| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 357, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n336\nSHEAR RESISTANCE OF CONCRETE (VRd,c)\nfck = 25 N mm−2\nCRd,c = 0.18/γC = 0.18/1.5 = 0.12 N mm−2\n \nk\nd\n \n \n .\n .\n=\n+\n=\n+\n=\n<\n1\n200\n1\n200\n500\n1 63\n2 0\nOK\n \nρ1\n1960\n300\n500\n0 013\n0 02\n \n \n \n .\n .\n=\n=\n×\n=\n<\nA\nb d\nsl\nw\nOK\nσcp = 0\nνmin = 0.035k3/2f ck\n1/2 = 0.035 × 1.633/2 × 251/2 = 0.364\nVRd,c = [CRd,ck(100ρ1fck)1/3 + k1σcp]bwd\n= [0.12 × 1.63(100 × 0.013 × 25)1/3]300 × 500 × 10−3\n= 93.6 kN ≥ (νmin + k1σcp)bwd = 0.364 × 300 × 500 × 10−3 = 54.6 kN\nSince VRd,c < VEd = 450 kN, shear reinforcement must be provided.\nCHECK COMPRESSION CAPACITY OF COMPRESSION STRUT, VRd,max, ASSUMING θ = 21.8°\nν1 = 0.6(1 − fck/250) = 0.6(1 − 25/250) = 0.54\nfcd = αccfck/γc = 0.85 × 25/1.5 = 14.2 N mm−2\n(Note αcc = 1.0 may be used.)\nVRd,max = bwzν1fcd/(cot θ + tan θ)\n= [300 × (0.9 × 500) × 0.54 × 14.2/(2.5 + 0.4)] × 10−3 = 357 kN\nSince VRd,max < VEd, strut angle > 21.8°. From equation 8.27\n \nθ . sin\n(\n/\n)\n.\n( \n/\n) . sin\n(\n \n/\n \n)\n.\n \n( \n/\n) \n.\n=\n−\n=\n×\n×\n×\n−\n=\n°\n−\n−\n0 5\n0 153\n1\n250\n0 5\n450\n10 300\n500\n0 153\n251\n25 250\n30 3\n1\n1\n3\nV\nb d\nf\nf\nRd,max\nw\nck\nck\nDIAMETER AND SPACING OF LINKS\nProvide shear reinforcement according to:\n \nV\nA\ns\nzf\nRd,s\nsw\nywd\nkN\n \n cot \n \n=\n=\nθ\n450\n⇒ \n \nA\ns\nsw ( . \n)(\n/ .\n) cot \n.\n .\n=\n×\n×\n° =\n450000\n0 9\n500 500 1 15\n30 3\n1 344\nMaximum spacing of links, smax, is\nsmax = 0.75d = 0.75 × 500 = 375 mm\nTherefore H12 at 150 mm centres (Asw/s = 1.507) would be suitable.\nNote. If longitudinal reinforcement is fully anchored at support and VEd = 450 kN ≤ 0.5bwdνfcd = 0.5 × 300 × 500\n× 0.54 × 14.2 × 10−3 = 575 kN (OK), shear force may be reduced to (av/2d) × 450 = 225 kN. In this case θ = 21.8°\nis OK and shear reinforcement is reduced by over 50 per cent.\nExample 8.5 continued\n\n[TABLES]\n[[{\"0\": \"Eurocodes 2: Design of concrete structures\"}, {\"0\": \"Example 8.5 continued\"}, {\"0\": \"SHEAR RESISTANCE OF CONCRETE (VRd,c)\"}, {\"0\": \"fck = 25 N mm−2\"}, {\"0\": \"CRd,c = 0.18/γC = 0.18/1.5 = 0.12 N mm−2\"}, {\"0\": \"200200=+=+=<OKk 1 1 1 63 2 0\"}, {\"0\": \"d500\"}, {\"0\": \"A1960sl===<ρOK 0 013 0 021\"}, {\"0\": \"×b d300 500\"}, {\"0\": \"σcp = 0\"}, {\"0\": \"ν1/2 = 0.035 × 1.633/2 × 251/2 = 0.364min = 0.035k3/2f ck\"}, {\"0\": \"σVRd,c = [CRd,ck(100ρ1fck)1/3 + k1cp]bwd\"}, {\"0\": \"= [0.12 × 1.63(100 × 0.013 × 25)1/3]300 × 500 × 10−3\"}, {\"0\": \"σ= 93.6 kN ≥ (νmin + k1cp)bwd = 0.364 × 300 × 500 × 10−3 = 54.6 kN\"}, {\"0\": \"Since VRd,c < VEd = 450 kN, shear reinforcement must be provided.\"}, {\"0\": \"CHECK COMPRESSION CAPACITY OF COMPRESSION STRUT, VRd,max, ASSUMING θ = 21.8°\"}, {\"0\": \"ν1 = 0.6(1 − fck/250) = 0.6(1 − 25/250) = 0.54\"}, {\"0\": \"(Note αfcd = αccfck/γc = 0.85 × 25/1.5 = 14.2 N mm−2cc = 1.0 may be used.)\"}, {\"0\": \"VRd,max = bwzν1fcd/(cot θ + tan θ)\"}, {\"0\": \"= [300 × (0.9 × 500) × 0.54 × 14.2/(2.5 + 0.4)] × 10−3 = 357 kN\"}, {\"0\": \"Since VRd,max < VEd, strut angle > 21.8°. From equation 8.27\"}, {\"0\": \"3××Vb d(/)(450 10/300 500)−−Rd,max11\"}, {\"0\": \"θ 0 5 sin 0 5 sin 30 3−×−ff0 153(1 /250)0 153 25 1 25 250)ckck\"}, {\"0\": \"DIAMETER AND SPACING OF LINKS\"}, {\"0\": \"Provide shear reinforcement according to:\"}, {\"0\": \"Asw==θV zf cot 450 kNRd,sywd\"}, {\"0\": \"s\"}, {\"0\": \"A450000==⇒ w 1 344\"}, {\"0\": \"ss××°(0 9 500)(500 1 15) cot 30 3\"}, {\"0\": \"Maximum spacing of links, smax, is\"}, {\"0\": \"smax = 0.75d = 0.75 × 500 = 375 mm\"}, {\"0\": \"Therefore H12 at 150 mm centres (A sw/s = 1.507) would be suitable.\"}, {\"0\": \"Note. If longitudinal reinforcement is fully anchored at support and VEd = 450 kN ≤ 0.5bwdνfcd = 0.5 × 300 × 500\"}, {\"0\": \"× 0.54 × 14.2 × 10−3 = 575 kN (OK), shear force may be reduced to (av/2d) × 450 = 225 kN. In this case θ = 21.8°is OK and shear reinforcement is reduced by over 50 per cent.\"}, {\"0\": \"336\"}, {\"0\": \"9780415467193_C083369/3/09, 2:55 PM\"}], [], []]\n\n[FORMULAS]\nc = 0; .18/γC = 0; .18/1.5 = 0; νmin = 0; 251/2 = 0; 3\n= 93; bwd = 0; 3 = 54; ν1 = 0; 25/1.5 = 14", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0357.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 358, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n337\nTable 8.11 Basic span/depth ratios for reinforced concrete members\nStructural system\nK\nHighly stressed concrete\nLightly stressed concrete\nρ = 1.5%\nρ = 0.5%\nSimply supported beams and slabs\n1.0\n14\n20\nEnd span of continuous beams and slabs\n1.3\n18\n26\nInterior spans of continuous beams and slabs\n1.5\n20\n30\nCantilever beams and slabs\n0.4\n6\n8\n \nB\n \nd\nK\nf\nf\n \n .\n .\n \n/\n=\n+\n\n\n\n+\n−\n\n\n\n\n\n\n\n\n\n\n\n\n11\n1 5\n3 2\n1\n3 2\nck\no\nck\no\nρ\nρ\nρ\nρ\nif ρ ≤ ρo\n(8.32)\n \nB\n \nd\nK\nf\nf\n \n .\n \n \n/\n=\n+\n−\n′\n\n\n\n+\n′\n\n\n\n\n\n\n\n\n\n\n\n\n11\n1 5\n1\n12\n1 2\nck\no\nck\no\nρ\nρ\nρ\nρ\nρ\nif ρ > ρo\n(8.33)\nwhere\nl/d\nis the limiting span/depth ratio\nK\nis the factor to take into account the\ndifferent structural systems, given in Table 8.11\nρo\n= fck \n×\n−\n10 3\nρ\nis the required tension reinforcement ratio\nρ′\nis the required compression reinforcement\nratio\nThe values in Table 8.11 assume the steel stress\nat the critical section, σs, is 310 N mm−2, corres-\nponding roughly to the stress under characteristic\nload when fyk = 500 N mm−2. Where other steel\nstresses are used, the values in the table can be\nmultiplied by 310/σs. It will normally be conserv-\native to assume that\n \nσ\nν\ns\nyk\ns,req\ns,pro\n \n= 310\n500\nf A\nA\n(8.34)\nwhere\nAs,req\nis the area of steel required\nAs,prov\nis the area of steels provided\nFor flanged sections where the ratio of the flange\nbreadth to the rib width, beff/bw, exceeds 3, the val-\nues in the table should be multiplied by 0.8. Lin-\near interpolation may be used to estimate values of\nl/d for intermediate values of beff/bw. For beams\nand slabs, other than flat slabs (see Fig. 3.52), where\nDesign of singly and doubly reinforced rectangular beams\n8.8.3 DEFLECTION\nCrack and deflection control are the two main\nserviceability checks required by EC 2. EC 2 also\nincludes a check on the stress in the concrete and\nsteel reinforcement under specified loadings but past\nperformance of concrete structures would seem to\nsuggest that this check is almost certainly unneces-\nsary in the UK and is therefore not discussed.\nLike BS 8110, EC 2 recommends that surface\ncrack widths should not generally exceed 0.3 mm\nin order to avoid poor durability. As previously\nmentioned, this is normally achieved by observing\nthe rules on minimum steel areas and either the\nmaximum bar spacing or the maximum bar dia-\nmeter. The rules for beams are presented in section\n8.8.4.\nEC 2 requirements for deflection control appear\nto be less onerous than those in BS 8110, namely\n1. the calculated sag of a beam, slab or cantilever\nunder quasi-permanent loading should not\nexceed span/250.\n2. the deflection occurring after construction of the\nelement should not exceed span/500.\nIt is true that the creep component of deflection\nis related to quasi-permanent loads and if appear-\nance is the only criterion it would be reasonable to\nbase the deflection check on this load combination.\nHowever, if function, including possible damage to\nfinishes, cladding and partitions is of concern, then\nthe use of characteristic load would seem more\nappropriate.\nAlthough it is possible to estimate actual deflec-\ntions of concrete elements in order to check com-\npliance it is more practical to use limiting span/\ndepth ratios as in BS 8110. Table 8.11 shows basic\nspan/depth ratio for commonly occurring reinforced\nconcrete members and support conditions. They\nhave been obtained using equations 8.32 and\n8.33.