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NATIONAL UNIVERSITY OF CIVIL ENGINEERING DEPARTMENT OF REINFORCED CONCRETE CONS TRUCTION REINFORCED CONCRETE STRUCTURES I MINI-PROJECT: RIBBED SLAB WITH ONE-WAY SLAB DESIGN SUPERVISOR: PHAM THANH TUNG STUDENT’S NAME: PHAM VAN TIEN STUDENT CODE: 205463 CLASS: 63XE2 DATE COMPLETED: 17.08.2021 Hanoi 2021 NATIONAL UNIVERSITY OF CIVIL ENGINEERING DEPARTMENT OF REINFORCED CONCRETE CONS TRUCTION REINFORCED CONCRETE STRUCTURES I MINI-PROJECT: RIBBED SLAB SYSTEM WITH ONE-WAY SLAB DESIGN LECTURER: PHAM THANH TUNG STUDENT’S NAME: PHAM VAN TIEN STUDENT CODE: 205463 CLASS: 63XE2 DATE COMPLETED: 17.08.2021 Hanoi, 08/2021 TABLE OF CONTENTS I GIVEN DATA II CALCULATION OF PLATE 2.1 Analysis 2.2 Pre-Select components dimensions 2.4 Design load Total 2.5 Design internal forces 2.6 Calculation of bending moment bearing reinforcement: III SECONDARY BEAM CALCULATION 10 3.1 Schematic Diagram 10 3.2 Design load 11 3.3 Design Internal force 11 3.4 Longitudinal reinforcement 13 3.5 Selection and arrangement of longitudinal reinforcement 15 3.6 Calculation of transverse reinforcement 15 3.7 Calculation and construction of the Material Resistance Envelope diagram: 18 3.8 Constructive reinforcement 21 IV CACULATION PRIMARY BEAM: 22 4.1 Schematic diagram 22 4.2 Design load 22 4.3 Design internal forces: 22 4.4 Longitudinal reinforcement: 28 4.5 Calculation of reinforcement for shear force 31 4.6 Calculation of hanging reinforcement: 34 4.7 Construct the Material Resistance Envelope 35 APPENDIX 40 APPENDIX 41 I GIVEN DATA Slab structure as presented in Figure 1.1 Length measured from the central axis of the beams and walls l1 = 2,55m; l2 = 6,8m Load bearing walls with thickness bt = 340mm Reinforced concrete columns: - Middle column: cross-sectional area bc× hc = 300x300mm - Columns inside walls on axis 1, 6: 300x300mm - Outer columns supporting the primary beams: 300x300mm Slab for civil buildings, which consists of layers as shown in Figure 1.1 Nominal live load Ptc = 8,05kN/m2; factor of reliability of live load n = 1,2 Materials: Concrete with Compressive Grade B20, plate reinforcement and stirrups class CI, longitudinal reinforcement class CII Figure 1.1 Plan of Slab and Slab details II CALCULATION OF PLATE 2.1 Analysis Ribbed slab supported by beams along directions The beams located on axis 2, 3, 4, are primary beams, perpendicular to which are secondary beams Dimensions of plate aperture: l1 = 2550mm, l2 = 6800mm, l2 > 2l1 so it has similar features of the One-way slab 2.2 Pre-Select components dimensions Thickness of plate: hb = D 1,1 l1 = 2550 = 80mm Assume hb = 80mm m 35 whereas D = 1,1 with moderate load, m = 35 with continuous plate Cross-sectional dimensions of secondary beam selection: 1 l2 = 6800 = 485,7 mm Assume hdp = 500mm, bdp = 220mm; mdp 14 hdp = Cross-sectional dimensions of primary beam selection: primary beam span equals to distance between columns, 7650mm hdc = 1 3l1 = 7650 = 695,5mm Assume hdc = 800mm, bdc = 300mm; mdc 11 2.3 Schematic diagram One-way slab, establish a simplified model of a plate strip with the width of b1=1m perpendicular to the secondary beams Consider the strip as a continuous beam Design span