FLAT SLAB DESIGN TO BS8110:PART 1:1997

Transcription

1 Consulting Company for App'd by FLAT SLAB DESIGN TO Slab geometry Span of slab in x-direction; Span of slab in y-direction; Column dimension in x-direction; Column dimension in y-direction; External column dimension in x-direction; External column dimension in y-direction; Edge dimension in x-direction; Edge dimension in y-direction; Effective span of internal bay in x direction; Effective span of internal bay in y direction; Effective span of end bay in x direction; Effective span of end bay in y direction; Span x = 7200 mm Span y = 7200 mm l x = 400 mm l y = 400 mm l x1 = 250 mm l y1 = 250 mm e x = l x1 / 2 = 125 mm e y = l y1 / 2 = 125 mm L x = Span x l x = 6800 mm L y = Span y l y = 6800 mm L x1 = Span x l x / 2 = 7000 mm L y1 = Span y l y / 2 = 7000 mm e x A B C Span x Span x 1 Span y e y l x1 l y1 l x l x 2 l y 3 Span y l y h Slab details Depth of slab; h = 250 mm 1

2 Consulting Company for App'd by Characteristic strength of concrete; f cu = 35 N/mm 2 Characteristic strength of reinforcement; f y = 500 N/mm 2 Characteristic strength of shear reinforcement; f yv = 500 N/mm 2 Material safety factor; γ m = 1.15 Cover to bottom reinforcement; c = 20 mm Cover to top reinforcement; c = 20 mm Loading details Characteristic dead load; G k = kn/m 2 Characteristic imposed load; Q k = kn/m 2 Dead load factor; γ G = 1.4 Imposed load factor; γ Q = 1.6 Total ultimate load; N ult = (G k γ G) + (Q k γ Q) = kn/m 2 Moment redistribution ratio; β b = 1.0 Ratio of support moments to span moments; i = 1.0 DESIGN SLAB IN THE X-DIRECTION SAGGING MOMENTS End bay A-B Effective span; Midspan moment; Support moment; L = 7000 mm d = 200 mm m = (N ult L 2 ) / (2 (1 + (1 + i)) 2 ) = knm/m m = i m = knm/m Design reinforcement K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 919 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 919 mm 2 /m Provide 20 dia 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 146 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = Allowable span to depth ratio; k 1 = Actual span to depth ratio; L / d =

3 Consulting Company for App'd by PASS - Span to depth ratio is OK Internal bay B-C Effective span; Midspan moment; Support moment; L = 6800 mm d = 202 mm m = (N ult L 2 ) / (2 ( (1 + i) + (1 + i)) 2 ) = knm/m m = i m = knm/m Design reinforcement K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 617 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 617 mm 2 /m Provide 16 dia 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1005 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 204 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = Allowable span to depth ratio; k 1 = Actual span to depth ratio; L / d = 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 d = 200 mm Support moment; m = 2 i m = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1996 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1996 mm 2 /m Provide 20 dia 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /m PASS - Support reinforcement is OK 3

4 Consulting Company for App'd by Internal column C3 Consider the reinforcement concentrated in half width strip over the support d = 200 mm Support moment; m = 2 i m = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1300 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1300 mm 2 /m Provide 20 dia 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /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 d = 200 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 2134 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 2134 mm 2 /m Provide 20 dia 125 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2513 mm 2 /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 d = 200 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = knm/m 4

5 Consulting Company for App'd by K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1383 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1383 mm 2 /m Provide 20 dia 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Support reinforcement is OK Corner column A1 d = 206 mm Total load on column; S = ((Span x / 2) + e x) ((Span y / 2) + e y) N ult = 247 kn Area of column head; A = l x l y1 = m 2 Support moment; m = S (1 (N ult A / S) 1/3 ) / 2 = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1211 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1211 mm 2 /m Provide 16 dia 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1340 mm 2 /m PASS - Support reinforcement is OK Edge column A2, A3 d = 202 mm Total load on column; S = Span x (Span y / 2 + e y) N ult = 477 kn Area of column head; A = l x1 l y = m 2 Support moment; m = S (1 (N ult A / S) 1/3 ) / 5.14 = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 956 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 956 mm 2 /m Provide 16 dia 175 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1149 mm 2 /m 5

6 Consulting Company for App'd by 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; A s_req = A sx1 / 2 = 1047 mm 2 /m Provide 16 dia 150 centres - 'U' bars with 1600 mm long legs Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1340 mm 2 /m PASS - Edge reinforcement is OK Distribution reinforcement Provide 12 dia 300 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 377 mm 2 /m DESIGN SLAB IN THE Y-DIRECTION SAGGING MOMENTS End bay 1-2 Effective span; Midspan moment; Support moment; L = 7000 mm d = 220 mm m = (N ult L 2 ) / (2 (1 + (1 + i)) 2 ) = knm/m m = i m = knm/m Design reinforcement K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 825 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 825 mm 2 /m Provide 20 dia 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 175 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = Allowable span to depth ratio; k 1 = Actual span to depth ratio; L / d = PASS - Span to depth ratio is OK Internal bay 2-3 Effective span; L = 6800 mm 6

