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Đồ án này thực hiện theo cuốn Sàn sườn bê tông toàn khối (GS. Nguyễn Đình Cống). Ngôn ngữ sử dụng là tiếng anh chuyên ngành xây dựng. Cuốn đồ án này giúp cho các bạn sinh viên có thêm một nguồn tham khảo cách trình bày đồ án, và cũng là một nguồn thú vị cung cấp các thuật ngữ tiếng anh chuyên ngành xây dựng
RIBBED SLAB SYSTEM WITH ONE WAY SLAB GIVEN DATA Plan of slab structure as shown in the bellow figure Plan of slab Primary beams are orientated axis 1, 2, 3, 4, Secondary beams are orientated axis A, B, C, D, E Slab structure ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) 1.1 Selection of dimensions of components of slab – Dimensions of aperture of plate + Distance between adjacent secondary beams: l1 = 1,9 m + Distance between adjacent primary beams: l2 = 5,45 m – Load bearing brick wall thickness: t = 34 cm Beams are directly supported on the wall – Dimensions of columns: bc = hc = 30 cm 1.2 Characteristics of the construction – Building construction slab which has levels as shown in the figure – Nominal live load: ptc = 9,75 kN/m2, reliability coefficient of live load: n = 1,2 1.3 Selection of materials – Concrete: Heavy concrete, Grade of concrete with compressive strength: grade B12,5 With Rb = 7,5 MPa, Rbt = 0,66 MPa – Reinforcement: + For plate: Constructive reinforcement C-I Main reinforcement C-I + For beam: Constructive reinforcement C-I Main reinforcement C-II DESIGN OF PLATE 2-way slab Have: l1 = 1,9 m, l2 = 5,45 m l2 > 2l1, thus design approximately as in 1-way slab 2.1 Selection of dimensions of components of slab – Thickness of the slab: hb = hb = Dl1 m , where: D = 1.4 with big load, m = 35 with 1-way slab Dl1 1,4 ×1900 = m 35 = 76 mm Thus, select hb = 80 mm – Dimensions of the secondary beams: hdp = 1 l2 = × 5450 = 454,2 mdp 12 mm Thus, select hdp = 450 mm, bdp = 180 mm – Dimensions of the primary beams: with span is 3l1 1 hdc = 3l1 = × × 1900 = 633,3 mdc mm Thus, select hdc = 650 mm, bdc = 250 mm ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) 2.2 The design scheme – One-way slab, establish the simplified model for a m wide strip of the plate which is perpendicular to the secondary beams, regard strip of slab like a continuous beam – The design span length of the slab: + Outer span length: bdp t 0,18 0,34 0,08 l0b = l1b − − + Cb = 1,9 − − + = 1,68 2 2 + Middle span length: l0 = l1 − bdp = 1,9 − 0,18 = 1,72 m m Span difference: l0 − l0b 1,72 − 1,68 = × 100% = 2,33 % < 10% l0b 1,72 2.3 The design load – Dead load is calculated as following: Layer Nominal value Tile 10 mm, γ = 20 kN/m3 0,01×20 = 0,200 Lining mortar 30 mm, γ = 18 kN/m3 0,03×18 = 0,540 Reinforced concrete 80 mm, γ = 25 kN/m3 0,08×25 = 2,000 Plaster mortar 10 mm, γ = 18 kN/m3 0,01×18 = 0,180 Total 2,920 Dead load: gb = 3,356 kN/m2 – Live load: pb = ptc×n = 9,75×1,2 = 11,7 kN/m2 – Total load: qb = gb + pb = 3,356 + 11,7 = 15,056 kN/m2 Calculate in strip of slab b1 = m: qb = 15,056×1 = 15,056 kN/m Reliability coefficient 1,1 1,3 1,1 1,3 Design value 0,220 0,702 2,200 0,234 3,356 2.4 Internal force Following the plastic hinge scheme model: – Bending moment at the outer span and the secondary support: ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) M nh = M g 2 15,056 × 1,682 qblob = ±3,863 =± = ± 11 11 kN.m – Bending moment at the middle spans and the middle supports: M nhg1 = M g1 = ± 15,056 × 1,722 qblb2 = ±2,784 = ± 16 16 kN.m – The magnitude of the maximum shear force: QBt = 0,6qblob = 0,6 × 15,056 × 1,68 = 15,176 kN Check the shear strength: QBt = 15,176 < 0,5Rbtb1h0 = 0,5 × 0,66 × 1× 65 = 21,45 kN Concrete is satisfied shear resistance Sheme of calculation and internal force diagram in slab 2.5 Bending moment reinforcement – Data: Concrete grade B12,5, Rb = 7,5 MPa Reinforced steel C-I, Rs = 225 MPa – Calculate the internal force based on plastic hinge scheme, with αpl = 