length of plate: Outer span: lob = l1b − bdp − bt 0, 22 0,34 + Cb = 2,55 − − + 0,5 0,08 = 2,31m 2 Middle span: lo = l1 − bdp = 2,55 − 0,22 = 2,33m Span differences: 2,33 − 2,31 100% = 0,86% 2,33 2.4 Design load Dead load is calculated and presented in Table 2.1 Table 2.1 Calculation of dead load Layers of plate Nominal Value (kN/m2) Factor of Reliabilit y Design Value (kN/m2) - Lining brick of 10mm, γ = 20kN/m3 0,01×20 = 0,200 1,1 0,220 0,03×18 = 0,540 1,3 0,702 0,08×25 = 2,000 1,1 2,200 0,01×18 = 0,180 1,3 0,234 - Lining mortar of 30mm, γ = 18kN/m3 - Reinforced concrete slab of 70mm, γ = 25kN/m3 - Plastering mortar, γ = 18kN/m3 Total 2,920 Round up gb = 3,36kN/m2 Live load pb = Ptc.n = 8,05×1,2 = 9,66kN/m2 Total load qb = gb+pb = 3,36+9,66 = 13,02kN/m2 Strip of slab with thickness of b1 = 1m has qb = 13,02×1 = 13,02kN/m 2.5 Design internal forces Following the plastic scheme model: - Bending moments at the outer span and the second support: q l 13,02 2,312 M nh = M g = b ob = = 6,314kNm 11 11 - Bending moments at the middle spans and middle supports: q l2 13,02 2, 232 M nhg1 = M g1 = b o = = 4, 416kNm 16 16 The value of the maximum shear force: QBT = 0,6qblob = 0,6×13,02×2,31 = 18,04kN 3,356 Figure 2.1 Schematic and internal force diagram in slab strip 2.6 Calculation of bending moment bearing reinforcement: Given data: Concrete grade B20, has Rb = 11,5MPa, reinforcement class CI, has Rs = 225MPa Calculate the internal force based on plastic scheme, with coefficient of limiting compression area pl = 0,255 Assume that a = 15mm for all section: h0 = hb – a = 80 – 15 = 65mm At the outer support and outer span, with M = 3,9kNm (convert into 3,9×106Nmm): m = M 6,314 106 = = 0,13 pl = 0, 255 Rbbh02 11,5 1000 652 According to Appendix in ‘Reinforced concrete structures’, ζ = 0,930; 6,314 106 As = = = 464mm Rs h0 225 0,930 65 M % = As 464 = 100 = 0,71% b1h0 1000 65 Select diameter of reinforcement 8mm, as = 50,27mm2, spacing of adjacent rebars is: s = b1as 1000 50, 27 = 108mm → Select ɸ8, s = 100mm As 464 At the middle supports and middle spans, with M = 4,416kNm, αm = 0,091; ζ = 0,952; As = 317mm2 Select diameter of reinforcement 8mm, as = 50,27mm2, design spacing of adjacent rebars is s = 158,6mm → Select ɸ8, s = 150mm Check the working height ho with covering thickness of 10mm: hot = 80 – 10 – 0,5×8 = 66mm > 65mm → Safety Reinforcement under negative moment: with pb/gb = 9,66/3,36 = 2,9 < 3, value ν = 0,25; extension from edge of the secondary beam is: νlo = 0,25×2,33= 0,583m, extension from axis of the secondary beam is: νlo+0,5bdp = 0,583+0,5×0,22 = 0,6925m Longitudinal reinforcement under negative moment is arranged alternatively, extension of the shorter reinforcement from the edge of secondary beam is: From the axis of secondary beam: 1 lo + 0,5 bdp = 2,33 + 0,5 0, 22 = 0,5m 6 Longitudinal reinforcement under positive moment is arranged alternatively, distance from the end of the shorter reinforcement to edge of the secondary beam is: 1 lo = 2,33 = 0, 291m 8 Check the shear strength: Qbmin = 0,625Rbtb1h0 = 0,625×0,9×1000×65 = 36562N = 36,56kN QBT = 18,4kN < Qbmin Concrete achieves shear resistance! 