\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"Table 8.11 Basic span/depth ratios for reinforced concrete members\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"Structural system\", \"1\": \"K\", \"2\": \"Highly stressed concrete\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"ρ = 1.5%\", \"3\": \"\"}, {\"0\": \"Simply supported beams and slabs\", \"1\": \"1.0\", \"2\": \"14\", \"3\": \"20\"}, {\"0\": \"End span of continuous beams and slabs\", \"1\": \"1.3\", \"2\": \"18\", \"3\": \"26\"}, {\"0\": \"Interior spans of continuous beams and slabs\", \"1\": \"1.5\", \"2\": \"20\", \"3\": \"30\"}, {\"0\": \"Cantilever beams and slabs\", \"1\": \"0.4\", \"2\": \"6\", \"3\": \"8\"}], [{\"0\": \"\", \"1\": \"Design of singly and doubly reinforced rectangular beams\"}, {\"0\": \"8.8.3 DEFLECTION\", \"1\": \"3 2B\"}, {\"0\": \"Crack and deflection control are the two main\", \"1\": \" ρ ρ ρ ρ oo=++− K f f 111 53 21ckckd\"}, {\"0\": \"serviceability checks required by EC 2. EC 2 also\", \"1\": \"\"}, {\"0\": \"includes a check on the stress in the concrete and\", \"1\": \"if ρ ≤ ρ(8.32)o\"}, {\"0\": \"steel reinforcement under specified loadings but past\", \"1\": \"\"}, {\"0\": \"performance of concrete structures would seem to\", \"1\": \"1 2\"}, {\"0\": \"suggest that this check is almost certainly unneces-\", \"1\": \" ρ ρ ρ′B1o=++ K f f111 5\"}, {\"0\": \"sary in the UK and is therefore not discussed.\", \"1\": \"ckckρρ−′d 12o\"}, {\"0\": \"Like BS 8110, EC 2 recommends that surface\", \"1\": \"\"}, {\"0\": \"crack widths should not generally exceed 0.3 mm\", \"1\": \"if ρ > ρ(8.33)o\"}, {\"0\": \"in order to avoid poor durability. As previously\", \"1\": \"\"}, {\"0\": \"mentioned, this is normally achieved by observing\", \"1\": \"where\"}, {\"0\": \"the rules on minimum steel areas and either the\", \"1\": \"l/dis the limiting span/depth ratio\"}, {\"0\": \"maximum bar spacing or the maximum bar dia-\", \"1\": \"Kis the factor to take into account the\"}, {\"0\": \"meter. The rules for beams are presented in section\", \"1\": \"different structural systems, given in Table 8.11\"}, {\"0\": \"8.8.4.\", \"1\": \"ρ ×10 3fck o=\"}, {\"0\": \"EC 2 requirements for deflection control appear\", \"1\": \"ρis the required tension reinforcement ratio\"}, {\"0\": \"to be less onerous than those in BS 8110, namely\", \"1\": \"ρ′is the required compression reinforcement\"}, {\"0\": \"1.the calculated sag of a beam, slab or cantilever\", \"1\": \"ratio\"}, {\"0\": \"under quasi-permanent loading should not\", \"1\": \"The values in Table 8.11 assume the steel stress\"}, {\"0\": \"exceed span/250.\", \"1\": \"at the critical section, σs, is 310 N mm−2, corres-\"}, {\"0\": \"2.the deflection occurring after construction of the\", \"1\": \"ponding roughly to the stress under characteristic\"}, {\"0\": \"element should not exceed span/500.\", \"1\": \"load when fyk = 500 N mm−2. Where other steel\"}, {\"0\": \"\", \"1\": \"stresses are used, the values in the table can be\"}, {\"0\": \"It is true that the creep component of deflection\", \"1\": \"multiplied by 310/σs. It will normally be conserv-\"}, {\"0\": \"is related to quasi-permanent loads and if appear-\", \"1\": \"ative to assume that\"}, {\"0\": \"ance is the only criterion it would be reasonable to\", \"1\": \"\"}, {\"0\": \"base the deflection check on this load combination.\", \"1\": \"fAyks,reqσ(8.34) = 310\"}, {\"0\": \"However, if function, including possible damage to\", \"1\": \"s500Aνs,pro\"}, {\"0\": \"finishes, cladding and partitions is of concern, then\", \"1\": \"\"}, {\"0\": \"the use of characteristic load would seem more\", \"1\": \"where\"}, {\"0\": \"appropriate.\", \"1\": \"is the area of steel requiredAs,req\"}, {\"0\": \"Although it is possible to estimate actual deflec-\", \"1\": \"is the area of steels providedAs,prov\"}, {\"0\": \"tions of concrete elements in order to check com-\", \"1\": \"\"}, {\"0\": \"pliance it is more practical to use limiting span/\", \"1\": \"For flanged sections where the ratio of the flange\"}, {\"0\": \"depth ratios as in BS 8110. Table 8.11 shows basic\", \"1\": \"breadth to the rib width, beff/bw, exceeds 3, the val-\"}, {\"0\": \"span/depth ratio for commonly occurring reinforced\", \"1\": \"ues in the table should be multiplied by 0.8. Lin-\"}, {\"0\": \"concrete members and support conditions. They\", \"1\": \"ear interpolation may be used to estimate values of\"}, {\"0\": \"have been obtained using equations 8.32 and\", \"1\": \"l/d for intermediate values of beff/bw. For beams\"}, {\"0\": \"8.33.\", \"1\": \"and slabs, other than flat slabs (see Fig. 3.52), where\"}, {\"0\": \"Table 8.11 Basic span/depth ratios for reinforced concrete members\", \"1\": \"\"}, {\"0\": \"Structural systemK\", \"1\": \"Highly stressed concreteLightly stressed concrete\"}, {\"0\": \"\", \"1\": \"ρ = 1.5%ρ = 0.5%\"}, {\"0\": \"Simply supported beams and slabs1.0\", \"1\": \"1420\"}, {\"0\": \"End span of continuous beams and slabs1.3\", \"1\": \"1826\"}, {\"0\": \"Interior spans of continuous beams and slabs1.5\", \"1\": \"2030\"}, {\"0\": \"Cantilever beams and slabs0.4\", \"1\": \"68\"}], [], []]\n\n[FORMULAS]\nρ = 0; pro\n \n= 310", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0358.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 359, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n338\n(a) Non-continuous member\n(b) Continuous member\nln\nln\nai = lesser of h/2 and t/2\nh\nt\nai = lesser of h/2\nand t/2\nt\nh\nspan lengths exceed 7 m, the values of l/d should\nbe multiplied by 7/leff.\nwhere\nleff\neffective span of the member = ln + a1 + a2\n(Cl. 5.3.2.2)\nin which\nln\nis the clear distance between the faces of\nsupports\na1,a2\nare the distances at each end of the span\nshown in Fig. 8.16\nFig. 8.16 Definition of a1 and a2 (a) Non-continuous member (b) Continuous member (based on Fig. 5.4, EC 2).\nExample 8.6 Deflection check for concrete beams (EC 2)\nCarry out a deflection check for Examples 8.2 and 8.3.\nFOR EXAMPLE 8.2\nfck = 25 N mm−2 ⇒ ρo = fck × 10−3 = 25 × 10−3 = 5 × 10−3\n \nρ \n \n \n .\n \n=\n=\n×\n=\n×\n−\nA\nbd\nsl\n1008\n275\n450\n8 15\n10 3\nSince ρ > ρo, use equation 8.33\n \nB\n \nd\nK\nf\nf\n \n .\n \n \n .\n .\n \n.\n \n \n \n.\n/\n=\n+\n−\n′\n\n\n\n+\n′\n\n\n\n\n\n\n\n\n\n\n\n\n=\n+\n×\n×\n−\n\n\n\n\n\n\n\n\n\n=\n−\n−\n11\n1 5\n1\n12\n1 0 11\n1 5 25\n5\n10\n8 15\n10\n0\n15 6\n12\n3\n3\nck\no\nck\no\nρ\nρ\nρ\nρ\nρ\n \nσ\nν\ns\nyk\ns,req\ns,pro\nNmm\n \n \n \n \n \n \n \n=\n=\n×\n×\n×\n=\n−\n310\n500\n310\n500\n1008\n500\n1260\n248\n2\nf A\nA\nAllowable span/depth ratio = basic l/d × 310/σs = 15.6 × 310/248 = 19.5\nActual span/depth ratio = 7000/450 = 15.6 < allowable OK\nFOR EXAMPLE 8.3\nfck = 25 N mm−2 ⇒ ρo = fck × 10−3 = 25 × 10−3 = 5 × 10−3\n \nρ \n \n \n \n.\n \n=\n=\n×\n=\n×\n−\nA\nbd\ns1\n1144\n230\n317\n15 69\n10 3\n \n′ =\n=\n×\n=\n×\n−\nρ \n \n \n .\n \nA\nbd\ns2\n291\n230\n317\n3 99\n10 3\n\n[TABLES]\n[[{\"0\": \"effective span of the member = ln + a1 + a2(Cl. 5.3.2.2)\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"ai = lesser of h/2and t/2\", \"1\": \"\", \"2\": \"\", \"3\": \"ln\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"ai = lesser of h/2 and t/2\", \"1\": \"\", \"2\": \"h\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], [], [{\"None\": \"\", \"\": \"\"}, {\"None\": \"\", \"\": \"\"}], [{\"\": \"\"}]]\n\n[FORMULAS]\n310/248 = 19", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0359.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 360, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n339\nSince ρ > ρo,\n \nB\n \nd\nK\nf\nf\n \n .\n \n \n \n/\n=\n+\n−\n′\n\n\n\n+\n′\n\n\n\n\n\n\n\n\n\n\n\n\n11\n1 5\n1\n12\n1 2\nck\no\nck\no\nρ\nρ\nρ\nρ\nρ\n \n=\n+\n×\n×\n−\n×\n\n\n\n+\n×\n×\n\n\n\n\n\n\n\n\n\n\n\n\n=\n−\n−\n−\n−\n−\n .\n .\n \n.\n \n .\n \n \n.\n \n \n \n \n.\n/\n1 0 11\n1 5 25\n5\n10\n15 69\n10\n3 99\n10\n1\n12 25 3 99\n10\n5\n10\n14 5\n3\n3\n3\n3\n3\n1 2\n \nσ\nν\ns\nyk\ns,req\ns,pro\nNmm\n \n \n \n \n \n \n \n=\n=\n×\n×\n×\n=\n−\n310\n500\n310\n500\n1144\n500\n1470\n241\n2\nf A\nA\nAllowable span/depth ratio = basic l/d × 310/σs × 7/span = 14.5 × 310/241 × 7/9 = 18.6\nActual span/depth ratio = 9000/317 = 28.4 > allowable Not OK\nExample 8.6 continued\n8.8.4 REINFORCEMENT DETAILS FOR BEAMS\nThis section outlines EC 2 requirements regarding\nthe detailing of beams with respect to:\n1. reinforcement percentages\n2. spacing of reinforcing bars\n3. anchorage lengths\n4. curtailment of reinforcement\n5. lap lengths.\nFor a brief discussion on each of these aspects\nrefer to section 3.9.1.6 of this book.\n8.8.4.1 Reinforcement percentages\n(Cl. 9.2.1.1, EC 2)\nThe cross-sectional area of the longitudinal tensile\nreinforcement, As1, in beams should not be less\nthan the following:\n \nA\nf\nf\nb d\nb d\ns\nctm\nyk\nt\nt\n1\n0 26\n0 0013\n .