0,255 Assume a = 15 mm for all section: h0 = hb – a = 80 – 15 = 65 mm – At the outer bearing and outer span, with M = 3,863 kN.m: M 3,863.103 αm = = = 0,122 < α pl = 0,255 Rbbh02 7,5 × 106 × 1× 0,0652 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) ζ = + − 2α m As = µ= = + − × 0,122 = 0,935 M 3,863.103 = Rsζh0 225.106 × 0,935 × 0,065 = 2,826.10-4 m2 = 282,6 mm2 As 282,6 = × 100% = 0,435% bh0 1000 × 65 Thus, Select diameter of reinforced steel mm, spacing of adjacent steels is 100 mm (ϕ6a100) – At the middle bearings and middle spans, with M = 2,784 kN.m: αm = ζ = M 2,784.103 = = 0,0879 < α pl = 0,255 Rbbh02 7,5 × 106 × 1× 0,0652 + − 2α m + − × 0,0879 = = 0,954 2 M 2,784.103 As = = Rsζh0 225.106 × 0,954 × 0,065 µ= = 1,995.10-4 m2 = 199,5 mm2 As 199,5 = × 100% = 0,307% bh0 1000 × 65 Thus, select diameter of reinforced steel mm, spacing of adjacent steels is 140 mm (ϕ6,a140) – Check the working height h0 with thickness of reinforcement protection cover c = 10 mm: h0 = hb – c – ϕ = 80 – 10 – ×6 = 67 mm > 65 mm – safety – Positive moment longitudinal reinforcement: is placed alternatively Distance from the end of the shorter reinforcement to edge of the secondary beam: 8 l0 = ×1720 = 215 mm – With apertures of slab which are pinned to beams by edges, the main reinforcement is allowed to reduce 20% area As 80% = 80% As = 80% × 199,5 = 159,6 mm Check reinforcement amount again: ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) µ80% = As80% 159,6 = × 100% = 0,246% > µ = 0,05% bh0 1000 × 65 Thus, select diameter of reinforced steel mm, spacing of adjacent steels is 170 mm (ϕ6,a170) – Negative moment longitudinal reinforcement: is placed alternatively pb 11,7 = = 3, 486 > ⇒ ν = gb 3,356 + Extension of reinforcement is: νl0 = ×1720 = 573,3 mm ≈ 575 mm (seen from edge of the secondary beam) 1 2 νl0 + bdp = ×1720 + ×180 ≈ 665 mm (seen from axis of the secondary beam) + Extension of the shorter reinforcement is: 1 6 l0 = ×1720 = 286,7 mm ≈ 290 mm (seen from edge of the secondary beam) 1 1 6 l0 + bdp = ×1720 + ×180 = 380 mm (seen from axis of the secondary beam) 2.6 Constructive reinforcement – Negative moment reinforcement is placed in direction which is perpendicular to the primary beams + Select ϕ6, s = 200 mm, area per m plate is 141 mm2, higher 50% than area of reinforcement at the middle bearing of plate which is 0,5×199,5 = 99,75 mm2 + Extension is: 1 4 l0 = ×1720 = 430 mm (seen from edge of the primary beam) 1 1 4 l0 + bdc = ×1720 + ×250 = 555 mm (seen from axis of the primary beam) + To slab can subjected to negative moment at the wall when slab is supported to the wall, we place reinforcement with an extension seen from edge of the wall: 1 6 l0 = ×1720 = 286,7 mm ≈ 290 mm ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) – Distribution reinforcement is placed in direction which is perpendicular to moment reinforcement: Select ϕ6, s = 250 mm, area per m plate is 131 mm 2, higher than 20% area of reinforcement at the outer bearing of plate which is 0,2×282,6 = 56,52 mm2, and higher than 20% area of reinforcement at the middle bearing of plate which is 0,2×199,5 = 39,9 mm Region which is allowed to reduce 20% reinforcement ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) Arrangement of reinforcement in slab Arrangement of reinforcement in slab Arrangement of reinforcement in slab ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) Some cross section in secondary beam ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) DESIGN OF SECONDARY BEAM 3.1 The design scheme – The secondary beams is symmetrically continuous beams with spans – The secondary beams is placed on wall Sd = 220 mm S d l2 ; 40 Cd = min( ) = min(110; 136,25) = 110 mm – The design span length of the secondary beam: + Outer span length: b t 0,25 0,34 l pb = l2 − dc − + Cd = 5,45 − − + 0,11 = 5,265 2 2 + Middle span length: l p = l2 − bdc = 5,45 − 0,25 = 5,2 m m Span difference: l pb − l p 5,265 − 5,2 = × 100% = 1,23% < 10% l pb 5,265 3.2 The design load – Dead load + Self-weight of the secondary beam (not include plate 80 mm): g0p = bdp(hdp – hb)γn where γ = 25 kN/m2, is unit weight of concrete n = 1,1, is reliability coefficient of self-weight of beam g0p = 0,18×(0,45 – 0,08)×25×1,1 = 1,8315 kN/m + Dead load transferred from the plate: gpl1 = 3,356×1,9 = 6,3764 kN/m – Live load pp = pbl1 = 11,7×1,9 = 22,23 kN/m – Total load: qp = 1,8315 + 6,3764 + 22,23 = 30.4379 ≈ 30,44 kN/m pp gp = 22,23 = 2,7084 1,8315 + 6,3764 3.3 Internal force Calculate the internal force based on plastic hinge scheme model a Bending moment 10 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) B M mg = 306,737 − 0,3 × (306,737 + 94,856) × 1,9 = 275,032 kN.m At bearing C: C M mg = Mg − hc ( M g + M ) 2l1 M4 = 109,669 kN.m C M mg = 254,015 − 0,3 × (254,015 + 109,669) × 1,9 = 225,303 kN.m b Shear force – Determine shear force diagram caused by dead load G QG = βG = β×52,179 kN.m – Determine moment diagram caused by live load Pi QP = βP = β×121,154 kN.m β is looked up in Appendix 12 Shear force at center of span is calculated by section method The results are shown in the bellow table: Shear force QG QP1 QP2 QP3 QP4 QP5 QP6 Qmax Qmin β Q β Q β Q β Q β Q β Q β Q Right of Center of Left of Right of Center of bearing A outer bearing B bearing B span span 0,714 -1,286 1,005 37,256 -14,923 -67,102 52,440 0,261 0,857 -1,143 0,048 103,829 -17,322 -138,479 5,815 -0,143 -0,143 1,048 -17,325 -17,325 -17,325 126,969 5,815 0,679 -1,321 1,274 82,264 -38,890 -160,044 154,350 33,196 -0,095 -0,095 0,810 -11,510 -11,510 -11,510 98,135 -23,019 0,810 -1,19 0,286 98,135 -23,019 -144,173 34,650 34,650 0,036 0,187 4,362 22,656 141,085 -26,433 -78,252 206,790 33,911 19,931 -53,813 -227,146 57,255 -22,758 Left of bearing C -0,995 -51,918 -0,952 -115,339 -0,726 -87,958 -1,19 -144,173 0,286 34,650 -17,268 -196,091 26 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) Shear force envelope diagram in primary beam 4.4 Calculation of longitudinal reinforcement – Data: Concrete grade B12,5, Rb = 7,5 MPa Reinforced steel C-II, Rs = 280 MPa, Rsc = 280 MPa With work condition factor γb2 = 1,0, look up in Appendix 9, αR = 0,442 a Negative moment – Design with rectangular cross section: b = 250 mm, h = 650 mm Assume a = 70 mm, h0 = 650 – 70 = 580 mm – At bearing B, αm = ζ = = 275,032 kN.m M 275,032.103 = = 0,436 < α R = 0,442 Rbbh02 7,5 × 106 × 0,25 × 0,582 + − 2α m + − × 0,436 = = 0,679 2 As = µ= B M mg M 275,032.103 = Rsζh0 280.106 × 0,679 × 0,58 = 2,4948.10-3 m2 = 2494,8 mm2 As 2494,8 = × 100% = 1,72% bh0 250 × 580 – At bearing C, αm = C M mg = 225,303 kN.m M 225,303.103 = = 0,357 < α R = 0,442 Rbbh02 7,5 × 106 × 0,25 × 0,582 27 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) ζ = + − 2α m + − × 0,357 = = 0,767 2 As = µ= M 225,303.103 = Rsζh0 280.106 × 0,767 × 0,58 = 1,8083.10-3 m2 = 1808,3 mm2 As 1808,3 = × 100% = 1,25% bh0 250 × 580 b Positive moment – Design with T-cross section, the flange is in compressive zone, flange thickness Assume a = 40 mm, h0 = 650 – 40 = 610 mm h 'f = 80 mm h 'f – Because =80 mm > 0,1×650 = 65 mm = 0,1hdc, the extension of flange Sf is not greater than bellow values: 1 lc = × 5700 6 + span of primary beam = = 950 mm + A haft of clear distance of two secondary beams 0,5×5450 = 2725 mm Thus, Sf = min(950; 2725) = 950 mm Flange breadth: b 'f = b + 2S f = 250 + × 950 = 2150 mm – Determine the location of the neutral axis M f = Rbb 'f h 'f (h0 − 0,5h 'f ) + M ma x = 7,5×2150×80×(610 – 0,5×80) = 735,3.106 N.mm = 267,405.106 N.mm + M f > M ma x , the neutral axis is in the flange – Design as in rectangular section with b = – At outer span, M = 268,291 kN.m αm = ζ = b 'f = 2150 mm, h = 650 mm, h0 = 610 mm M 268,291.103 = = 0,0447 < α R = 0,442 Rbbh02 7,5 × 106 × 2,15 × 0,612 + − 2α m + − × 0,0447 = = 