2.7 Constructive reinforcement - Reinforcement under negative moment is placed in direction which is perpendicular to the primary beam and wall bracing: Select ɸ6, s = 150mm with cross-sectional area of 188,5mm2 per 1m, higher than 50% of the area of reinforcement at the middle support of plate which is 0,5×317mm2 = 159mm2, using capping reinforcement, extension from edge of the primary beam is: 1 lo = 2,33 = 0,5825m 4 From axis of the primary beam: 1 lo = 2,33 = 0,39m 6 1 lo + 0,5 bdc = 2,33 + 0,5 0,30 = 0,7325m 4 - Distribution reinforcement is placed in direction which is perpendicular to load bearing reinforcement: select ɸ6, s = 200 have area per 1m is 141mm2, guaranteed to be higher than 30% of the area of assumed reinforcement at the span (outer span 0,3×464=139mm2, middle span 0,3×317=95,1mm2) Figure 2.2 Reinforcement arrangement in slab III SECONDARY BEAM CALCULATION 3.1 Schematic Diagram Given secondary beam type is a symmetrical 5-span continuous beam We consider the left half of the beam (figure 1.4) There is an overlap of Sd between the beam and the wall, which is equal to the thickness of the wall, Sd = 340mm Cd = (Sd/2 và l2/40); (Sd/2) = 170mm = (l2/40) = 170mm So Cd = 170mm Design span length of secondary beam: Outer span length: l pb = l2 − bdc bt 0,30 0,34 − + Cd = 6,8 − − + 0,17 = 6,65m 2 2 Middle span length: l p = l2 − bdc = 6,80 − 0,30 = 6,50m Span differences: 6,65 − 6,50 100% = 2, 26% 10% 6,65 Figure 3.1 Schematic diagram and Internal force diagrams for secondary beam b Determine the envelope diagram of Shear Force: Ordinate in the envelope diagram of shear force: - Due to the effect of dead load G: QG = βG = β×90,619 (kN) - Due to the effect of live load Pi: QPi = βiP = βi×167,504 (kN) Whereas the coefficient β is according to Appendix 12 in ‘Reinforced concrete structures’, cases are shown in Figure 4.3, the result are shown in Table 4.2 Table 4.2 Calculation and Combination of the Shear Forces Right of support A Middle of the outer span Left of support B Right of support B Middle of Span Left of support C β 0,714 … -1,286 1,005 … -0,995 Q 64,70 -25,92 -116,54 91,07 0,45 -90,17 β 0,857 … -1,143 0,048 … Q 143,55 -23,95 -191,46 8,04 … … β -0,143 … -0,143 1.048 … -0,952 Q -23,95 -23,95 -23,95 175,54 8,04 -159,46 β 0,679 … -1,321 1.274 … -0,726 Q 113,74 -53,77 -221,27 213,40 45,90 -121,61 β -0,095 -0,095 0,810 … -1,190 Q -15,91 -15,91 -15,91 135,68 -31,83 -199,33 β 0,810 … -1,190 0,286 … 0,286 Q 135,68 -31,83 -199,33 47,91 47,91 47,91 β … … 0,036 … … -0.143 Q 6,030 6,030 6,030 -29,95 -29,95 -29,95 Qmax 208,25 -41,83 -110,51 304,47 48,36 -42,26 Qmin 40,75 -79,69 -337,81 99,91 -31,37 -289,50 Shear force (kN) QG QP1 QP2 QP3 QP4 QP5 QP6 In the middle of spans, the shear force Q is determined using the method of section, apply the equation of equilibrium for spans For example, in the middle of the outer span: Q = QA – G = 64,70 – 90,619 = –25,92kN The envelope diagram of shear force is shown in Figure 4.6 Figure 4.6 Envelope diagram of shear force 4.4 Longitudinal reinforcement: Concrete with Compressive Grade B20 with Rb = 11,5MPa; Rbt = 0,9MPa; reinforcement class CII with Rs = 280MPa, Rsc = 280MPa According to Appendix 8, with working condition coefficient of concrete γb2 = 1,0; coefficient of limiting