_ten = Asy1e = 2094 mm /m 2 2 Column perimeter; uc = 2 × (lx + ly) = 1600 mm Average effective depth of reinforcement; d = h – c - φp = 214 mm Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.417 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4168 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 9601 mm2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.847 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.312 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 872 mm Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 5452 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 12559 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.847 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.003 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 382 mm Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 6736 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 15516 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 15
- 16. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Date Calc. by Dr. C. Sachpazis Chk'd by Date 18/01/2014 App'd by Date 2 vc = 0.847 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.812 N/mm v < vc no shear reinforcement required Internal central column B3 Design shear transferred to column; Vt = (1.05 × Spanx) × Spany × Nult = 969 kN Design effective shear transferred to column; Veff = 1.15 × Vt = 1114 kN Area of tension steel in x-direction; Asx_ten = Asx1i = 2094 mm /m 2 2 Area of tension steel in y-direction; Asy_ten = Asye = 1570 mm /m Column perimeter; uc = 2 × (lx + ly) = 1600 mm Average effective depth of reinforcement; d = h – c - φp = 214 mm Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.254 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4168 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 7636 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.785 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.249 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 872 mm Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 5452 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 9988 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = Veff / (u × d) = 0.955 N/mm Shear reinforcement required at perimeter; Asv_req = (v - vc) × u × d / (0.95 × fyv) = 418 mm vc < v <= 1.6 × vc 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 6736 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) 16
- 17. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Date Calc. by Dr. C. Sachpazis Chk'd by Date 18/01/2014 As_ten = 12340 mm App'd by Date 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.785 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.773 N/mm v < vc no shear reinforcement required Internal edge column C1 Design shear transferred to column; Vt = Spanx × ((0.45 × Spany) + ey) × Nult = 431 kN Design effective shear transferred to column; Veff = 1.4 × Vt = 604 kN Area of tension steel in x-direction; Asx_ten = Asxe = 1570 mm /m Area of tension steel in y-direction; Asy_ten = Asy_edge = 1148 mm /m Column perimeter; uc = lx + (2 × ly1) = 900 mm 2 2 (Library item: Flat slab shear map C1) d = h – c - φp = 214 mm Average effective depth of reinforcement; Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.135 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2184 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 2989 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = Veff / (u × d) = 1.292 N/mm Shear reinforcement required at perimeter; Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 945 mm 1.6 × vc < v <= 2 × vc 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 2826 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 3862 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = Veff / (u × d) = 0.998 N/mm Shear reinforcement required at perimeter; Asv_req = (v - vc) × u × d / (0.95 × fyv) = 365 mm vc < v <= 1.6 × vc 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) 17
- 18. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 3468 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 4734 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.712 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.814 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 159 mm Shear reinforcement at a perimeter of 3.75d - (803 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4110 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 5607 mm2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.711 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.686 N/mm v < vc no shear reinforcement required Internal central column C2 Design shear transferred to column; Vt = Spanx × (1.05 × Spany) × Nult = 969 kN Design effective shear transferred to column; Veff = 1.15 × Vt = 1114 kN Area of tension steel in x-direction; Asx_ten = Asxe = 1570 mm /m Area of tension steel in y-direction; Asy_ten = Asy1i = 2094 mm /m 2 2 Column perimeter; uc = 2 × (lx + ly) = 1600 mm Average effective depth of reinforcement; d = h – c - φp = 214 mm Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.254 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4168 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 7636 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.785 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.249 N/mm 18
