Unreinforced masonry diaphragm walls - Concrete Block Association

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Geometrical properties of diaphragm walls TIED BLOCKWORK WALLS block depth rib area I value Z value width (m) (m) spacing (m) (m2/m) (m4/m) (m3/m) 0.10 0.44 0.45 0.253 0.0062 0.0282 0.10 0.44 0.68 0.235 0.0061 0.0278 0.10 0.44 0.90 0.227 0.0061 0.0276 0.10 0.44 1.13 0.221 0.0060 0.0275 0.10 0.44 1.35 0.218 0.0060 0.0274 0.10 0.66 0.45 0.302 0.0176 0.0535 0.10 0.66 0.68 0.268 0.0170 0.0516 0.10 0.66 0.90 0.251 0.0167 0.0508 0.10 0.66 1.13 0.241 0.0166 0.0502 0.10 0.66 1.35 0.234 0.0164 0.0498 0.10 0.89 0.45 0.353 0.0375 0.0842 0.10 0.89 0.68 0.301 0.0354 0.0795 0.10 0.89 0.90 0.277 0.0344 0.0773 0.10 0.89 1.13 0.261 0.0338 0.0759 0.10 0.89 1.35 0.251 0.0334 0.0751 Table 3 BONDED WALLS block depth rib area I value Z value width (m) (m) spacing (m) (m2/m) (m4/m) (m3/m) 0.10 0.44 0.73 0.233 0.0061 0.0277 0.10 0.44 1.18 0.220 0.0060 0.0275 0.10 0.55 0.73 0.248 0.0108 0.0392 0.10 0.55 1.18 0.230 0.0106 0.0385 0.10 0.78 0.73 0.279 0.0255 0.0654 0.10 0.78 1.18 0.249 0.0247 0.0632 Table 4 QUION BONDED WALLS block depth rib area I value Z value width (m) (m) spacing (m) (m2/m) (m4/m) (m3/m) 0.10 0.44 0.45 0.253 0.0062 0.0282 0.10 0.44 0.90 0.227 0.0061 0.0276 0.10 0.44 1.35 0.218 0.0060 0.0274 0.10 0.67 0.45 0.304 0.0183 0.0547 0.10 0.67 0.90 0.252 0.0174 0.0519 0.10 0.67 1.35 0.235 0.0171 0.0509 Table 5 Design Of freestanding diaphragm walls Design of single storey propped diaphragm wall Choose trial section and calculate loading at Ultimate Limit State Choose trial section and calculate loading at Ultimate Limit State Has the wall got a membrane dpc No Yes Calculate moment of resistance from equation 2 Calculate the flexural moment within the span of the wall assuming the moment at the base of the wall is the lesser of: Calculate the moment of resistance from equation 1 and 2 and adopt the greater value ■ Cracked section moment of resistance (equation 2) ■ Elastic section moment as if propped cantilever Not adequate No No Moment of resistance > actual moment Yes Calculate shear stress from equation 3 and confirm < allowable Calculate the tie coefficient from equation 4 and choose a suitable wall tie from table 1 Check horizontal bending in flange using equations 1 & 5 Steel tie connection flange/rib Flange/rib Block bonded connection Block bonded connection Flange/rib Check the flexural moment of resistance > actual flexural moment Yes Calculate the shear stress from equation 3 and confirm < allowable Check horizontal bending in flange using equation 1 & 5 No No Steel tie connection flange/rib Calculate the tie coefficient from equation 4 and choose a suitable wall tie from table 1 Adequate Not adequate Adequate Wall structurally adequate for shear and bending Fig 4 Wall structurally adequate for shear and bending Fig 3