compression area when internal forces are determined using elastic diagram is ξR = 0,623; αR = 0,429 a For negative moment: For the rectangular section with dimensions b = 300mm, h = 800mm At supports, reinforcement of primary beam must be put underneath the top layer of reinforcement in secondary beam, so a is considerably large Assume a = 80mm, h0 = 800 – 80 = 720mm At support B, with Mmg = 562,73kNm m = = M 562,73 106 = = 0,315 R = 0, 429 Rbbh02 11,5 300 7202 + − 2 m + − 0,315 = = 0,804 2 As = M Rs h0 = 562,728 106 = 3470mm 280 0,804 720 Check % = As 3470 = 100 = 1,61% bdc h0 300 720 At support C, with Mmg = 456,00kNm; αm = 0,255; ζ = 0,850; As = 2661mm2, μ = 1,23% b For positive moment: For T-section when the flange is compressed with hf = 80mm Assume a = 70mm, h0 = 800 – 70 = 730mm Effective breadth Sf is minimum value of following value: + (1/6)ld = (1/6)×7,65 = 1,275m + Half of clearance between two adjacent primary beams: 0,5l=0,5×6,5= 3,25m (due to h’f > 0,1h; with h = 800mm and transverse beams are secondary beams with clearance of 2,55m) So, Sf ≤ min(1,275; 3,25) = 1,275m Assume Sf = 1,275m = 1275mm Flange width b’f = b + 2Sf = 300 + 2×1275 = 2850mm Mf = Rb b’f h’f (h0 - 0,5h’f ) = 11,5×2850×80×(680 - 0,5×80) = 1678,08×106Nmm Mmax = 531,473kNm < Mf = 1678.08kNm → Neutral axis is within the flange For rectangular section with dimensions b = b’f = 2850mm, h = 800mm, a = 70mm, h0 = 720mm At outer span, with M = 531,473kNm M 531, 473 106 m = = = 0,03 R = 0, 429 Rbb f h02 11,5 2850 7302 = + − 2 m + − 0,03 = = 0,985 2 As = Check % = M Rs h0 = 531, 473 106 = 2641mm 280 0,985 730 As 2641 = 100 = 1, 21% bdc h0 300 720 At middle span, with M = 361,42kNm; αm = 0,021, ζ = 0,990, As = 1787mm2, μ = 0,82% Table 4.3 Selection of longitudinal reinforcement for primary beam Section Outer span Support B Second span Support C Design As 2641mm2 3470mm2 1787mm2 2661mm2 Rebars 3Φ25+2Φ28 4Φ28+2Φ28 2Φ25+2Φ22 4Φ25+2Φ22 Rebars area 2704mm2 3695mm2 1722mm2 2723mm2 Longitudinal reinforcement arrangement at the main sections is presented in Figure 4.7 Figure 4.7 Longitudinal reinforcement arrangement at the main sections Recheck ho: + For reinforcement at the outer span: select covering thickness c = 30mm > ɸmax, arrange layers with vertical spacing t = 30mm: - Bottom layer of 2ɸ25+1ɸ25, As1 = 1472,6mm2; a1 = c + 0,5d1 = 30 + 0,5×25 = 42,5mm - Upper layer of 2ɸ28, As2 = 1231,5mm2; a2 = c + d1 + t + 0,5d2 = 30 + 25 + 30 + 0,5×28 = 99mm → a= 1472,6 42,5 + 1231,5 99 = 68,23mm ~ 68,5mm 1472,6 + 1231,5 → hot = 800 – 68,5 = 731,5mm > ho = 730mm (assumed value for calculations above) So, it is safe with a = 70mm! + For reinforcement at the middle span: select covering thickness c = 30mm, arrange layer of reinforcement: 2ɸ25+2ɸ22, As = 1742mm2: a = c + 0,5d = 30 + 0,5×25 = 42,5mm →hot = 800 – 42,5 = 757,5 > ho = 730mm So, it is safe with a = 70mm! + For reinforcement under negative bending moment: select covering thickness (measured to the inner edge of the reinforcement arranged in the secondary beam) c = 47mm (thickness of covering of secondary beam is 25mm and maximum reinforcement diameter is ɸ22) - At support C, arrange layers with vertical spacing t = 30mm: ● Top layer of 4ɸ25, As1 = 