0,977 2 28 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) M 268,291.103 = Rsζh0 280.106 × 0,977 × 0,61 As = µ= = 1,6076.10-3 m2 = 1607,6 mm2 As 1607,6 = × 100% = 1,05% bh0 250 × 610 – At middle spans, M = 186,322 kN.m αm = ζ = + − 2α m + − × 0,0310 = = 0,985 2 As = µ= M 186,322.103 = = 0,0310 < α R = 0,442 Rbbh02 7,5 × 106 × 2,15 × 0,612 M 186,322 103 = Rsζh0 280.106 × 0,985 × 0,61 = 1,1084.10-3 m2 = 1108,4 mm2 As 1108,4 = × 100% = 0,73% bh0 250 × 610 4.5 Selection and arrangement of longitudinal reinforcement Section As (mm2) Diameter of rein Area of rein Outer span Outer span 1607,6 2ϕ25 + 2ϕ22 1742,1 Bearing B Bearing B Middle span Bearing C 2494,8 1108,4 1808,3 4ϕ22 + 2ϕ25 2ϕ22 + 1ϕ22 4ϕ20+ 2ϕ20 2502,3 1140,4 1885,0 Middle span Bearing C Arrangement of reinforcement in primary beam – Check h0 again + At outer span, select thickness of protection cover c = 25 mm 29 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) h0 = 650 – 25 – 25/2 = 612,5 mm > 610 mm – safety + At bearing B, thickness of protection cover calculated to edge of reinforcement of secondary beam is 20 + 18 = 38 mm Constructing the primary beam reinforcement into layers which are laid under the secondary beam reinforcement, the upper layer 4ϕ22 (As1 = 1520,5; a1 = 38 + 22/2 = 49 mm), the lower layer 2ϕ25 (As2 = 981,8; a2 = 38 + 22 + 30 + 25/2 = 102,5 mm) a= 49 × 1520,5 + 102,5 × 981,8 = 69,99 2502,3 mm h0 = h – a = 650 – 70 = 580 mm ≥ 580 mm – safety + At middle span, select thickness of protection cover c = 25 mm h0 = 650 – 25 – 22/2 = 614 mm > 610 mm – safety + At bearing C, thickness of protection cover calculated to edge of reinforcement of secondary beam is 20 + 18 = 38 mm Constructing the primary beam reinforcement into layers which are laid under the secondary beam reinforcement, the upper layer 4ϕ20 (As1 = 1256,6; a1 = 38 + 20/2 = 48 mm), the lower layer 2ϕ20 (As2 = 628,3; a2 = 38 + 20 + 30 + 20/2 = 98 mm) 48× 1256,6 + 98 × 628,3 a= = 64,7 1885 h0 = 650 – 64,7 = 585,3 ≈ 585 mm > 580 mm – safety – Check clear distance between adjacent bars (4ϕ25 in a layer) Thickness of protection cover c = 25 mm, clear distance: b − 2c − 4φ 250 − × 25 − × 25 t= = = 50 2 mm > 30 mm – satisfaction 4.6 Design of lateral reinforcement – From shear force diagram of primary beam, have: + Right side of bearing A: + Left side of bearing B: QAp QBt + Right side of bearing B: = 141,085 kN, shear force is constant in segment l1 = 227,146 kN, shear force is constant in segment l1 QBp = 205,790 kN, shear force is constant in segment l1 QCt + Left side of bearing C: = 196,091 kN, shear force is constant in segment l1 – Materials: Rb = 7,5 MPa, Rbt = 0,66 MPa, Rsw = 175 MPa QAp a Calculate with the shear force in the right of bearing A: = 141,085 kN Dimensions of the beam: b = 250 mm, h = 650 mm, h0 = 612,5 mm – Condition of calculating shear reinforcement 30 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) + Check condition to resist principal compressive stress of beam’s web: Q = 141,085 kN < 0,3Rbbh0 = 0,3×7,5.103×0,25×0,6125 = 344,531 kN – satisfaction + Check shear strength of concrete: Qbmin = 0,5Rbtbh0 = 0,5×0,66.103×0,25×0,6125 = 50,531 kN < Q = 141,085 kN Concrete is not capable to resist all of shear force, need to design stirrups – Calculation of stirrups without inclined reinforcement + Assume c ≤ 2h0 With assumption that concrete and stirrups resist all of shear force: Q = 141,085 kN = qsw = QDB = 4,5Rbtbh02 qsw Q2 141,085 = 4,5 Rbtbh02 4,5 × 0,66.103 × 0,25 × 0,61252 = 71,458 N/mm Recalculate c0 with qsw = 71,458 N/mm: Rbtbh02 × 0,66 × 250 × 612,52 c0 = = = 1316,248 qsw 71,458 mm c0 = 1316,248 mm < 2h0 = 2×612,5 = 1225 mm, not agree with the assumption + Assume c ≥ 2h0, take c = c0 = 2h0 1,5Rbtbh02 + 0,75qsw 2h0 2h0 Q = QDB = Q − 0,75Rbt bh0 141085 − 0,75 × 0,66 × 250 × 612,5 q sw = = 1,5h0 1,5 × 612,5 = 71,062 N/mm qswmin = 0,25Rbtb = 0,25×0,66×250 = 41,25 