\n .\n=\n≥\nwhere\nbt\nis the breadth of section\nd\neffective depth\nfctm\nmean tensile strength of concrete\n(see Table 8.2)\nfyk\ncharacteristic yield stress of reinforcement\n(= 500 N mm−2)\nThe area of the tension, As1, and of the com-\npression reinforcement, As2, should not be greater\nthan the following other than at laps:\nAs1, As2 ≤ 0.04Ac\nwhere\nAc\nis the cross-sectional area of concrete.\n8.8.4.2 Spacing of reinforcement (Cl. 8.2, EC 2)\nThe clear horizontal or vertical distance between\nreinforcing bars should not be less than the following:\n(a) maximum bar diameter\n(b) maximum aggregate size, dg + 5 mm\n(c) 20 mm\nTo ensure that the maximum crack width does\nnot exceed 0.3 mm EC 2 recommends that the\nminimum reinforcement necessary to control crack-\ning should be provided and either the maximum\nbar spacing or the maximum bar diameter should\nnot exceed the values shown in Table 8.12. The\nsteel stress, σs, can conservatively be estimated\nusing equation 8.35.\n \nσ\nγ\nν\ns\nyk\ns,req\ns\ns,pro\n \n= f mA\nnA\n(8.35)\nwhere\nfyk\n= 500 N mm−2\nγs\n= 1.15\nm\nquasi-permanent load (= gk + Ψ2qk)\nn\ndesign ultimate load (= γGgk + γQqk)\nAs,req\narea of steel required\nAs,prov\narea of steel provided\n8.8.4.3 Anchorage (Cl. 8.4.3, EC 2)\nEC 2 distinguishes between basic and design anch-\norage lengths. The basic anchorage length, lb,rqd, is\nthe length of bar required to resist the maximum\nforce in the reinforcement, As.σsd. Assuming a con-\nstant bond stress of fbd the basic anchorage length\nof a straight bar is given by:\nlb,rqd = (φ/4)(σsd/fbd)\n(8.36)\nDesign of singly and doubly reinforced rectangular beams\n\n[TABLES]\n[[{\"0\": \"Actual span/depth ratio = 9000/317 = 28.4 > allowable Not OK\", \"1\": \"\"}, {\"0\": \"8.8.4 REINFORCEMENT DETAILS FOR BEAMS\", \"1\": \"8.8.4.2 Spacing of reinforcement (Cl. 8.2, EC 2)\"}, {\"0\": \"This section outlines EC 2 requirements regarding\", \"1\": \"The clear horizontal or vertical distance between\"}, {\"0\": \"the detailing of beams with respect to:\", \"1\": \"reinforcing bars should not be less than the following:\"}, {\"0\": \"1.reinforcement percentages\", \"1\": \"(a) maximum bar diameter\"}, {\"0\": \"2.spacing of reinforcing bars\", \"1\": \"(b) maximum aggregate size, dg + 5 mm(c)20 mm\"}, {\"0\": \"3.anchorage lengths\", \"1\": \"\"}, {\"0\": \"4.curtailment of reinforcement\", \"1\": \"To ensure that the maximum crack width does\"}, {\"0\": \"5.lap lengths.\", \"1\": \"not exceed 0.3 mm EC 2 recommends that the\"}, {\"0\": \"For a brief discussion on each of these aspects\", \"1\": \"minimum reinforcement necessary to control crack-\"}, {\"0\": \"refer to section 3.9.1.6 of this book.\", \"1\": \"ing should be provided and either the maximum\"}, {\"0\": \"\", \"1\": \"bar spacing or the maximum bar diameter should\"}, {\"0\": \"8.8.4.1 Reinforcement percentages\", \"1\": \"not exceed the values shown in Table 8.12. The\"}, {\"0\": \"(Cl. 9.2.1.1, EC 2)\", \"1\": \"steel stress, σs, can conservatively be estimated\"}, {\"0\": \"The cross-sectional area of the longitudinal tensile\", \"1\": \"using equation 8.35.\"}, {\"0\": \"reinforcement, As1, in beams should not be less\", \"1\": \"f mAyks,reqσ\"}, {\"0\": \"than the following:\", \"1\": \"(8.35) =sγnAνss,pro\"}, {\"0\": \"fctm=≥ A 0 26b d 0 0013s1tt\", \"1\": \"where\"}, {\"0\": \"fyk\", \"1\": \"= 500 N mm−2fyk\"}, {\"0\": \"where\", \"1\": \"γ= 1.15s\"}, {\"0\": \"is the breadth of sectionbt\", \"1\": \"mquasi-permanent load (= gk + Ψ2qk)\"}, {\"0\": \"deffective depth\", \"1\": \"ndesign ultimate load (= γGgk + γQqk)\"}, {\"0\": \"mean tensile strength of concretefctm\", \"1\": \"area of steel requiredAs,req\"}, {\"0\": \"(see Table 8.2)\", \"1\": \"area of steel providedAs,prov\"}, {\"0\": \"characteristic yield stress of reinforcementfyk\", \"1\": \"\"}, {\"0\": \"(= 500 N mm−2)\", \"1\": \"8.8.4.3 Anchorage (Cl. 8.4.3, EC 2)\"}, {\"0\": \"The area of the tension, A s1, and of the com-\", \"1\": \"EC 2 distinguishes between basic and design anch-\"}, {\"0\": \"pression reinforcement, A s2, should not be greater\", \"1\": \"orage lengths. The basic anchorage length, l b,rqd, isthe length of bar required to resist the maximum\"}, {\"0\": \"than the following other than at laps:\", \"1\": \"force in the reinforcement, As.σsd. Assuming a con-\"}, {\"0\": \"A s1, A s2 ≤ 0.04Ac\", \"1\": \"stant bond stress of fbd the basic anchorage lengthof a straight bar is given by:\"}, {\"0\": \"where\", \"1\": \"\"}, {\"0\": \"is the cross-sectional area of concrete.\", \"1\": \"(8.36)l b,rqd = (φ/4)(σ\"}], [{\"0\": \"Design of singly and doubly reinforced rectangular beams\"}, {\"0\": \"Example 8.6 continued\"}, {\"0\": \"Since ρ > ρo,\"}, {\"0\": \"1 2ρ′B1\"}, {\"0\": \" ρ ρ o=++ K f f 111 5ckckρρ−′d 12\"}, {\"0\": \"o\"}, {\"0\": \"1 2−−33××5 1013 99 10\"}, {\"0\": \" =+=+ 1 0 11 1 525 25 14 5−−−333×−××15 69 10 3 99 10125 10\"}, {\"0\": \"××310fA3105001144 yks,req−2σ===241 Nmms\"}, {\"0\": \"×ν500A5001470 s,pro\"}, {\"0\": \"Allowable span/depth ratio = basic l/d × 310/σs × 7/span = 14.5 × 310/241 × 7/9 = 18.6\"}, {\"0\": \"Actual span/depth ratio = 9000/317 = 28.4 > allowable Not OK\"}, {\"0\": \"8.8.4.2 Spacing of reinforcement (Cl. 8.2, EC 2)8.8.4 REINFORCEMENT DETAILS FOR BEAMS\"}, {\"0\": \"The clear horizontal or vertical distance betweenThis section outlines EC 2 requirements regarding\"}, {\"0\": \"reinforcing bars should not be less than the following:the detailing of beams with respect to:\"}, {\"0\": \"(a) maximum bar diameter1.reinforcement percentages\"}, {\"0\": \"(b) maximum aggregate size, dg + 5 mm2.spacing of reinforcing bars(c)20 mm\"}, {\"0\": \"3.anchorage lengths\"}, {\"0\": \"4.curtailment of reinforcementTo ensure that the maximum crack width does\"}, {\"0\": \"5.lap lengths.not exceed 0.3 mm EC 2 recommends that the\"}, {\"0\": \"minimum reinforcement necessary to control crack-For a brief discussion on each of these aspects\"}, {\"0\": \"ing should be provided and either the maximumrefer to section 3.9.1.6 of this book.\"}, {\"0\": \"bar spacing or the maximum bar diameter should\"}, {\"0\": \"not exceed the values shown in Table 8.12. The8.8.4.1 Reinforcement percentages\"}, {\"0\": \"steel stress, σs, can conservatively be estimated(Cl. 9.2.1.1, EC 2)\"}, {\"0\": \"using equation 8.35.The cross-sectional area of the longitudinal tensile\"}, {\"0\": \"reinforcement, As1, in beams should not be lessf mAyks,reqσ\"}, {\"0\": \"than the following:(8.35) =sγnAνss,pro\"}, {\"0\": \"fctm=≥ A 0 26b d 0 0013wheres1tt\"}, {\"0\": \"fyk = 500 N mm−2fyk\"}, {\"0\": \"γ= 1.15wheres\"}, {\"0\": \"mis the breadth of sectionquasi-permanent load (= gk + Ψ2qk)bt\"}, {\"0\": \"ndesign ultimate load (= γdeffective depthGgk + γQqk)\"}, {\"0\": \"area of steel requiredmean tensile strength of concreteAs,reqfctm\"}, {\"0\": \"area of steel provided(see Table 8.2)As,prov\"}, {\"0\": \"characteristic yield stress of reinforcementfyk\"}, {\"0\": \"(= 500 N mm−2)8.8.4.3 Anchorage (Cl. 8.4.3, EC 2)\"}, {\"0\": \"EC 2 distinguishes between basic and design anch-The area of the tension, A s1, and of the com-\"}, {\"0\": \"orage lengths. The basic anchorage length, l b,rqd, ispression reinforcement, A s2, should not be greaterthe length of bar required to resist the maximum\"}, {\"0\": \"than the following other than at laps:force in the reinforcement, As.σsd. Assuming a con-\"}, {\"0\": \"A s1, A s2 ≤ 0.04Acstant bond stress of fbd the basic anchorage lengthof a straight bar is given by:\"}, {\"0\": \"where\"}, {\"0\": \"is the cross-sectional area of concrete.(8.36)A cl b,rqd = (φ/4)(σsd/fbd)\"}], [], []]\n\n[FORMULAS]\n7/9 = 18; γs\n= 1", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0360.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 361, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n340\nHatched zone: ‘poor’ bond conditions\n300\n250\nDirection of concreting\nDirection of concreting\nDirection of concreting\nh < 250\nh > 600\nh > 250\n(a)\n(b)\n(c)\nTable 8.12\nMaximum bar spacing and bar diameter for crack control (based on Table 7.2N and\n7.3N, EC 2)\nSteel Stress (N mm−2)\nMaximum bar spacing (mm)\nMaximum bar size (mm) φs\nwk = 0.3 mm\nwk = 0.3 mm\n160\n300\n32\n200\n250\n25\n240\n200\n16\n280\n150\n12\n320\n100\n10\n360\n50\n8\nNote: The values in the table assume cnom = 25 mm and fck = 30 N mm−2\nTable 8.13\nUltimate bond stress, fbd (N mm−2), assuming good bond conditions\nfck (N mm−2)\n12\n16\n20\n25\n30\n35\n40\n45\n50\nfbd where φ ≤ 32 mm\n1.65\n2.00\n2.32\n2.69\n3.04\n3.37\n3.68\n3.98\n4.35\nFig. 8.17 Definition of bond conditions (Fig. 8.2, EC 2).