7 Consulting Company for App'd by Midspan moment; Support moment; d = 222 mm m = (N ult L 2 ) / (2 ( (1 + i) + (1 + i)) 2 ) = knm/m m = i m = knm/m Design reinforcement K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 561 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 561 mm 2 /m Provide 16 dia 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1005 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 186 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = Allowable span to depth ratio; k 1 = Actual span to depth ratio; L / d = 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 d = 220 mm Support moment; m = 2 i m = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1758 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1758 mm 2 /m Provide 20 dia 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /m PASS - Support reinforcement is OK Internal column C3 Consider the reinforcement concentrated in half width strip over the support d = 220 mm 7

8 Consulting Company for App'd by Support moment; m = 2 i m = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1160 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1160 mm 2 /m Provide 20 dia 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /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 d = 220 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1875 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1875 mm 2 /m Provide 20 dia 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /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 d = 220 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = knm/m K = m / (d 2 f cu) =

9 Consulting Company for App'd by z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1233 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1233 mm 2 /m Provide 20 dia 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Support reinforcement is OK Edge column B1, C1 d = 222 mm Total load on column; S = (Span x / 2 + e x) Span y N ult = 477 kn Area of column head; A = l y1 l x = m 2 Support moment; m = S (1 (N ult A / S) 1/3 ) / 5.14 = knm/m K = m / (d 2 f cu) = z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 859 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 859 mm 2 /m Provide 16 dia 175 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1149 mm 2 /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; A s_req = A sy1 / 2 = 785 mm 2 /m Provide 16 dia 200 centres - 'U' bars with 1600 mm long legs Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1005 mm 2 /m PASS - Edge reinforcement is OK PUNCHING SHEAR Corner column A1 Design shear transferred to column; V t = ((0.45 Span x) + e x) ((0.45 Span y) + e y) N ult = 202 kn Design effective shear transferred to column; V eff = 1.25 V t = 252 kn Area of tension steel in x-direction; A sx_ten = A scorner = 1340 mm 2 /m Area of tension steel in y-direction; A sy_ten = A scorner = 1340 mm 2 /m Column perimeter; u c = l x1 + l y = 650 mm Average effective depth of reinforcement; d = h c - φ p = 214 mm 9

10 Consulting Company for App'd by Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 1292 mm A s_ten = 1731 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 119 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 1613 mm A s_ten = 2161 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 16 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 1934 mm A s_ten = 2592 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Penultimate edge column A2 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = ((0.45 Span x) + e x) (1.05 Span y) N ult = 453 kn V eff = 1.4 V t = 634 kn Area of tension steel in x-direction; A sx_ten = A sx_edge = 1148 mm 2 /m 10

11 Consulting Company for App'd by Area of tension steel in y-direction; A sy_ten = A sy1e = 2094 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = (2 l x1)+ l y = 900 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 3588 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 947 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 4628 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 372 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 5669 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 154 mm 2 11

12 Consulting Company for App'd by Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 6710 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Internal edge column A3 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = ((0.45 Span x) + e x) Span y N ult = 431 kn V eff = 1.4 V t = 604 kn Area of tension steel in x-direction; A sx_ten = A sx_edge = 1148 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sye = 1570 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = (2 l x1)+ l y = 900 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 2989 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 945 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 3862 mm 2 v c = N/mm 2 12

13 Consulting Company for App'd by Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 365 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 4734 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 159 mm 2 Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 5607 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Penultimate edge column B1 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = (1.05 Span x) ((0.45 Span y) + e y) N ult = 453 kn V eff = 1.4 V t = 634 kn Area of tension steel in x-direction; A sx_ten = A sx1e = 2513 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy_edge = 1148 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = l x + (2 l y1) = 900 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 4066 mm 2 13

14 Consulting Company for App'd by v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 789 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 5241 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 331 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 6416 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 104 mm 2 Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 7592 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 v < v c no shear reinforcement required 14

15 Consulting Company for App'd by Penultimate central column B2 Design shear transferred to column; Design effective shear transferred to column; V t = (1.05 Span x) (1.05 Span y) N ult = 1017 kn V eff = 1.15 V t = 1170 kn Area of tension steel in x-direction; A sx_ten = A sx1e = 2513 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy1e = 2094 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 9601 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 872 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 382 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm A s_ten = mm 2 15