N/mm < qsw = 71,062 N/mm – Select stirrups: diameter ϕ8, number of legs n = Cross-sectional area of stirrups perpendicular to member axis: nπφ w2 × 3,14 × 82 Asw = = = 100,53 4 mm2 Calculation spacing of stirrups: R A 175 × 100,53 stt = sw sw = = 247,446 qsw 71,062 mm Structural spacing of stirrups: sct < min(1/3h; 300) = min(1/3×650; 300) = 216,667 mm Maximum spacing of stirrups: 31 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) sma x = Rbtbh02 0,66 × 250 × 612,52 = = 438,748 Q 141085 mm The final spacing of stirrups is chosen as minimum of {stt; sct; smax}: s1 = min(247,446; 216,667; 438,748) = 216,667 mm Take s1 = 210 mm for segment near the bearing Spacing of stirrups at remaining segment: s2 ≤ min(0,75h; 500) = min(0,75×650; 500) = 487,5 mm Take s2 = 250 mm for remaining segment QBt b Calculate with the shear force in the left of bearing B: = 227,146 kN Dimensions of the beam: b = 250 mm, h = 650 mm, h0 = 580 mm – Condition of calculating shear reinforcement + Check condition to resist principal compressive stress of beam’s web: Q = 227,146 kN < 0,3Rbbh0 = 0,3×7,5.103×0,25×0,580 = 326,25 kN – satisfaction + Check shear strength of concrete: Qbmin = 0,5Rbtbh0 = 0,5×0,66.103×0,25×0,580 = 47,85 kN < Q = 227,146 kN Concrete is not capable to resist all of shear force, need to design stirrups – Calculation of stirrups without inclined reinforcement + Assume c ≤ 2h0 With assumption that concrete and stirrups resist all of shear force: Q = 227,146 kN = qsw = QDB = 4,5Rbtbh02 qsw Q2 227,1462 = 4,5 Rbtbh02 4,5 × 0,66.103 × 0,25 × 0,5802 = 206,565 N/mm qswmin = 41,25 < qsw = 206,565 N/mm Recalculate c0 with qsw = 206,565 N/mm: Rbtbh02 × 0,66 × 250 × 5802 c0 = = = 733,089 qsw 206,565 mm c0 = 733,089 mm < 2h0 = 2×580 = 1160 mm, agree with the assumption – Select stirrups: diameter ϕ8, number of legs n = Cross-sectional area of stirrups perpendicular to member axis: nπφw2 × 3,14 × 82 Asw = = = 201,06 4 mm2 Calculation spacing of stirrups: 32 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) stt = Rsw Asw 175 × 201,06 = = 170,336 qsw 206,565 mm Maximum spacing of stirrups: Rbtbh02 0,66 × 250 × 5802 sma x = = = 244,363 Q 227146 mm The final spacing of stirrups is chosen as minimum of {stt; sct; smax}: s1 = min(170,336; 216,667; 244,363) = 170,336 mm Take s1 = 170 mm for segment near the bearing QBp c Calculate with the shear force in the right of bearing C: = 206,790 kN Dimensions of the beam: b = 250 mm, h = 650 mm, h0 = 580 mm – Condition of calculating shear reinforcement + Check condition to resist principal compressive stress of beam’s web: Q = 206,790 kN < 0,3Rbbh0 = 0,3×7,5.103×0,25×0,580 = 326,250 kN – satisfaction + Check shear strength of concrete: Qbmin = 0,5Rbtbh0 = 0,5×0,66.103×0,25×0,580 = 47,85 kN < Q = 206,790 kN Concrete is not capable to resist all of shear force, need to design stirrups – Calculation of stirrups without inclined reinforcement + Assume c ≤ 2h0 With assumption that concrete and stirrups resist all of shear force: Q = 206,790 kN = qsw = QDB = 4,5Rbtbh02 qsw Q2 206,7902 = 4,5Rbtbh02 4,5 × 0,66.103 × 0,25 × 0,5802 = 171,201 N/mm qswmin = 41,25 < qsw = 171,201 N/mm Recalculate c0 with qsw = 171,201 N/mm: c0 = Rbtbh02 × 0,66 × 250 × 5802 = = 805,252 qsw 171,201 mm c0 = 805,252 mm < 2h0 = 2×580 = 1160 mm, agree with the assumption – Select stirrups: diameter ϕ8, number of legs n = Cross-sectional area of stirrups perpendicular to member axis: Asw = nπφ w2 × 3,14 × 82 = = 100,53 4 mm2 Calculation spacing of stirrups: 33 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) stt = Rsw Asw 175 × 100,53 = = 102,761 qsw 171,201 mm Maximum spacing of stirrups: s ma x = Rbt bh02 0,66 × 250 × 580 = = 269,343 Q 206079 mm The final spacing of stirrups is chosen as minimum of {stt; sct; smax}: s1 = min(102,761; 216,667; 269,343) = 102,761 mm Take s1 = 100 mm for segment near the bearing QCt d Calculate with