\nwhere\nφ\ndiameter of bar to be anchored\nσsd\ndesign stress of the bar at the position from\nwhere the anchorage is measured\nfbd\nultimate bond stress = 2.25η1η2 fctd\n(8.37)\nwhere\nη1\ncoefficient related to the quality of the bond\nbetween the concrete and steel. EC 2 states\nthat bond conditions are considered to be\ngood for:\n(a) all bars in members with an overall\ndepth (h) of less than or equal to\n250 mm (Fig. 8.17a)\n(b) all bars located within the lower\n250 mm depth in members with an\noverall depth exceeding 250 mm\n(Fig. 8.17b)\n(c) all bars located at a depth greater than\nor equal to 300 mm in members with\nan overall depth exceeding 600 mm\n(Fig. 8.17c).\nη1\n= 1.0 when good bond exists and 0.7 for all\nother conditions\nη2\nis related to bar diameter and is 1.0 for\nφ ≤ 32 mm and (132 − φ)/100 for φ > 32 mm\nfctd\nis the design value of the concrete tensile\nstrength = αctfctk,0.05/γC\nin which\nαct\n= 1.0\nfctk,0.05\nis the 5% fractile characteristic axial\ntensile strength of concrete (see Table 8.2)\nTable 8.13 shows values of the ultimate bond\nstress for a range of concrete strengths assuming\n\n[TABLES]\n[[{\"0\": \"\"}, {\"0\": \"\"}, {\"0\": \"\"}, {\"0\": \"(c)\"}, {\"0\": \"\"}, {\"0\": \"\"}], [{\"0\": \"\"}, {\"0\": \"\"}, {\"0\": \"(b)\"}, {\"0\": \"\"}, {\"0\": \"\"}, {\"0\": \"\"}], [{\"0\": \"\"}, {\"0\": \"\"}, {\"0\": \"(a)Hatched zone: ‘poor’ bond conditions\"}, {\"0\": \"\"}, {\"0\": \"\"}], [], [], [{\"\": \"\"}, {\"\": \"\"}], [{\"None\": \"\", \"\": \"\"}, {\"None\": \"\", \"\": \"\"}], [{\"\": \"\"}]]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0361.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 362, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n341\ngood bond conditions and φ ≤ 32 mm. When bond\nconditions are poor the values in the table should\nbe multiplied by 0.7.\nThe design anchorage length, lbd, is used to deter-\nmine the curtailment of longitudinal reinforcement\nin beams as discussed in section 8.8.4.4 and is\ngiven by:\nlbd = α1α2α3α4α5lb,rqd ≥ lb,min\n(8.38)\nwhere\nα1–α5\nare coefficients which take account of,\nrespectively, shape of bar, concrete\ncover/spacing between bars, transverse\nreinforcement not welded to the main\nreinforcement, transverse reinforcement\nwelded to the main reinforcement and\npressure transverse to the reinforcement.\nThe product of α1–α5 can conservatively\nbe taken as 1.\nlb,min\nminimum anchorage length given by:\nlb,min = maximum of {0.3.lb,rqd; 10φ;\n100 mm} (for anchorages in tension)\nlb,min = maximum of {0.6.lb,rqd; 10φ;\n100 mm} (for anchorages in compression)\nAs a simplification for the shapes shown in\nFig. 8.18(b) to (d) an equivalent anchorage length,\nlb,eq, may be used where lb,eq is given by\nlb,eq = α1lb,rqd\n(8.39)\nin which\nα1\n= 1.0 for straight and curved bars in\ncompression\n= 1.0 for straight bars in tension\n= 0.7 for curved bars in tension if cd > 3.0φ\nwhere\ncd\nis as shown in Fig. 8.19 for straight and\ncurved bars\nφ\nis the diameter of the bar to be anchored\nFig. 8.18 Common methods of anchorage. Equivalent\nanchorage lengths for bends, hooks and loops (based on\nFig. 8.1, EC 2)\nTable 8.14\nDesign anchorage lengths, lbd, as multiples of bar diameters assuming φ ≤ 32 mm and\ngood bond conditions for a range of concrete strengths\nSteel Grade\nConcrete Class\nC20/25\nC25/30\nC30/37\nC35/45\nC40/50\nC45/55\nC50/60\nDeformed bars\nStraight bars in tension\n47\n41\n36\n33\n30\n28\n25\ntype 2\nor compression\nfyk = 500 N mm−2\nCurved bars in tension*\n33\n29\n25\n23\n21\n20\n18\nCurved bars in\n47\n41\n36\n33\n30\n28\n25\ncompression\n*In the anchorage region, cd > 3φ\nDesign of singly and doubly reinforced rectangular beams\nBbd\nφ\nBb\nBb.eq\nBb.eq\nBb.eq\n(a) Basic tension anchorage\nlength, Bb, for any shape\nmeasured along the\ncentre line\n(b) Bend\n(d) Loop\n(c) Hook\nTable 8.14 shows the design anchorage lengths\nfor straight and curved bars as multiples of bar\ndiameters for high yield bars ( fyk = 500 N mm−2)\nembedded in a range of concrete strengths. Note\nthat the values in the table apply to good bond\n(a) cd = lesser of a/2, c1 and c\n(b) cd = lesser of a/2 and c1\nc1\na\na\nc\nc1\nFig. 8.19 Values of cd for beams and slabs (a) straight\nbars (b) bent or hooked bars (based on Fig. 8.3, EC 2).\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}], [], []]\n\n[FORMULAS]\nlbd = α1α2α3α4α5lb; eq = α1lb", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0362.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 363, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n342\nconditions and to bar sizes less than or equal to 32\nmm. Where bond conditions are poor the values in\nthe table should be divided by 0.7 and where the\nbar diameter exceeds 32 mm the values should\nbe divided by [(132 − φ)/100]. The minimum radii\nto which reinforcement may be bent is shown in\nTable 8.15 whilst the anchorage lengths for straights,\nhooks, bends and loops are shown in figure 8.18.\n8.8.4.4 Curtailment of bars (Cl. 9.2.1.3, EC 2)\nThe curtailment length of bars in beams is ob-\ntained from Fig. 8.20. The theoretical cut-off point\nis based on the design bending moment curve which\nhas been horizontally displaced by an amount al\nin the direction of decreasing bending moment, a\nconsequence of the additional longitudinal force in\nthe tension steel due to the applied shear force.\nThis is sometimes referred to as the ‘shift rule’.\nIt is also necessary to ensure that the bar extends\nan anchorage length beyond the point where it\nis fully required for bending and shear. Thus in\nrelation to the bending moment envelope, bar\nshould extend (a1 + lbd) beyond the point where\nfully required and a1 beyond the point where no\nlonger required.\nIf the shear resistance is calculated according to\nthe ‘variable strut inclination method’ the shift, al,\nis given by:\na1 = \n \nz\n2 (cot θ − cot α)\n(8.40)\nwhere\nθ\nis the angle of the concrete strut to the\nlongitudinal axis\nα\nis the angle of the shear reinforcement to the\nlongitudinal axis.\nz\n≈ 0.9d\nExample 8.7 Calculation of anchorage lengths (EC 2)\nCalculate the anchorage lengths for straight and curved bars in tension as multiples of bar size assuming:\n1. The bars are high yield of diameter, φ ≤ 32 mm\n2. Concrete strength class is C20/25\n3. Bond conditions are good\n4. cd > 3φ.\nSTRAIGHT BARS\nHigh yield bars, fyk\n= 500 N mm−2\nConcrete strength, fck\n= 20 N mm−2\nUltimate bond stress, fbd\n= 2.32 (Table 8.13)\nDesign strength of bars, fyd = fyk/γS = 500/1.15\nThe basic anchorage length, lb,rqd, is\nlb,rqd = (φ/4)(fyd/fbd)\n= (φ/4)(500/1.15)/2.32 ≈ 47φ\nHence, the anchorage length, lbd, is\nlbd = α1α2α3α4α5lb,rqd = 1.0 × 47φ = 47φ\n(see Table 8.14)\nCURVED BARS\nThe calculation is essentially the same for this case except that α1 = 0.7 for curved bars and therefore, lbd, is\nlbd = lb,eq = α1lb,rqd = 0.7 × 47φ ≈ 33φ\n(see Table 8.14)\nTable 8.15\nMinimum diameter of bends, hooks\nand loops (Table 8.1N, EC 2)\nBar diameter\nMinimum mandrel diameter\nφ ≤ 16 mm\n4φ\nφ > 16 mm\n7φ\n\n[TABLES]\n[[{\"0\": \"has been horizontally displaced by an amount al\", \"1\": \"is given by:\"}, {\"0\": \"in the direction of decreasing bending moment, a\", \"1\": \"\"}, {\"0\": \"consequence of the additional longitudinal force in\", \"1\": \"z 2(cot θ − cot α)a1 =\"}, {\"0\": \"the tension steel due to the applied shear force.\", \"1\": \"\"}, {\"0\": \"This is sometimes referred to as the ‘shift rule’.\", \"1\": \"whereθ\"}, {\"0\": \"It is also necessary to ensure that the bar extends\", \"1\": \"is the angle of the concrete strut to the\"}, {\"0\": \"an anchorage length beyond the point where it\", \"1\": \"longitudinal axisα\"}, {\"0\": \"is fully required for bending and shear. Thus in\", \"1\": \"is the angle of the shear reinforcement to the\"}, {\"0\": \"relation to the bending moment envelope, bar\", \"1\": \"longitudinal axis.