16 Consulting Company for App'd by v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Internal central column B3 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = (1.05 Span x) Span y N ult = 969 kn V eff = 1.15 V t = 1114 kn Area of tension steel in x-direction; A sx_ten = A sx1i = 2094 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sye = 1570 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 7636 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 872 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = 9988 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 418 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm 16

17 Consulting Company for App'd by A s_ten = mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Internal edge column C1 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = Span x ((0.45 Span y) + e y) N ult = 431 kn V eff = 1.4 V t = 604 kn Area of tension steel in x-direction; A sx_ten = A sxe = 1570 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy_edge = 1148 mm 2 /m Column perimeter; u c = l x + (2 l y1) = 900 mm (Library item: Flat slab shear map C1) Average effective depth of reinforcement; d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 2989 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 945 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 3862 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 365 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) 17

18 Consulting Company for App'd by u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 4734 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 159 mm 2 Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 5607 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Internal central column C2 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = Span x (1.05 Span y) N ult = 969 kn V eff = 1.15 V t = 1114 kn Area of tension steel in x-direction; A sx_ten = A sxe = 1570 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy1i = 2094 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 7636 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 18

19 Consulting Company for App'd by Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 872 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = 9988 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 418 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm A s_ten = mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Internal column C3 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = Span x Span y N ult = 923 kn V eff = 1.15 V t = 1061 kn Area of tension steel in x-direction; A sx_ten = A sxi = 1570 mm 2 /m Area of tension steel in y-direction; A sy_ten = A syi = 1570 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 6544 mm 2 19

20 Consulting Company for App'd by v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 834 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = 8560 mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 403 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm A s_ten = mm 2 v c = N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = N/mm 2 CURTAILMENT OF REINFORCEMENT Internal column v < v c no shear reinforcement required Radius of circular yield line; r = (l x l y / π) 1/2 (1.05 Span x 1.05 Span y / (l x l y)) 1/3 = 1601 mm Minimum curtailment length in x-direction; Minimum curtailment length in y-direction; Corner column l int_x = Max(r + 12 D, 0.25 Span x) = 1841 mm l int_y = Max(r + 12 D, 0.25 Span y) = 1841 mm Radius of yield line; r = (l x1 l y / π) 1/2 ((0.45 Span x + e x) (0.45 Span y + e y)/ (l x1 l y)) 1/3 Minimum curtailment length in x-direction; r = 863 mm l corner_x = Max(r + 12 D, 0.2 Span x) = 1440 mm 20

21 Consulting Company for App'd by Minimum curtailment length in y-direction; l corner_y = Max(r + 12 D, 0.2 Span y) = 1440 mm Edge columns Radius of yield line in x-direction; r = (l x1 l y / π) 1/2 ((0.45 Span x + e x) (1.05 Span y) / (l x1 l y)) 1/3 Minimum curtailment length in x-direction; r = 1130 mm l edge_x = Max(r + 12 D, 0.2 Span x) = 1440 mm Radius of yield line in y-direction; r = (l x l y1 / π) 1/2 ((0.45 Span y + e y) (1.05 Span x) / (l x l y1)) 1/3 Minimum curtailment length in y-direction; r = 1130 mm l edge_y = Max(r + 12 D, 0.2 Span y) = 1440 mm 21

22 Consulting Company for App'd by A B C e x 1 e y l y1 n l x1 n Span x s r e Span x s r f s 0.2 x Span y c x 2 Span y Span y q p j q d a j e x b l f ly 0.5 x Span y p g h 3 k k m q l x 0.5 x Span x 0.2 x Span x When the effective span in the x direction, L x, is greater than the effective span in the y direction, L y, 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 Consulting Company for App'd by REINFORCEMENT KEY a = 20 dia 150 centres - (2094 mm 2 /m); b = 16 dia 200 centres - (1005 mm 2 /m) c = 20 dia 200 centres - (1570 mm 2 /m); d = 16 dia 200 centres - (1005 mm 2 /m) e = 20 dia 125 centres - (2513 mm 2 /m); f = 20 dia 200 centres - (1570 mm 2 /m) g = 20 dia 150 centres - (2094 mm 2 /m); h = 20 dia 200 centres - (1570 mm 2 /m) j = 20 dia 150 centres - (2094 mm 2 /m); k = 20 dia 200 centres - (1570 mm 2 /m) l = 20 dia 150 centres - (2094 mm 2 /m); m = 20 dia 200 centres - (1570 mm 2 /m) n = 16 dia 150 centres - (1340 mm 2 /m) p = 16 dia 175 centres - (1148 mm 2 /m); q = 16 dia 150 centres - (1340 mm 2 /m) r = 16 dia 175 centres - (1148 mm 2 /m); s = 16 dia 200 centres - (1005 mm 2 /m) Distribution bars = 12 dia 300 centres - (377 mm 2 /m) Shear reinforcement is required - Refer to output above for details. 23