Design example • Design an 8 metre tall diaphragm wall to a single storey building. • The characteristic wind pressure is 0.70 KN/m 2 and the blockwork unit weight may be taken as 16.68 kN/m 3 . • Assume the weight of the roof structure and capping beam balances any wind uplift. Reference Calculation Output Wall loading w = 0.98 kN/m 2 BS 5628 Ultimate limit state wind loading = 0.70 x 1.4 = 0.98 kN/m 2 Elastic moment at the base of a propped cantilever M = wh 2 /8 = 0.98 x 8 2 /8 = 7.84 kNm M = 7.84 kNm Try 660 x 900mm tied diaphragm wall constructed from 100mm7N/mm 2 blocks set in class III mortar Table 3 A = 0.251m 2 Z = 0.508m 3 D = 0.66m t w = 100mm BS 5268 f k = 6.4 N/mm 2 f kxperp = 0.6 N/mm 2 f kxpar = 0.25 N/mm 2 ϒ m = 3.5 ϒ mv = 2.5 BS 5268 Vertical load at base due to self weight = 0.251 x 8 x 16.68 = 33.49 kN At ultimate limit state vertical load = 33.49 x 0.9 = 30.14 kN R = 30.14 kN Cracked section moment of resistance Equation 2 M r = (R/2) {D – (Rϒ m / 1.1f k )} = (30.14/2) {0.66 – (30.14 x 3.5/1.1 x 6.4 x 10 3 )} = 9.72 kNm M = 9.72 kNm Since 7.84 kNm < 9.72 kNm adopt 7.84 kNm as base moment Base moment = 7.84 kNm Vertical flexural bending moment Prop force at head of wall = 0.5 (8 x 0.98) – (7.84 / 8) = 2.94 kN Position of zero shear force = 2.94 / 0.98 = 3m Max flexural bending moment = (3 x 2.94) – (3 2 x 0.98 / 2) = 4.41 kNm Max flexural M = 4.41 kNm Vertical flexural resistance Vertical load at max flexural moment = 3 / 8 x 30.14 = 11.3 kN g d = 11.3 / 0.251 x 10 3 = 0.045 N/mm 2 Equation 1 M r = [(f kx/ ϒ m ) + g d ] Z = [(0.25/3.5) + 0.045] x 0.0508 x 10 3 = 5.91 kNm M r = 5.91 kNm 5.91 kNm > 4.41 kNm therefore wall adequate in vertical bending Check the stability of the wall assuming stability entirely from gravity. Design wind pressure = 0.70 x 1.0 = 0.70kN/m 2 Elastic bending moment at base = 0.70 x 8 2 / 8 = 5.60kNm Since the elastic moment at the base is less than the cracked section moment of resistance at the base take the base moment equal to the elastic moment. Prop force at roof level = (0.70 x 8 / 2) – (5.6 / 8) = 2.1Kn Position of zero shear = 2.1 / 7.0 = 3m Maximum moment = (2.1 x 3) – (0.70 x 3 2 / 2) = 3.15kNm Design weight of wall at position of zero shear = 0.251 x 3 x 16.68 x 1.0 = 12.56 kN/m M r = (12.56 / 2) (0.66 – (12.56 x 3.5 / (1.1 x 6.4 x 10 3 ))) = 4.1kNm Since M r > 3.15 kNm the wall is stable under gravity alone Shear capacity Maximum shear per metre = 0.5 (8 x 0.98) + (7.84/8) = 4.9 kN Shear per rib V = 0.9 x 4.9 = 4.41 kN V = 4.41 kN Equation 3 v = V/Dt w = 4.41 x 10 3 / (660 x 100) = 0.07 N/mm 2 v = 0.07 N/mm 2 BS 5628 f v / ϒ mv = 0.35/2.5 = 0.14N/mm > 0.07 N/mm 2 therefore shear adequate Tie design Assume vertical tie spacing = 225mm ϒ m = 1.15 f y = 250 N/mm 2 Equation 4 Table 1 K v = t w v s ϒ m / f y = 100 x 0.07 x 225 x 1.15/250 = 7.25 For 20 x 5 mild steel ties K v = 8.3 > 7.25 therefore OK K v = 7.25 Use 20 x 5 mild steel ties at 225mm vertical centres Horizontal flexural bending Equation 5 M = wB 2 / 10 = 0.98 x 0.9 2 / 10 = 0.08 kNm M = 0.08 kNm Z = 1000 x 1002 / 6 = 1.667 x 10 6 mm 3 g d = 0 Equation 1 M r = (f kx / ϒ m ) Z = (0.60 / 3.5) x 1.667 = 0.29 kNm M r = 0.29 kNm 0.29 kNm > 0.08 kNm therefore horizontal flexural bending adequate Conclusions – Wall adequate Table 6
Page 1 and 2: Cl/SfB Ff2 May 2010 Data Sheet 10 A
Page 3: Equation 5 ECCENTRICITY AT TOP OF S

Unreinforced masonry diaphragm walls - Concrete Block Association

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