\"}, {\"0\": \"should extend (a1 + lbd) beyond the point where\", \"1\": \"z≈ 0.9d\"}], [{\"0\": \"Eurocodes 2: Design of concrete structures\"}, {\"0\": \"conditions and to bar sizes less than or equal to 32Table 8.15 Minimum diameter of bends, hooks\"}, {\"0\": \"mm. Where bond conditions are poor the values inand loops (Table 8.1N, EC 2)\"}, {\"0\": \"the table should be divided by 0.7 and where the\"}, {\"0\": \"Bar diameterMinimum mandrel diameterbar diameter exceeds 32 mm the values should\"}, {\"0\": \"be divided by [(132 − φ)/100]. The minimum radii\"}, {\"0\": \"φ ≤ 16 mm4φto which reinforcement may be bent is shown in\"}, {\"0\": \"φ > 16 mm7φTable 8.15 whilst the anchorage lengths for straights,\"}, {\"0\": \"hooks, bends and loops are shown in figure 8.18.\"}, {\"0\": \"Example 8.7 Calculation of anchorage lengths (EC 2)\"}, {\"0\": \"Calculate the anchorage lengths for straight and curved bars in tension as multiples of bar size assuming:\"}, {\"0\": \"1.The bars are high yield of diameter, φ ≤ 32 mm\"}, {\"0\": \"2. Concrete strength class is C20/25\"}, {\"0\": \"3. Bond conditions are good\"}, {\"0\": \"4.cd > 3φ.\"}, {\"0\": \"STRAIGHT BARS\"}, {\"0\": \"= 500 N mm−2High yield bars, fyk\"}, {\"0\": \"= 20 N mm−2Concrete strength, fck\"}, {\"0\": \"= 2.32 (Table 8.13)Ultimate bond stress, fbd\"}, {\"0\": \"Design strength of bars, fyd = fyk/γS = 500/1.15\"}, {\"0\": \"The basic anchorage length, lb,rqd, is\"}, {\"0\": \"lb,rqd = (φ/4)(fyd/fbd)\"}, {\"0\": \"= (φ/4)(500/1.15)/2.32 ≈ 47φ\"}, {\"0\": \"Hence, the anchorage length, lbd, is\"}, {\"0\": \"αααα(see Table 8.14)lbd = α12345lb,rqd = 1.0 × 47φ = 47φ\"}, {\"0\": \"CURVED BARS\"}, {\"0\": \"The calculation is essentially the same for this case except that α1 = 0.7 for curved bars and therefore, lbd, is\"}, {\"0\": \"(see Table 8.14)lbd = lb,eq = α1lb,rqd = 0.7 × 47φ ≈ 33φ\"}, {\"0\": \"8.8.4.4 Curtailment of bars (Cl. 9.2.1.3, EC 2)fully required and a1 beyond the point where no\"}, {\"0\": \"The curtailment length of bars in beams is ob-longer required.\"}, {\"0\": \"tained from Fig. 8.20. The theoretical cut-off pointIf the shear resistance is calculated according to\"}, {\"0\": \"is based on the design bending moment curve whichthe ‘variable strut inclination method’ the shift, a l,\"}, {\"0\": \"is given by:has been horizontally displaced by an amount al\"}, {\"0\": \"in the direction of decreasing bending moment, a\"}, {\"0\": \"z 2consequence of the additional longitudinal force in(cot θ − cot α)(8.40)a1 =\"}, {\"0\": \"the tension steel due to the applied shear force.\"}, {\"0\": \"This is sometimes referred to as the ‘shift rule’.whereθ\"}, {\"0\": \"It is also necessary to ensure that the bar extendsis the angle of the concrete strut to the\"}, {\"0\": \"an anchorage length beyond the point where itlongitudinal axisα\"}, {\"0\": \"is fully required for bending and shear. Thus inis the angle of the shear reinforcement to the\"}, {\"0\": \"relation to the bending moment envelope, barlongitudinal axis.\"}, {\"0\": \"z≈ 0.9dshould extend (a1 + lbd) beyond the point where\"}, {\"0\": \"342\"}], [], []]\n\n[FORMULAS]\nfbd\n= 2; rqd = 1; eq = α1lb; rqd = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0363.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 364, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n343\nFor members with vertical links al will be equal\nto:\na1 = \n \nz\n2 (cot θ − cot 90°) = \n \nz\n2 cot θ\nSince, in practice, the strut angle often takes the\nvalue cot θ = 2.5 the shift, a1, is given by\na1 = 1.25z\nEC 2 also recommends the following rules for\nthe curtailment of reinforcement in beams subjected\nto predominantly uniformly distributed loads.\n(i) Top reinforcement at end supports. In\nmonolithic construction, even where simple sup-\nports have been assumed in design, the section at\nsupports should be designed for a bending moment\narising from partial fixity of at least 25 per cent of\nthe maximum bending moment in the span.\n(ii) Bottom reinforcement at end supports.\nWhen there is little or no fixity at end supports,\nat least 25 per cent of the reinforcement provided\nat mid-span should be taken into the support and\nanchored as shown in Fig. 8.21. The tensile force\nto be anchored, FE, is given by:\nFE = |VEd|.a1/z + NEd\n(8.41)\nwhere\nVEd\nis the shear force at the end support\na1\nis obtained from equation 8.40\nNEd\nis the axial force in the member\n(iii) Bottom reinforcement at intermediate\nsupports. At least 25 per cent of the reinforcement\nrequired at mid-span should extend to intermedi-\nate supports. The anchorage length should not be\nless than 10φ for straight bars.\nThe reinforcement required to resist possible\npositive moments due to, for example, accidental\nactions such as explosions and collisions, should\nbe continuous which may be achieved by means of\nlapped bars as shown in Fig 8.22a or 8.22b.\nFig. 8.23 shows the simplified rules for curtail-\nment of reinforcement in simply supported and\ncontinuous beams recommended in The Concrete\nCentre publication ‘How to design concrete struc-\ntures using Eurocode 2’.\n8.8.4.5 Lap lengths (Cl. 8.7.3, EC 2)\nLap length, lo, is given by:\nlo = α1α2α3α5α6lb,rqd ≥ lo,min = 0.3α6lb,rqd\n(8.42)\n- Envelope of MEdlz + NEd\nA\nA\nB\nC\nIbd\n- acting tensile force Fs\nB\n- resisting tensile force FRs\nC\nIbd\nIbd\nIbd\nIbd\na1\na1\nIbd\nIbd\nIbd\nFtd\nFtd\nFig. 8.20 Envelope line for curtailment of reinforcement in flexural members (based on Fig. 9.2, EC 2).\nDesign of singly and doubly reinforced rectangular beams\nFig. 8.21 Curtailments and anchorages of bottom\nreinforcement at end supports (based on Fig. 9.3, EC 2).\n(a)\n(b)\n100%\n25%\nb\nlbd ≥ 10φ\nlbd ≥ 10φ\n0.08l\n25%\n100%\n0.08l\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"Ibd\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [{\"0\": \"\", \"1\": \"AB\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"\", \"21\": \"\", \"22\": \"\", \"23\": \"\", \"24\": \"\", \"25\": \"\", \"26\": \"\", \"27\": \"\", \"28\": \"\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"\", \"34\": \"\", \"35\": \"\", \"36\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"\", \"21\": \"\", \"22\": \"\", \"23\": \"\", \"24\": \"\", \"25\": \"\", \"26\": \"\", \"27\": \"\", \"28\": \"\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"\", \"34\": \"\", \"35\": \"\", \"36\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"\", \"21\": \"\", \"22\": \"\", \"23\": \"\", \"24\": \"\", \"25\": \"\", \"26\": \"\", \"27\": \"\", \"28\": \"Ibd\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"Ibd\", \"34\": \"\", \"35\": \"\", \"36\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"\", \"21\": \"\", \"22\": \"\", \"23\": \"\", \"24\": \"\", \"25\": \"\", \"26\": \"\", \"27\": \"\", \"28\": \"\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"\", \"34\": \"\", \"35\": \"\", \"36\": \"\"}, {\"0\": \"\", \"1\": \"I bd\", \"2\": \"\", \"3\": \"a1I bd\", \"4\": \"\", \"5\": \"\", \"6\": \"Ftd\", \"7\": \"\", \"8\": \"C\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"\", \"21\": \"\", \"22\": \"a1\", \"23\": \"I bd\", \"24\": \"\", \"25\": \"Ftd\", \"26\": \"\", \"27\": \"\", \"28\": \"\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"\", \"34\": \"\", \"35\": \"Ibd\", \"36\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"\", \"21\": \"\", \"22\": \"\", \"23\": \"\", \"24\": \"\", \"25\": \"\", \"26\": \"\", \"27\": \"\", \"28\": \"\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"\", \"34\": \"\", \"35\": \"\", \"36\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"I bd\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"I bd\", \"21\": \"\", \"22\": \"\", \"23\": \"\", \"24\": \"\", \"25\": \"\", \"26\": \"\", \"27\": \"\", \"28\": \"\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"\", \"34\": \"\", \"35\": \"\", \"36\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\", \"19\": \"\", \"20\": \"\", \"21\": \"\", \"22\": \"\", \"23\": \"\", \"24\": \"\", \"25\": \"\", \"26\": \"\", \"27\": \"\", \"28\": \"\", \"29\": \"\", \"30\": \"\", \"31\": \"\", \"32\": \"\", \"33\": \"\", \"34\": \"\", \"35\": \"\", \"36\": \"\"}], [{\"0\": \"l bd ≥ 10φ\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"25%\", \"2\": \"100%\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"0.08lb\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"(b)\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [], [], [{\"None\": \"C\", \"B\": \"\"}], [{\"None\": \"\", \"td\": \"\"}], [{\"\": \"F\\ntd\"}], [], [{\"\": \"\", \"25%\": \"\", \"None\": \"\"}, {\"\": \"\", \"25%\": \"\", \"None\": \"\"}], [{\"\": \"\"}]]\n\n[FORMULAS]\nlo = α1α2α3α5α6lb; min = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0364.