- 19. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Date Calc. by Dr. C. Sachpazis Chk'd by Date 18/01/2014 App'd by Date vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 872 mm Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 5452 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 9988 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = Veff / (u × d) = 0.955 N/mm Shear reinforcement required at perimeter; Asv_req = (v - vc) × u × d / (0.95 × fyv) = 418 mm vc < v <= 1.6 × vc 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 6736 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 12340 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.785 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.773 N/mm v < vc no shear reinforcement required Internal column C3 Design shear transferred to column; Vt = Spanx × Spany × Nult = 923 kN Design effective shear transferred to column; Veff = 1.15 × Vt = 1061 kN Area of tension steel in x-direction; Asx_ten = Asxi = 1570 mm /m Area of tension steel in y-direction; Asy_ten = Asyi = 1570 mm /m Column perimeter; uc = 2 × (lx + ly) = 1600 mm 2 2 Average effective depth of reinforcement; d = h – c - φp = 214 mm Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm Design shear stress at column perimeter; v0 = Veff / (uc × d) = 3.099 N/mm 2 2 PASS - Maximum concrete shear stress not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 4168 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 6544 mm 2 19
- 20. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Date Chk'd by Date 18/01/2014 App'd by Date Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.746 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 1.190 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 834 mm Shear reinforcement at a perimeter of 2.25d - (482 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 5452 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 8560 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.746 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.910 N/mm vc < v <= 1.6 × vc Shear reinforcement required at perimeter; 2 Asv_req = (v - vc) × u × d / (0.95 × fyv) = 403 mm Shear reinforcement at a perimeter of 3.00d - (642 mm) Length of shear perimeter; u = uc + (2 × (kx × ky) × k × d) = 6736 mm Area of tension steel at shear perimeter; As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) × Asx_ten) As_ten = 10576 mm 2 Design concrete shear stress; 1/3 1/3 1/4 vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25 2 vc = 0.746 N/mm Nominal design shear stress at perimeter; 2 v = Veff / (u × d) = 0.736 N/mm v < vc no shear reinforcement required CURTAILMENT OF REINFORCEMENT Internal column Radius of circular yield line; 1/2 r = (lx × ly / π) 1/3 × (1.05 × Spanx × 1.05 × Spany / (lx × ly)) = 1601 mm Minimum curtailment length in x-direction; lint_x = Max(r + 12 × D, 0.25 × Spanx) = 1841 mm Minimum curtailment length in y-direction; lint_y = Max(r + 12 × D, 0.25 × Spany) = 1841 mm Corner column Radius of yield line; ly)) r = (lx1 × ly / π) 1/2 × ((0.45 × Spanx + ex) × (0.45 × Spany + ey)/ (lx1 × 1/3 r = 863 mm Minimum curtailment length in x-direction; lcorner_x = Max(r + 12 × D, 0.2 × Spanx) = 1440 mm 20
- 21. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Calc. by Dr. C. Sachpazis Minimum curtailment length in y-direction; Date Chk'd by Date 18/01/2014 App'd by Date lcorner_y = Max(r + 12 × D, 0.2 × Spany) = 1440 mm Edge columns Radius of yield line in x-direction; ly)) r = (lx1 × ly / π) 1/2 × ((0.45 × Spanx + ex) × (1.05 × Spany) / (lx1 × 1/3 r = 1130 mm Minimum curtailment length in x-direction; ledge_x = Max(r + 12 × D, 0.2 × Spanx) = 1440 mm Radius of yield line in y-direction; r = (lx × ly1 / π) 1/2 × ((0.45 × Spany + ey) × (1.05 × Spanx) / (lx × 1/3 ly1)) r = 1130 mm Minimum curtailment length in y-direction; ledge_y = Max(r + 12 × D, 0.2 × Spany) = 1440 mm 21
- 22. Job Ref. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Date Calc. by Dr. C. Sachpazis Chk'd by Date 18/01/2014 B A ex 1 Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis ey Span x n l y1 e s n Date C Span x l x1 App'd by f s r 0.2 x Spany s r Span y c a b q e x 2 j j Span y x f ly 0.5 x Spany p l q d p g h 3 k m k q 0.2 x Spanx lx 0.5 x Spanx When the effective span in the x direction, Lx, is greater than the effective span in the y direction, Ly, the reinforcement in the outer layer is assumed to be that in the x direction otherwise it is assumed to be that in the y direction. 22
- 23. Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997 Project: Section Civil & Geotechnical Engineering Consulting Company for Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info Sheet no./rev. 1 Civil & Geotechnical Engineering GEODOMISI Ltd. - Dr. Costas Sachpazis Date Calc. by Dr. C. Sachpazis Job Ref. Chk'd by 18/01/2014 Date App'd by Date REINFORCEMENT KEY 2 a = 20 dia bars @ 150 centres - (2094 mm /m); 2 b = 16 dia bars @ 200 centres - (1005 mm /m) 2 c = 20 dia bars @ 200 centres - (1570 mm /m); 2 d = 16 dia bars @ 200 centres - (1005 mm /m) 2 e = 20 dia bars @ 125 centres - (2513 mm /m); 2 f = 20 dia bars @ 200 centres - (1570 mm /m) 2 g = 20 dia bars @ 150 centres - (2094 mm /m); 2 h = 20 dia bars @ 200 centres - (1570 mm /m) 2 j = 20 dia bars @ 150 centres - (2094 mm /m); 2 k = 20 dia bars @ 200 centres - (1570 mm /m) 2 l = 20 dia bars @ 150 centres - (2094 mm /m); 2 m = 20 dia bars @ 200 centres - (1570 mm /m) 2 n = 16 dia bars @ 150 centres - (1340 mm /m) 2 p = 16 dia bars @ 175 centres - (1148 mm /m); 2 q = 16 dia bars @ 150 centres - (1340 mm /m) 2 r = 16 dia bars @ 175 centres - (1148 mm /m); 2 s = 16 dia bars @ 200 centres - (1005 mm /m) 2 Distribution bars = 12 dia bars @ 300 centres - (377 mm /m) Shear reinforcement is required - Refer to output above for details. 23

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