the shear force in the right of bearing C: = 196,091 kN Dimensions of the beam: b = 250 mm, h = 650 mm, h0 = 585 mm – Condition of calculating shear reinforcement + Check condition to resist principal compressive stress of beam’s web: Q = 196,091 kN < 0,3Rbbh0 = 0,3×7,5.103×0,25×0,585 = 329,232 kN – satisfaction + Check shear strength of concrete: Qbmin = 0,5Rbtbh0 = 0,5×0,66.103×0,25×0,5853 = 48,287 kN < Q = 196,091 kN Concrete is not capable to resist all of shear force, need to design stirrups – Calculation of stirrups without inclined reinforcement + Assume c ≤ 2h0 With assumption that concrete and stirrups resist all of shear force: Q = 195,441 kN = QDB = 4,5Rbtbh02 qsw Q2 196,0912 qsw = = 4,5Rbtbh02 4,5 × 0,66.103 × 0,25 × 0,5852 = 151,169 N/mm qswmin < qsw = 151,169 N/mm Recalculate c0 with qsw = 151,169 N/mm: c0 = Rbtbh02 × 0,66 × 250 × 5852 = = 864,333 qsw 151,169 mm c0 = 864,333 mm < 2h0 = 2×585,3 = 1170,6 mm, agree with the assumption – Select stirrups: diameter ϕ8, number of legs n = Cross-sectional area of stirrups perpendicular to member axis: Asw = nπφw2 × 3,14 × 82 = = 100,53 4 mm2 Calculation spacing of stirrups: 34 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) stt = Rsw Asw 175 × 100,53 = = 116,378 qsw 151,169 mm Maximum spacing of stirrups: Rbtbh02 0,66 × 250 × 5852 sma x = = = 288,259 Q 196091 mm The final spacing of stirrups is chosen as minimum of {stt; sct; smax}: s1 = min(116,378; 216,667; 288,259) = 116,378 mm Take s1 = 110 mm for segment near the bearing 4.7 Design of local stirrups – At the place where the primary beam supports the secondary beam, need to lay local stirrups to strengthen for primary beam – Concentrated load from secondary beam acting on primary beam: P1 = P + G1 = 121,154 + 44,733 = 165,887 kN – The local stirrups is similar stirrups, area of local stirrups: Asw = h − hdp 165,887.103 × 1 − 612,5 − 450 P1 1 − h0 612,5 = = 696,4 Rsw 175 mm2 Arrangement of local stirrups hs = 612,5 – 450 = 162,5 mm – Select local stirrups: diameter ϕw = mm, number of legs n = asw = nwπφ sw × 3,14 × 82 = = 100,53 4 mm2 35 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) Asw 696,4 = = 6,9 a sw 100,53 m= Select local stirrups, arrange steels in each side Distance between steels is 40 mm 4.8 Calculation and drawing moment capacity envelop diagram a Calculation of moment capacity – At outer span, positive moment, T-section with the flange is in compressive zone, the flange b 'f breadth b = = 2150 mm, reinforced steels 2ϕ22 + 2ϕ25, As = 1742,1 mm2, h0 = 612,5 mm RA 280 × 1742,1 ξ= s s = = 0,049 Rb bh0 7,5 × 2150 × 612,5 h 'f x = ξh0 = 0,049×612,5 = 29,9 mm < = 80 mm – the neutral axis is in flange ζ = – 0,5ξ = – 0,5×0,049 = 0,975 Mtd = RsAsζh0 = 280×1742,1×0,975×612,5 = 291,392.106 N.mm = 291,392 kN.m – At bearing B, negative moment, rectangular section b×h = 250×650 mm, reinforced steels 4ϕ22 + 2ϕ25, As = 2502,3 mm2, h0 = 580 mm RA 280 × 2502,3 ξ= s s = = 0,644 Rbbh0 7,5 × 250 × 580 ζ = – 0,5ξ = – 0,5×0,644 = 0,678 Mtd = RsAsζh0 = 280×2502,3×0,678×580 = 275,466.106 N.mm = 275,466 kN.m The results of calculations of bearing capacity are shown in the bellow table: Section Quantity and area reinforcement (mm2) h0 mm Center of outer span 2ϕ25 + 2ϕ22, As = 1742,1 612,5 Outer span adjacency Cut 2ϕ25, remain 2ϕ22, As = 760,3 614 Center of bearing B 4ϕ22 + 2ϕ25, As = 2502,3 580 Left bearing B adjacency Cut 2ϕ25, remain 4ϕ22, As = 1520,5 601 Left bearing B adjacency Cut 2ϕ22, remain 2ϕ22, As = 760,3 601 Right bearing B adjacency Cut 2ϕ25, remain 4ϕ22, As = 1520,5 601 Right bearing B adjacency Cut 2ϕ22, remain 2ϕ22, As = 760,3 601 Center of middle span 2ϕ22 +1ϕ22, As = 1140,4 614 Middle span adjacency Cut 1ϕ22, remain 2ϕ22, As = 760,3 614 Center of bearing C 4ϕ20+ 2ϕ20, As = 1885,0 585 Bearing C adjacency Cut 2ϕ20, remain 4ϕ20, As = 1256,6 602 ξ ζ 0,049 0.022 0,644 0,378 0,189 0,378 0,189 0,032 0,022 0,481 