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 365, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n344\nFig. 8.22 Anchorage at intermediate supports (Fig. 9.4, EC 2).\nl ≥ 10φ\nφ\nφ\nlbd\nlbd\nwhere\nlb,rqd\nanchorage length\nlo,min\nminimum lap length which should be not\nless than 15φ or 200 mm\nα1–α5\nare as for equation 8.38\nα6\n= (ρ1/25)0.5: 1.0 ≤ α6 ≤ 1.5 (Table 8.15)\nin which\nρ1\npercentage of reinforcement lapped within\n0.65lo from the centre of the lap length\nconsidered\nTable 8.16 shows the compression and tension\nlap lengths as multiples of bar size for grade 500\nbars ( fyk = 500 N mm−2) embedded in a range\nof concrete strengths. Note that the values in the\ntable apply to good bond conditions and to bar\nsizes less than or equal to 32 mm. Where the bond\nconditions are poor the values in the table should\nbe divided by 0.7 and where the bar diameter\nexceeds 32 mm the values should be divided by\n[(132 − φ)/100].\n(a)\n(b)\n30%\n100%\n60%\n30%\n100%\n0.3l – al\n0.15l + al ≥ lbd\n0.3l + al\nFace of support\n25%\n100%\nlbd\n0.08l\nFig. 8.23 Simplified detailing rules for beams (a) simple\nend supports (b) internal supports of continuous beams.\nTable 8.16\nLap lengths as multiples of bar size for grade 500 ( fyk = 500 N mm−2) deformed bars\n% of lapped bars\nα6\nConcrete class\nrelative to the total\nx-section area, ρ1\nC20/25\nC25/30\nC30/37\nC35/45\nC40/45\nCompression\n< 25%\n1\n47\n41\n36\n33\n30\n& tension laps\n 33%\n1.15\n54\n48\n42\n38\n35\n 50%\n1.4\n66\n58\n51\n47\n42\n> 50%\n1.5\n71\n62\n54\n50\n45\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"100%25%\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [], [], [{\"None\": \"\", \"\": \"\"}], [{\"\": \"\"}, {\"\": \"\"}], [{\"\": \"\", \"None\": \"\"}], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0365.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 366, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n345\ngk = 25.5 kN m–1\nqk = 20 kNm–1\nb = 300 mm\nh = 600 mm\n6 m\nExample 8.8 Design of a simply supported beam (EC 2)\nDesign the longitudinal and link reinforcement for the beam shown in Fig. 8.24 assuming Class XC1 exposure. The\nbeam is made of class C25/30 concrete of bulk density 25 kNm−3 and grade 500 steel.\nFig. 8.24\nDESIGN MOMENT (M)\nLoading\nPermanent\nSelf weight of beam = 0.6 × 0.3 × 25 = 4.5 kNm−1\nTotal permanent load, gk = 4.5 + 25.5 = 30 kNm−1\nVariable\nTotal variable load, qk = 20 kNm−1\nUltimate load\nTotal ultimate load, ω = 1.35gk + 1.5qk = 1.35 × 30 + 1.5 × 20 = 70.5 kNm−1\nDesign moment\n \nMEd\nkNm\n \n \n. \n \n.\n \n=\n=\n×\n=\nωB 2\n2\n8\n70 5\n6\n8\n317 25\nULTIMATE MOMENT OF RESISTANCE, MRd\nEffective depth\nMinimum cover, cmin = max{cmin,dur; cmin,b; 10 mm}\nFor Class XC1 exposure, cmin,dur = 15 mm (Table 8.10)\nAssume diameter of longitudinal reinforcement is 25 mm, hence cmin,b = 25 mm\nNominal cover to main steel = cmin,b + ∆cdev = 25 + 10 = 35 mm\nAssuming diameter of links = 8 mm, nominal cover to links = 35 − 8 = 27 mm > nominal cover for durability (= 15\n+ 10 = 25 mm). Hence, effective depth, d, is\nd = 600 − 27 − 8 − 25/2 = 552 mm\n[Note that, in practice, spacers in multiples of 5 mm are fixed to links. Nominal cover to links will be either 25 mm\nor 30 mm (to ensure at least 35 mm to main bars).]\nUltimate moment\nMRd = 0.167fcubd2 = 0.167 × 25 × 300 × 5522 × 10−6 = 381.6 kNm\nSince MRd > MEd design as a singly reinforced beam.\nDesign of singly and doubly reinforced rectangular beams\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"Design of singly and doubly reinforced rectangular beams\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Example 8.8 Design of a simply supported beam (EC 2)\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Design the longitudinal and link reinforcement for the beam shown in Fig. 8.24 assuming Class XC1 exposure. The\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"beam is made of class C25/30 concrete of bulk density 25 kNm−3 and grade 500 steel.\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"gk = 25.5 kN m–1qk = 20 kNm–1\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"h = 600 mm\"}, {\"0\": \"\", \"1\": \"6 m\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"b = 300 mm\", \"10\": \"\"}, {\"0\": \"Fig. 8.24\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"DESIGN MOMENT (M)\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Loading\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Permanent\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Self weight of beam = 0.6 × 0.3 × 25 = 4.5 kNm−1\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Total permanent load, gk = 4.5 + 25.5 = 30 kNm−1\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Variable\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Total variable load, qk = 20 kNm−1\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Ultimate load\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Total ultimate load, ω = 1.35gk + 1.5qk = 1.35 × 30 + 1.5 × 20 = 70.5 kNm−1\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Design moment\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"=\", \"2\": \"ωB 2\", \"3\": \"=\", \"4\": \"70 5\", \"5\": \"×\", \"6\": \"6\", \"7\": \"2\", \"8\": \"=\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"MEd\", \"1\": \"\", \"2\": \"8\", \"3\": \"\", \"4\": \"\", \"5\": \"8\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"317 25\", \"10\": \"kNm\"}, {\"0\": \"\", \"1\": \"ULTIMATE MOMENT OF RESISTANCE, M Rd\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Effective depth\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Minimum cover, cmin = max{cmin,dur; cmin,b; 10 mm}\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"For Class XC1 exposure, cmin,dur = 15 mm (Table 8.10)\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Assume diameter of longitudinal reinforcement is 25 mm, hence cmin,b = 25 mm\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Nominal cover to main steel = cmin,b + ∆cdev = 25 + 10 = 35 mm\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Assuming diameter of links = 8 mm, nominal cover to links = 35 − 8 = 27 mm > nominal cover for durability (= 15\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"+ 10 = 25 mm). Hence, effective depth, d, is\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"d = 600 − 27 − 8 − 25/2 = 552 mm\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"[Note that, in practice, spacers in multiples of 5 mm are fixed to links. Nominal cover to links will be either 25 mm\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"or 30 mm (to ensure at least 35 mm to main bars).]\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"Ultimate moment\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"MRd = 0.167fcubd 2 = 0.167 × 25 × 300 × 5522 × 10−6 = 381.6 kNm\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}, {\"0\": \"\", \"1\": \"Since MRd > MEd design as a singly reinforced beam.\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\"}], [], [], []]\n\n[FORMULAS]\n25 = 4; gk = 4; ω = 1; 20 = 70; 6 = 381", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0366.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 367, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n346\nExample 8.8 continued\nMAIN STEEL, AS1\n \nK\nM\nf bd\no\nck\n=\n=\n×\n×\n×\n=\n \n \n.\n \n \n \n .\n2\n6\n2\n317 25\n10\n25\n300\n552\n0 139\nz = d[0.5 + ( .\n \n/ . )\n0 25\n3\n3 4\n−K o\n]\n= 552[0.5 + ( .\n .\n/ . )\n0 25\n3\n0 139 3 4\n−\n×\n] = 473 mm\n \nA\nM\nf z\ns\nyk\nmm\n1\n6\n2\n0 87\n317 25\n10\n0 87\n500\n473\n1542\n \n.\n \n.\n \n.\n \n \n \n \n=\n=\n×\n×\n×\n=\nTherefore, from Table 3.10 provide 4H25 (As1 = 1960 mm2)\nCHECK MAXIMUM AND MINIMUM STEEL AREAS AND MINIMUM BAR SPACING\n0.04Ac ≥ As ≥ \n \n0 26\n.\nf\nf\nb d\nctm\nyk\nt ≥ 0.0013btd\n0.04Ac = 0.04 × bh = 0.04 × 300 × 600 = 7200 mm2 > As1\nOK\n \n0 26\n0 26 0 3\n25\n500\n300\n552\n221\n2 3\n2\n.\n .