1963,6mm2; a1 = c + 0,5d1 = 47 + 0,5×25 = 59,5mm ● Lower layer of 2ɸ22, As2 = 760,3mm2; a2 = c + d1 + t + 0,5d2 = 47 + 25 + 30 + 0,5×22 = 113mm → a= 1963,6 59,5 + 760,3 113 = 74, 4mm ~74,5mm 1963,6 + 760,3 → hot = 800 – 74,5=725,5mm > ho = 720mm, → So, it is safe with a = 80mm! - At support B, arrange layers with vertical spacing t = 30mm: ● Top layer of 4ɸ28, As1 = 2463mm2; a1 = c + 0,5d1 = 47 + 0,5×28 = 61mm ● Lower layer of 2ɸ28, As2 = 1231,5mm2; a2 = c + d1 + t + 0,5d2 = 47 + 28 + 30 + 0,5×28 = 119mm; → a= 2463 61 + 1231,5 119 = 80,33mm ~80,5mm 2463 + 1231,5 → hot = 800 – 80,5 = 719,5mm < ho = 720mm However, it’s not necessary to recalculate the h0, because the difference between h0t and h0 is quite small and the chosen rebars area is larger than the designed rebars area + Check spacing of reinforcement: (1 row 4ɸ25) Select covering at the side is c = 30mm > ɸ Reinforcement is arranged equidistantly, horizontal spacing is: t= b − 2c − 4 300 − 30 − 28 = = 42,67mm (Larger than the requirement of 3 30mm) The inner bars should be arranged near the bars in the corner to widen the gap between them, which makes it easier to use compactors in concrete casting phase 4.5 Calculation of reinforcement for shear force From shear force diagram of primary beam: Right edge of support A: QAP = 208,25kN, shear force is constant through the length l1 Left edge of support B: QBT = 337,81kN, shear force is constant through the length l1 Right edge of support B: QBP = 304,47kN, shear force is constant through the length l1 Left edge of support C: QCT = 289,50kN, shear force is constant through the length l1 ❖ Design stirrups for shear force at the right of support A: QAP= 208,25kN Beam dimensions: b = 300mm, h = 800mm, ho = 744mm Determine: 0,3Rbbho = 0,3×11,5×300×742,5 = 757102N = 757,102kN Determine: Qbmin = 0,5Rbtbho = 0,5×0,9×300×731,5 = 98753N = 98,753kN Qbmin = 98,753kN < Q = 208,25kN < 0,3Rbbho = 757,102kN Satisfied the condition (12)! Calculation of stirrups (without diagonal reinforcement): Given: Q = 208,25kN = Qđb = 4,5 Rbt bh02 qsw (Which also means that all shear forces are applied to concrete and stirrups) qsw = Q2 4,5 Rbt bh02 = 2082502 4,5 0,9 300 731,52 = 66, N/mm qswmin = 0,25Rbtb = 0,25×0,9×300 = 67,5N/mm qswmin = 66,7N/mm < qsw = 67,5N/mm + Check Co by qsw = 67,5N/mm: Co = 1,5Rbt bh02 1,5 0,9 300 731,52 = = 2069mm 0,75qsw 0,75 67,5 Co = 2069mm > 2ho = 2×742,5mm = 1485mm Assume Co= 2ho=1485mm + Given: Q = 208, 25kN = QÐB = => qsw = Q − 0, 75 Rbt bh0 1,5Rbt bh02 + 0,75qsw 2ho 2ho = 208250 − 0, 75 0, 300 731, 1, 731, 1, 5h0 = 54,8 N / mm Clearly, qsw < qswmin = 67,5N/mm Design stirrups based on qswmin = 67,5N/mm + Select the diameter of the stirrups 10mm (ɸ w = 10mm), legs (n = 2) Area of a layer of stirrups: Asw = nw2 = 3,14 10 = 157, 08mm The design spacing according to calculations of Stt between the stirrup layers: Stt = Rsw Asw 175 157,08 = = 407mm qsw 67,5 The spacing between layers of stirrups according to structure: Sct = (0,5ho = 372mm; 300mm) →select Sct = 300mm The greatest spacing between layers of stirrups Smax : Smax Rbt bh02 0,75 300 731,52 = = = 578mm Q 208250 The stirrups is arranged with spacing s = (Stt; Sct; Smax); select s = Sct = 300mm ❖ Calculation with shear force at the left side of support B: QBT= 337,81kN Dimensions of beam: b = 300mm, h = 800mm, ho = 723mm Determine: 0,3Rbbho = 0,3×11,5×300×723 = 748305N = 748,305kN Determine: Qbmin = 0,5Rbtbho = 0,5×0,9×300×723 = 97605N = 97,605kN Qbmin = 0,5Rbtbho = 97,605kN < Q = 337,81kN < 0,3Rbbho = 748,305kN Satisfied the conditions (12)! + Select stirrups that satisfies structural requirements: Select ɸ10, legs, s = 150mm Determine qsw (the force that the stirrups can bear) qsw = Rsw Asw 175 157,08 = = 182, 21 N/mm s 150 qswmin = 0,25Rbtb = 0,25×0,9×300 = 67,5N/mm < qsw = 182,21N/mm Co = 1,5 Rbt bh02 0, 75qsw = 1,5 0,9 300 719,52 0, 75 182, 21 = 1239mm Co = 1239mm < 2ho = 2×723 = 1439mm Qđb = 4,5 Rbt bh02 qsw = 4,5 0,9 300 719,52 182, 21 = 338536 N = 338,536 kN Qđb = 338,536kN > Q = 337,81kN Therefore, no diagonal reinforcement needed ❖ Calculation with the shear force at the right side of support B: QBP= 304,74kN Arranged stirrups with ɸ10, legs, s = 180mm: qsw = Rsw Asw s = 175 157, 08 180 = 152, 72 N/mm qswmin = 0,25Rbt b = 0,25×0,9×300 = 67,5N/mm < qsw = 152,72N/mm 1,5 Rbt bh02 Co = 0, 75qsw = 1,5 0,9 300 719,52 0, 75 152, 72 = 1352mm Co = 1352mm < 2ho = 1439mm Qđb = 4,5 Rbt bh02 qsw = 4,5 0,9 300 719,52 152,72 = 309932 N = 309,932kN Qđb = 309,932kN > QBP = 304,74kN ❖ Calculation with the shear force at the left side of support C: QCT= 289,50kN Select reinforcement ɸ10, legs, s =200mm qsw = Rsw Asw 175 157,08 = = 137, 445 N/mm s 200 qswmin = 0,25Rbtb = 0,25×0,9×300 = 67,5N/mm < qsw = 137,445N/mm Co = 1,5 Rbt bh02 0, 75qsw = 1,5 0,9 300 7252 0, 75 137, 455 = 1436mm Co = 1436mm < 2ho = 2×725 = 1450mm Qđb = 4,5 Rbt bh02 qsw = 4,5 0,9 300 7252 137, 445 = 296272 N = 296, 272 kN Qđb = 296,272kN > Q = 289,50kN 4.6 Calculation of hanging reinforcement: At the position where the secondary beams overlap the primary beams, hanging reinforcement needs to be arranged to reinforce the primary beams The concentrated force transverses from the secondary beam to the primary beam: P1 = P+G1 = 167,504 + 90,619 = 256,123kN Hanging reinforcement is stirrups with area of: h 231,5 P1 1 − s 256,123 103 1 − h0 731,5 = 1000, 4mm2 Asw = = Rsw 175 hs= hodc – hdp = 731,5 – 500 = 231,5mm Use stirrups ɸ10, asw = 78,5mm2, number of legs ns = 2, number of stirrups necessary: m= Asw 1000,38 = = 6,377 ns as 78,5 Select stirrups, arrange on each side of the secondary beam Spacing between stirrups is 60mm, the distance from innermost stirrups to the edge of the secondary beam is 50mm 4.7 Construct the Material Resistance Envelope a Calculate bearing capacity + At outer span: For positive moment, cross-section is T-section with neutral axis at the flange, flange effective breath b = b’f = 2850mm, arranged reinforcement of 2Φ28+3Φ25, reinforcement area As = 2704mm2, h0 = 735mm as calculated = Rs As 280 2704 = = 0,0316 Rbbh0 11,5 2850 731,5 x = ξh0 = 0,0313×731,5 = 23,1mm < h’f = 80mm – Neutral axis is within the flange ζ = – 0,5ξ = – 0,5×0,0316 = 0,9842 Mtd = Rs Asζ h0 = 280×2704×0,9842×731,5 = 545,088×106Nmm = 545,088kNm + At support B: Negative moment, rectangular section b× h = 300mm×800mm, arranged top layer of 4Φ28, lower layer of 2Φ28; area As = 3694,5mm2; h0 = 719,5mm as calculated = Rs As 280 3694,5 = = 0, 417 R = 0,623 Rbbh0 11,5 300 719,5 ζ = 1-0,5ξ = – 0,5×0,417 = 0,792 Mtd = Rs As ζ h0 = 280×3694,5×0,792×719,5 = 589,206×106Nmm = 589,206kNm Results of bearing capacity are presented in Table 4.4, all sections are calculated following the singly-reinforced beam case (section with positive moment is b’f instead of b) = Rs As ; = − 0,5 ; M td = Rs As h0 Rbbh0 Table 4.4 Bearing capacity of sections Section Amount and area of reinforcement (mm2) h0 (mm) ξ ζ Mtd (kNm) Middle of outer span 2Φ25+1Φ25+2Φ28 – As = 2704 731,5 0,0316 0,9842 545,088 Edge of outer span Cut 2Φ28, remains 2Φ25+1Φ25 – As = 1472,6 757,5 0,0166 0,9917 309,745 Edge of outer span Cut 1Φ25, remains 2Φ25 – As = 981,8 757,5 0,0111 0,9945 207,087 Support B 2Φ28+2Φ28+2Φ28 – As = 3694,5 Edge of support B Edge of support B 719,5 0,417 0,792 589,206 Cut 2Φ28, remains 4Φ28 – A s= 2463 739 0,2705 0,865 440,716 Cut 2Φ28, remains 2Φ28 – As = 1231,5 739 0,135 0,932 237,590 757,5 0,0196 0,9902 365,849 Middle of second span 2Φ25+2Φ22 – As = 1742 Edge of second span Cut 2Φ22, remains 2Φ25 – As = 981,8 757,5 0,0111 0,9945 207,087 Support C 2Φ25+2Φ25+2Φ22 – As = 2724 725,5 0,305 0,848 469,000 Edge of support C Cut 2Φ22, remains 4Φ25– As =1963,6 740,5 0,2152 0,8924 363,323 Edge of support C Cut 2Φ25, remains 2Φ25 – As = 981,8 740,5 0.1076 0,9462 192,614 b Determine the theoretical sections for cutting of reinforcement Reinforcement no.4 (left edge, near support B): after cutting reinforcement no.3, section at the middle of second span still have reinforcement no.3 (2Φ25) at lower fiber, bearing capacity of lower fiber is 207,087kNm Material diagram intersects envelope diagram of moment at point H, this is the theoretical cutting section of reinforcement no.8 By geometric relation between congruent triangles OBD, OGH and OEF, the distances from H to B is 2143mm (Figure 4.8) Figure 4.8 Schematic diagram to determine the theoretical cut-off section Determine the extension length of reinforcement no.4 (at the left) – W4 t: Q is the slope of moment diagram: Q = 318,74 − 207,08 = 274,33kN 0, 407 In the cutting section of reinforcement no.4 there is no diagonal reinforcement, so In this area, stirrup is ɸ10, s = 150, so qsw = → W8l = Rsw Asw 175 157,08 = = 152,72 N / mm s 180 Q − Qs.inc 274,33 − + 5 = + 0,022 = 1,008m 20 = 20 0,022m = 0, 44m 2qsw 152,72 Round up W4l = 1010mm Similar calculation applied for other rebars, results are presented in Table 4.5 Table 4.5 Theoretical section of reinforcement Reinforcement Theoretical position for cutting of reinforcement Extension length Reinforcement no.1a (left) Distance to support A: 994mm W1a l = 1260mm Reinforcement no.1a (right) Distance to support B: 1746mm W1a r = 1070mm Reinforcement no.2 (left) Distance to support A: 1486mm W2 l = 1280mm Reinforcement no.2 (right) Distance to support B: 2087mm W2 r = 980mm Reinforcement no.4 (left) Distance to support B: 2143mm W4 l = 1010mm Reinforcement no.4 (right) Distance to support C: 1910mm W4 r = 1020mm Reinforcement no.5 Distance to support B: 3007mm W5 = 500mm Reinforcement no.5a (left) Distance to support B: 1101mm W5al = 1060mm Reinforcement no.5a (right) Distance to support B: 1389mm W5a r = 595mm