0,975 0,989 0,678 0,811 0,906 0,811 0,906 0,984 0,989 0,760 Mtd kN.m 291,392 129,306 275,466 207,535 115,858 207,535 115,858 192,896 129,306 234,635 0,321 0,844 178.800 36 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) Bearing C adjacency Cut 2ϕ20, remain 2ϕ20, As = 628.3 602 0,156 0.922 97.653 b Determine theoretical cutting position and extension length W – Reinforcement no.2 (the right part) After cutting reinforced steels no.2 (2ϕ25), section near the bearing B, the outer span remains reinforced steels no.1 (2ϕ22) Bearing capacity is 129,306 kN.m The moment capacity envelope diagram meet the moment envelope diagram at a point which is theoretical cutting position of the reinforced steels no.2 By geometrical relation between similar triangles, distance from this point to the outer bearing is 916 mm At this area, stirrups are arranged ϕ8a170 206,889 + 183,815 Q= = 1,9 The magnitude of shear force at this point: 205,634 kN Determine the extension W: qsw = W2p = Rsw Asw 175× 201,106 = s 170 Q − Qs , inc 2qsw + 5φ = = 206,974 N/mm 205,634 − + × 0,025 × 206,974 = 0,622 m > 20ϕ = 20×0,025 = 0,500 m W2p Take = 622 mm – The results of determining theoretical cutting position and extension length are shown in the bellow table: Reinforced steel RS no.2 – right part RS no.3 – left part RS no.3 – right part RS no.4 – left part RS no.4 – right part RS no.5 – left part RS no.5 – right part RS no.6 – left part Theoretical cutting position Distances axis of B-bearing 1523 mm Distances axis of B-bearing 437 mm Distances axis of B-bearing 465 mm Distances axis of B-bearing 841 mm Distances axis of B-bearing 1110 mm Distances axis of B-bearing 2620 mm Distances axis of B-bearing 1370 mm Distances axis of B-bearing 1702 mm Extension length W2p W3t W3p W4t W4p W5t W5p W6t = 623 mm = 673 mm = 735 mm = 1059 mm = 440 mm = 440 mm = 440 mm = 632 mm 37 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) RS no.6 – right part RS no.8 – left part RS no.9 – left part Distances axis of C-bearing 1548 mm Distances axis of C-bearing 393 mm Distances axis of C-bearing 837 mm W6p W8t W9t = 618 mm = 697 mm = 400 mm 4.9 Checking for anchorage of reinforcement After cutting, amount of remaining positive reinforcement when anchoring to the bearing must be greater 1/3 of area of positive at the center of span Outer span: 2ϕ22 + 2ϕ25, cut 2ϕ25 and remain 2ϕ22, reach 43,66% Middle span: 2ϕ22 + 1ϕ22, cut 1ϕ22 and remain 2ϕ22, reach 66,7% 38 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) Moment capacity and arrangement of reinforcement in primary beam Some cross sections 39 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) - The end - 40 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) [...]... 10 3,829 -1 7 ,322 -1 3 8,479 5, 815 -0 ,14 3 -0 ,14 3 1, 048 -1 7 ,325 -1 7 ,325 -1 7 ,325 12 6,969 5, 815 0,679 -1 , 3 21 1,274 82,264 -3 8,890 -1 6 0,044 15 4,350 33 ,19 6 -0 ,095 -0 ,095 0, 810 -1 1 , 510 -1 1 , 510 -1 1 , 510 98 ,13 5 -2 3, 019 0, 810 -1 , 19 0,286 98 ,13 5 -2 3, 019 -1 4 4 ,17 3 34,650 34,650 0,036 0 ,18 7 4,362 22,656 14 1,085 -2 6,433 -7 8,252 206,790 33, 911 19 ,9 31 -5 3, 813 -2 27 ,14 6 57,255 -2 2,758 Left of bearing C -0 ,995 - 51, 918 -0 ,952 -1 1 5,339... shown in the below table: Moment 1 2 B 3 4 C MG + MP1 268,2 91 206,889 -1 8 3, 815 -6 4,207 -4 3.640 -1 2 1 ,11 5 MG + MP2 37,638 -2 3,074 -1 8 3, 815 16 5,755 18 6,322 -1 2 1 .11 5 MG + MP3 227,087 12 4,940 -3 06,737 94,856 16 7, 217 -8 9,658 MG + MP4 49,378 -0 ,975 -1 5 0,667 14 4 ,11 7 10 9,669 -2 54, 015 MG + MP5 257,242 18 5,2 51 - 216 ,272 -4 2 ,10 9 33, 014 9,095 MG + MP6 79,073 59 ,10 5 -6 0,2 01 7 ,12 2 -2 4,549 -1 5 5,263 Because the beam is symmetrical,... -0 ,14 3 -9 8,753 -0 ,3 21 -2 21, 675 -0 ,095 -6 5,605 -0 ,19 0 -1 3 1, 210 3 0,079 23,496 -0 ,12 7 -8 7,703 0,206 14 2,259 4 0 ,11 1 33, 014 -0 ,11 1 -7 6,654 0,222 15 3,308 71, 360 13 4,203 12 0,6 21 76,655 -6 5,605 0 C -0 ,19 0 -5 6, 510 -0 ,095 -6 5,605 -0 ,095 -6 5,605 -0 ,048 -3 3 ,14 8 -0 ,286 -1 9 7,505 0,095 65,605 23 ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) MP6 Mmax Mmin α M 8,287 268,2 91. .. × 1, 9 × 0, 415 2 1 + 1 − 2α m 2 = 1 + 1 − 2 × 0,0 31 = 0,984 2 M 76, 786 .10 3 As = = Rsζ h0 280 .10 6 × 0,984 × 0, 415 µ= = 6, 715 .10 -4 m2 = 6 71, 5 mm2 As 6 71, 5 = × 10 0% = 0,899% bdp h0 18 0 × 415 – At middle spans, M = 51, 444 kN.m: M 51, 444 .10 3 αm = = = 0,0 21 < α pl = 0,255 Rbbh02 7,5 × 10 6 × 1, 9 × 0, 415 2 ζ = 1 + 1 − 2α m 1 + 1 − 2 × 0,0 21 = = 0,989 2 2 As = µ= M 51, 444 .10 3 = Rsζh0 280 .10 6 × 0,989 × 0, 415 ... UNIVERSITY OF CIVIL ENGINEERING (congdaihvp@yahoo.com) M 268,2 91. 103 = Rsζh0 280 .10 6 × 0,977 × 0, 61 As = µ= = 1, 6076 .10 -3 m2 = 16 07,6 mm2 As 16 07,6 = × 10 0% = 1, 05% bh0 250 × 610 – At middle spans, M = 18 6,322 kN.m αm = ζ = 1 + 1 − 2α m 1 + 1 − 2 × 0,0 310 = = 0,985 2 2 As = µ= M 18 6,322 .10 3 = = 0,0 310 < α R = 0,442 Rbbh02 7,5 × 10 6 × 2 ,15 × 0, 612 M 18 6,322 10 3 = Rsζh0 280 .10 6 × 0,985 × 0, 61 = 1, 1084 .10 -3 ... 33 ,14 8) − 33 ,14 8 = 13 4, 203 3 kN.m Similarly for diagram of MP4, MP5 and MP6, calculate moment at sections 1, 2, 3, 4 The results are shown in the below table: Moment (kN.m) MG α M MP1 α M MP2 α M MP3 α M MP4 α M MP5 α M 1 0,238 70,786 0,286 19 7,505 -0 ,048 -3 3 ,14 8 2 0 ,14 3 42,5 31 0,238 16 4,358 -0 ,095 -6 5,605 15 6,3 01 -0 ,0 31 - 21, 408 82,409 -0 ,063 -4 3,506 18 6,456 14 2,720 B -0 ,286 -8 5,062 -0 ,14 3 -9 8,753 -0 ,14 3... Center of outer span Outer span adjacency 2 18 + 1 16 , As = 710 Cut 1 16 , remain 2 18 , As = 508,9 Center of bearing 2 2 16 + 1 16 , As = 603,2 Bearing 2 adjacency Cut 1 16 , remain 2 16 , As = 402 ,1 Center of middle span 2 14 + 1 14 , As = 4 61, 8 Middle span adjacency Cut 1 14 , remain 2 14 , As = 307,9 h0 mm 4 21 ξ ζ 0,033 0,983 Mtd kN.m 82,308 4 21 422 0,024 0,296 0,988 0,852 59,277 60,709 422 423 0 ,19 8 0,0 21. .. 594,9 = × 10 0% = 0,796% bh0 18 0 × 415 – At bearing 3, M = 51, 444 kN.m M 51, 444 .10 3 αm = = = 0, 2 21 < α pl = 0, 255 Rb bh02 7,5 × 10 6 × 0 ,18 × 0, 415 2 ζ = 1 + 1 − 2α m 2 As = µ= = 1 + 1 − 2 × 0, 2 21 = 0,873 2 M 51, 444 .10 3 = Rsζ h0 280 .10 6 × 0,873 × 0, 415 = 5,069 .10 -4 m2 = 506,9 mm2 As 506,9 = × 10 0% = 0,678% bh0 18 0 × 415 b Longitudinal reinforcement subjected to positive moment – Design with T-section,... section Value 1, β2 1 Ordinate of moment M (kN.m) β2 Outer span Bearing 1 0 0 1 0,065 54,847 2 0,09 75,942 0,425l 0,0 91 76,786 3 0,075 63.285 4 0,02 16 .876 Bearing 2 – section 5 –0,0 715 –58.852 Secondary span 6 0, 018 –0,0338 14 . 816 –27.8 21 7 0,058 –0, 013 7 47.740 11 .276 0,5l 0,0625 8 0,058 –0, 011 1 47.740 –9 .13 6 9 0, 018 –0,0278 14 . 816 –22.882 51. 444 Bearing 3 – section 10 –0,0625 – 51. 444 – The section... = β 12 1 ,15 4 kN.m β is looked up in Appendix 12 Shear force at center of span is calculated by section method The results are shown in the bellow table: Shear force QG QP1 QP2 QP3 QP4 QP5 QP6 Qmax Qmin β Q β Q β Q β Q β Q β Q β Q Right of Center of Left of Right of Center of bearing A outer bearing B bearing B span 2 span 0, 714 -1 , 286 1, 005 37,256 -1 4 ,923 -6 7 ,10 2 52,440 0,2 61 0,857 -1 , 143 0,048 10 3,829 ... 82,264 -38,890 -16 0,044 15 4,350 33 ,19 6 -0,095 -0,095 0, 810 -11 , 510 -11 , 510 -11 , 510 98 ,13 5 -23, 019 0, 810 -1, 19 0,286 98 ,13 5 -23, 019 -14 4 ,17 3 34,650 34,650 0,036 0 ,18 7 4,362 22,656 14 1,085 -26,433... span span 0, 714 -1, 286 1, 005 37,256 -14 ,923 -67 ,10 2 52,440 0,2 61 0,857 -1, 143 0,048 10 3,829 -17 ,322 -13 8,479 5, 815 -0 ,14 3 -0 ,14 3 1, 048 -17 ,325 -17 ,325 -17 ,325 12 6,969 5, 815 0,679 -1, 3 21 1,274 82,264... Moment B C MG + MP1 268,2 91 206,889 -18 3, 815 -64,207 -43.640 -12 1 ,11 5 MG + MP2 37,638 -23,074 -18 3, 815 16 5,755 18 6,322 -12 1 .11 5 MG + MP3 227,087 12 4,940 -306,737 94,856 16 7, 217 -89,658 MG + MP4