\n. \n \n \n \n \n/\nf\nf\nb d\nctm\nyk\nt\nmm\n=\n×\n=\n×\n=\n≥ 0.0013btd = 0.0013 × 300 × 552 = 215 mm2 < As1\nOK\nClear distance between bars = 1/3[b − (2 × cover) − 4φ]\n= 1/3[300 − 2 × 35 − 4 × 25] = 43 mm\n> min{φ; dg (say 20 mm) + 5 mm; 20 mm]\nOK\nSHEAR REINFORCEMENT\nW\nVEd\nVEd\nRB\nRA\nDesign shear force, VEd\nVEd = 0.5(ωB) = 0.5 × 70.5 × 6 = 211.5 kN\nShear resistance of concrete (VRd,c)\nfck = 25 N mm−2\nCRd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2\n \nk\nd\n \n \n . .\n=\n+\n=\n+\n=\n<\n1\n200\n1\n200\n552\n1 6\n2 0\nOK\n\n[TABLES]\n[[{\"0\": \"Eurocodes 2: Design of concrete structures\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Example 8.8 continued\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"MAIN STEEL, AS1\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"6×317 2510M ===\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"0 139K o22××25300552fbd ck\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"−(0 25 3/3 4)]K oz = d[0.5 +\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"−×] = 473 mm0 25 3 0 139 3 4)= 552[0.5 + (\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"6×317 25 102===\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"M f 1542 Amms1××0 870 87 500 473zyk\", \"2\": \"\"}, {\"0\": \"Therefore, from Table 3.10 provide 4H25 (A s1 = 1960 mm2)\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"CHECK MAXIMUM AND MINIMUM STEEL AREAS AND MINIMUM BAR SPACING\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"fctm0 26.b d0.04A c ≥ A s ≥ ≥ 0.0013btd\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"fyk\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"0.04A c = 0.04 × bh = 0.04 × 300 × 600 = 7200 mm2 > A s1\", \"2\": \"OK\"}, {\"0\": \"\", \"1\": \"2 3×0 3 25ctm2==×=\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"f fb d0 26 0 26 mm300552221500yk\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"≥ 0.0013b td = 0.0013 × 300 × 552 = 215 mm2 < A s1\", \"2\": \"OK\"}, {\"0\": \"Clear distance between bars = 1/3[b − (2 × cover) − 4φ]\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"= 1/3[300 − 2 × 35 − 4 × 25] = 43 mm\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"OK> min{φ; dg (say 20 mm) + 5 mm; 20 mm]\", \"2\": \"\"}, {\"0\": \"SHEAR REINFORCEMENT\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"W\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"RBRA\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"VEd\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"VEd\", \"2\": \"\"}, {\"0\": \"Design shear force, VEd\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"VEd = 0.5(ωB) = 0.5 × 70.5 × 6 = 211.5 kN\", \"2\": \"\"}, {\"0\": \"Shear resistance of concrete (VRd,c)\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"fck = 25 N mm−2\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"CRd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"200200=+=+=<OKk 1 1 1 6 2 0\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"d552\", \"2\": \"\"}, {\"0\": \"346\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\nbh = 0; 6 = 211; c = 0; .18/γc = 0; .18/1.5 = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0367.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 368, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n347\nExample 8.8 continued\n \nρ1\n1960\n300\n552\n0 0118\n0 02\n \n \n \n .\n .\n=\n=\n×\n=\n<\nA\nb d\nsl\nw\nOK\nσcp = 0\nνmin = 0.035k3/2f ck\n1/2 = 0.035 × 1.63/2 × 251/2 = 0.354 N mm−2\nVRd,c = [CRd,ck(100ρ1fck)1/3 + k1σcp]bwd\n= [0.12 × 1.6(100 × 0.0118 × 25)1/3]300 × 552\n= 98243 N ≥ (νmin + k1σcp)bwd = 0.354 × 300 × 552 = 58 622 N\nOK\nSince VRd,c < VEd, shear reinforcement must be provided.\nCheck compression capacity of compression strut, VRd,max, assuming ~ = 21.8°\nν = 0.6(1 − fck/250) = 0.6(1 − 25/250) = 0.54\nfcd = αccfck/γc = 0.85 × 25/1.5 = 14.2 N mm−2\nVRd,max = bwzνfcd/(cot θ + tan θ)\n= [300 × (0.9 × 552) × 0.54 × 14.2 / (2.5 + 0.4)] × 10−3\n= 394.2 kN > VEd\nOK\nDiameter and spacing of links\nDesign of singly and doubly reinforced rectangular beams\nWhere VEd < VRd,c, provide minimum shear reinforcement, ρw,min, according to\nρw,min = (0.08 fck )/fyk = (0.08 25)/500 = 8 × 10−4\nρw,min = Asw/sbwsin α\n⇒ Asw/s = 8 × 10−4 × 300 × 1 = 0.24\nMaximum spacing of links, smax, is\nsmax = 0.75d = 0.75 × 552 = 414 mm\nHence, from Table 3.13 provide H8 at 300 mm centres, Asw/s = 0.335\nVRd,s = 0.335 × 497 × (500/1.15) × 2.5 × 10−3 = 181 kN\nWhere VEd > 181 kN provide shear reinforcement according to:\n \nV\nA\ns\nzf\nRd,s\nsw\nywd\nN\n \n cot \n \n=\n=\nθ\n211500\nVEd\n3 m\n3 m\nVRd,s (with minimum links) = 181 kN\nx\nx\nVEd = 211500 N\nx = VRd,s/ω = 181/70.5\n = 2.567 m\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"Design of singly and doubly reinforced rectangular beams\"}, {\"0\": \"Example 8.8 continued\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"A1960sl===<ρ 0 0118 1\", \"1\": \"0 02OK\", \"2\": \"\"}, {\"0\": \"×b d300 552\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"σcp = 0\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"νmin = 0.035k 3/2f ck\", \"1\": \"1/2 = 0.035 × 1.63/2 × 251/2 = 0.354 N mm−2\", \"2\": \"\"}, {\"0\": \"σVRd,c = [CRd,ck(100ρ1fck)1/3 + k1cp]bwd\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"= [0.12 × 1.6(100 × 0.0118 × 25)1/3]300 × 552\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"σ= 98243 N ≥ (νmin + k1\", \"1\": \"cp)bwd = 0.354 × 300 × 552 = 58 622 N\", \"2\": \"OK\"}, {\"0\": \"Since VRd,c < VEd, shear reinforcement must be provided.\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"Check compression capacity of compression strut, VRd,max, assuming ~ = 21.8°\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"ν = 0.6(1 − fck/250) = 0.6(1 − 25/250) = 0.54\", \"2\": \"\"}, {\"0\": \"fcd = αccfck/γ\", \"1\": \"c = 0.85 × 25/1.5 = 14.2 N mm−2\", \"2\": \"\"}, {\"0\": \"VRd,max = bwzνfcd/(cot θ + tan θ)\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"= [300 × (0.9 × 552) × 0.54 × 14.2 / (2.5 + 0.4)] × 10−3\", \"2\": \"\"}, {\"0\": \"OK= 394.2 kN > VEd\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Diameter and spacing of links\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"3 m3 m\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"VEdx\", \"1\": \"VEd = 211500 Nx = VRd,s/ω = 181/70.5\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"= 2.567 m\", \"2\": \"\"}, {\"0\": \"x\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"VRd,s (with minimum links) = 181 kN\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Where VEd < VRd,c, provide minimum shear reinforcement, ρ\", \"1\": \"w,min, according to\", \"2\": \"\"}, {\"0\": \"ρw,min = (0.08\", \"1\": \"fck )/fyk = (0.08 25 )/500 = 8 × 10−4\", \"2\": \"\"}, {\"0\": \"ρw,min = A sw/sbwsin α\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"⇒ A sw/s = 8 × 10−4 × 300 × 1 = 0.24\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Maximum spacing of links, smax, is\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"smax = 0.75d = 0.75 × 552 = 414 mm\", \"2\": \"\"}, {\"0\": \"Hence, from Table 3.13 provide H8 at 300 mm centres, A sw/s = 0.335\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"VRd,s = 0.335 × 497 × (500/1.15) × 2.5 × 10−3 = 181 kN\", \"2\": \"\"}, {\"0\": \"Where VEd > 181 kN provide shear reinforcement according to:\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"A\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\nνmin = 0; 251/2 = 0; ν = 0; 25/1.5 = 14; 3\n= 394; Asw/s = 0; s = 0; x = VRd; s/ω = 181/70; .5\n = 2", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0368.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 369, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n348\nExample 8.8 continued\nHowever, it can be seen that VEd = 181 kN occurs at x = 2.567 m which is less than d from the face of the support,\nand there is therefore no need to reduce the link spacing beyond this point.\nDEFLECTION\nfck = 25 N mm−2 ⇒ ρo = fck \n \n \n \n×\n=\n×\n=\n×\n−\n−\n−\n10\n25\n10\n5\n10\n3\n3\n3\n \nρ \n \n \n . \n=\n=\n×\n=\n×\n−\nA\nbd\ns1\n1542\n300\n552\n9 3\n10 3\nSince ρ > ρo, \n \nB\n \nd\nK\nf\nf\n \n .\n \n \n .\n .