Reinforcement no.6(left) Distance to support B: 500mm W6l = 1060mm Reinforcement no.6(right) Distance to support B: 540mm W6 r = 1135mm Reinforcement no.7a (left) Distance to support C: 1104mm W7a = 1180mm Reinforcement no.8 (left) Distance to support C: 479mm W8 = 1165mm Therefore, the extension length at the left-hand side of 1a Reinforcement is too large, it went through the bearing wall So, to guarantee the bearing capacity on the diagonal section, cutting reinforcement is not allowed at this section Check the reinforcement anchor The reinforcement on the lower fiber after being cut, the leftover when being pulled into the support must have the area greater than 1/3 of the area of reinforcement at the middle span + The outer span reinforcement is cut twice, initial rebars 2Φ25+1Φ25+2Φ28 cut 2Φ28, the remaining area (of 2Φ25+1Φ25) is 54,46% when pulled into the support Then from 2Φ25+1Φ25 cut 1Φ25, the remaining area (of 2Φ25) is 36,04% when pulled into the support + The middle span reinforcement from 2Φ25+2Φ22 cut 2Φ22, the remaining area (of 2Φ25) is 56,4% when pulled into the support + At support A: QA = 208,25kN > Qbmin = 0,5Rbt bho = 98,753kN Anchor length of longitudinal reinforcement = lan = 28Φ = 28×25 = 700mm (with bar no.1) + At support B: Anchor length of longitudinal reinforcement = 20Φ = 20×25 = 500mm (with bar no.1, reinforcement is not under compression – except for double reinforcement) Anchor length of longitudinal reinforcement = 20Φ = 20×25 = 500mm (with bar no.3, reinforcement is not under compression – except for double reinforcement) + At support C: Anchor length of longitudinal reinforcement = 20Φ = 20×25 = 500mm with bar no.3, reinforcement is not under compression – except for double reinforcement) 10 Constructive reinforcement a Reinforcement no.10 (2Φ14): This reinforcement is used as steel strip reinforcement at the outer span, where there is no negative moment The area is 308mm2, no less than 0,1%bh0 = 0,001×300×758 = 227,2mm2 b Reinforcement no.11 (2ɸ14): Theoretically, when the height of the beam is more than 700mm, bars of constructive longitudinal reinforcement need to be added at the upper fiber along the beam length but in most cases, they are arranged when h ≥ 600mm Primary beam Envelope diagram of material and rebars arrangement are presented in Figure 4.9 (Appendix 2) APPENDIX Figure 3.9 Resistance Envelope diagram of material and rebars arrangement of secondary beam APPENDIX Figure 4.9 Resistance Envelop diagram of material and rebars arrangement of primary beam ... 17 5,54 8,04 -15 9,46 β 0,679 … -1, 3 21 1.274 … -0,726 Q 11 3,74 -53,77 -2 21, 27 213 ,40 45,90 -12 1, 61 β -0,095 -0,095 0, 810 … -1, 190 Q -15 , 91 -15 , 91 -15 , 91 135,68 - 31, 83 -19 9,33 β 0, 810 … -1, 190 0,286... -243,468 -12 1,734 0,036 12 1,734 -0 ,14 3 MP6 M 15 ,377 30,754 46 ,13 1 -30,754 -10 7,638 -18 4,523 Mmax 5 31, 473 404 ,10 8 -15 2 ,13 4 318 ,736 3 61, 422 -9,9 81 Mmin 10 3,482 -22,6 01 -609,597 -10 7,973 -65,287 -498 ,19 7... MP3: M1 = 427 ,13 6 – 411 ,332× (1/ 3) = 290,025kNm; M2 = 427 ,13 6 – 411 ,322×(2/3) = 15 2, 915 kNm; M3 = 427 ,13 6 – ( 411 ,332 – 61, 5076)×(2/3) – 61, 5076 = 13 2, 412 kNm; M4 = 427 ,13 6 – ( 411 ,332 - 39,699)× (1/ 3)