\n \n. \n \n \n/\n=\n+\n−\n′\n\n\n\n+\n′\n\n\n\n\n\n\n\n\n\n\n\n\n=\n+\n×\n×\n−\n\n\n\n\n\n\n\n\n\n=\n−\n−\n11\n1 5\n1\n12\n1 0 11\n1 5 25\n5\n10\n9 3\n10\n0\n15\n12\n3\n3\nck\no\nck\no\nρ\nρ\nρ\nρ\nρ\n \nσ\nν\ns\nyk\ns,req\ns,pro\nNmm\n \n \n \n \n \n \n. \n=\n=\n×\n×\n×\n=\n−\n310\n500\n310\n500\n1542\n500\n1960\n243 9\n2\nf A\nA\nAllowable span/depth ratio = basic 1/d × 310/σs = 15 × 310/243.9 = 19\nActual span/depth ratio = 6000/552 = 10.9 < allowable\nOK\nREINFORCEMENT DETAILS\nThe sketches below show the main reinforcement requirements for the beam. For reasons of buildability the actual\nreinforcement details may well be slightly different.\n300 mm\n300 mm\n20H8-300\nFE = VEdal/z = (VEdcot θ)/2 = 1.25VEd for cot θ = 2.5\n⇒FE = 1.25 × 211.5 = 264.4 kN\nAs,req = (264.4 × 103)/(500/1.15) = 608 mm2\nlbd = (41 × 25) × 608/1960 = 318 mm measured from front face of support to end of bar\n ≥ 0.3lbd,req = 0.3 × (41 × 25) = 308 mm OK\n2H12\nH8 links\n4H25\n300 mm\n552 mm\n48 mm\n2H12\nH8 links\nH25\n125 mm\n152.5\n87.5\n5700 mm\n\n[TABLES]\n[[{\"0\": \"300 mm\", \"1\": \"5700 mm\"}, {\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"H8 links\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"H2587.5\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"152.5\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"2H12\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"H8 linksH2587.5152.5\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [{\"0\": \"H8 links\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"4H25\", \"2\": \"\"}], [], [], [{\"\": \"\", \"None\": \"\"}], [], [{\"\": \"5700 mm\"}]]\n\n[FORMULAS]\n.9 = 19; FE = 1; 211.5 = 264; req = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0369.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 370, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n349\nx = 0.616d\n200 kNmm–2\nεS = 0.002175 = εy\nfS = 500/1.15\nd\nεy = 0.002175\nεcu = 0.0035\nExample 8.9 Analysis of a singly reinforced beam (EC 2)\nCalculate the maximum variable load that the beam shown below can carry assuming that the load is (i) uniformly\ndistributed or (ii) occurs as a point load at mid-span.\nMAXIMUM UNIFORMLY DISTRIBUTED LOAD, qK\nPermanent load, gk = 0.5 × 0.3 × 25 = 3.75 kNm−1\nTotal ultimate load, ω = 1.35gk + 1.5qk\n= 1.35 × 3.75 + 1.5 × qk\nDesign of singly and doubly reinforced rectangular beams\nMOMENT CAPACITY OF SECTION\nEffective depth, d, is\nd = h − cover − φ/2 = 500 − 40 − 25/2 = 447 mm\n \nK\nM\nf bd\nM\no\nck\n=\n=\n×\n×\n \n \n \n \n2\n2\n25\n300\n447\nz = d[0.5 + ( .\n \n/ . )\n0 25\n3\n3 4\n−K o\n]\n= 447[0.5 + ( .\n \n/ . \n \n \n)\n0 25\n3\n3 4\n25\n300\n4472\n−\n×\n×\n×\n×\nM\n]\n(1)\n \nA\nM\nf z\nM\nz\ns\nyk\nmm\n \n.\n \n.\n \n \n \n \n=\n=\n×\n×\n=\n0 87\n0 87\n500\n1960\n2\n(2)\nSolving equations (1) and (2) simultaneously gives\nM = 295.6 kNm and z = 346.7 mm\n⇒ \n \nx\nd\nz\nd\n \n.\n \n \n.\n.\n \n. \n .\n=\n−\n=\n−\n=\n≤\n0 4\n447\n346 7\n0 4\n250 7\n0 616\nmm\n(to ensure section is under-reinforced)\n= 0.616 × 447 = 275 mm\nOK\nfck = 25 Nmm–2\nb = 300\nh = 500\n40 mm cover\n4H25 (As1 = 1960 mm2)\n7 m\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Design of singly and doubly reinforced rectangular beams\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"Example 8.9 Analysis of a singly reinforced beam (EC 2)\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"Calculate the maximum variable load that the beam shown below can carry assuming that the load is (i) uniformly\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"distributed or (ii) occurs as a point load at mid-span.\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"b = 300\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"h = 500\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"fck = 25 Nmm–2\", \"7\": \"\"}, {\"0\": \"7 m\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"4H25 (As1 = 1960 mm2)\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"40 mm cover\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"MOMENT CAPACITY OF SECTION\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"Effective depth, d, is\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"d = h − cover − φ/2 = 500 − 40 − 25/2 = 447 mm\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"MM==K o\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"2××25300fbd ck\", \"1\": \"2447\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"−(0 25 3/K oz = d[0.5 +\", \"1\": \"3 4)\", \"2\": \"]\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"−(0 25 3 = 447[0.5 +\", \"1\": \"× M\", \"2\": \"/3 4\", \"3\": \"×\", \"4\": \"25\", \"5\": \"×\", \"6\": \"×]300 4472)\", \"7\": \"(1)\"}, {\"0\": \"M fM==A s\", \"1\": \"\", \"2\": \"=\", \"3\": \"1960\", \"4\": \"\", \"5\": \"mm\", \"6\": \"2\", \"7\": \"(2)\"}, {\"0\": \"×0 87z0 87 500yk\", \"1\": \"×\", \"2\": \"z\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"Solving equations (1) and (2) simultaneously gives\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"M = 295.6 kNm and z = 346.7 mm\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"−−d z447 346 7===⇒ x 250 7 mm\", \"1\": \"≤\", \"2\": \"0 616d\", \"3\": \"\", \"4\": \"(to ensure section is under-reinforced)\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"0 40 4\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"= 0.616 × 447 = 275 mmOK\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"εcu = 0.0035\", \"7\": \"\"}, {\"0\": \"fS = 500/1.15\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"x = 0.616d\", \"7\": \"\"}, {\"0\": \"200 kNmm–2\", \"1\": \"d\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"εS = 0.002175 = εy\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"εy = 0.002175\", \"7\": \"\"}, {\"0\": \"MAXIMUM UNIFORMLY DISTRIBUTED LOAD, qK\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"Permanent load, gk = 0.5 × 0.3 × 25 = 3.75 kNm−1\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"Total ultimate load, ω = 1.35gk + 1.5qk\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"= 1.35 × 3.75 + 1.5 × qk\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"349\"}, {\"0\": \"9780415467193_C08349\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"9/3/09, 3:00 PM\", \"7\": \"\"}], [], []]\n\n[FORMULAS]\n.002175 = εy; fS = 500/1; gk = 0; 25 = 3; ω = 1; b = 300; h = 500", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0370.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 371, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n350\nExample 8.9 continued\nSubstituting into\n \nM \n= ωB2\n8\n \n295 6\n1 35\n3 75\n1 5\n7\n8\n2\n. ( .\n .\n .\n) \n=\n×\n+\n×\nqk\nHence\nqk = 28.8 kNm−1\nMAXIMUM POINT LOAD, QK\nFactored permanent load, WP = (1.35gk) span\n= (1.35 × 3.75) 7 = 35.44 kN\nFactored variable load, Wv = 1.5Qk\nSubstituting into\n \nM\nW\nW\n \n \n=\n+\nP\nV\nB\nB\n8\n4\n \n295 6\n35 44\n7\n8\n1 5\n7\n4\n. \n.\n .\n \n=\n×\n+\n×\nQk\nHence Qk = 100.8 kN\n8.9 Design of one-way solid slabs\n8.9.1 DEPTH, BENDING, SHEAR\nEC 2 requires a slightly different approach to BS\n8110 for designing one-way spanning solid slabs.\nTo calculate the depth of the slab, the designer must\nfirst estimate the percentage of steel required in\nthe slab for bending. Generally, most slabs will be\nlightly reinforced, i.e. ρ < 0.5%. The percentage of\nreinforcement can be used, together with the sup-\nport conditions, to calculate an appropriate span/\ndepth ratio using equations 8.32 and 8.33. The\neffective depth of the slab is obtained by multiplying\nthis ratio by the span (see Example 8.10). The actual\narea of steel required in the slab for bending is\ncalculated using the equations developed in section\n8.8.1. Provided this agrees with the assumed value,\nthe calculated depth of the slab is acceptable.\nThe designer will also need to check that the\nslab will not fail in shear. The shear resistance of\nthe slab can be calculated as for beams (see section\n8.8.2). According to Cl. 6.2.1(3), where the design\nshear force (VEd) is less than the design shear\nresistance of the concrete alone (VRd,c) no shear\nreinforcement is necessary. The same requirement\nappears in BS 8110.\nWhere VEd > VRd,c, all the shear force must be\nresisted by the shear reinforcement. The actual area\nof shear reinforcement in slabs can be calculated\nas for beams (section 8.8.2). This should not be\nless than the minimum value, ρw,min, given by\nρw,min = (0.08 fck )/fyk\nFor members with vertical shear reinforcement, the\nmaximum longitudinal spacing of successive series\nof shear links, smax, should not exceed\nsmax = 0.75d\nMoreover, the maximum transverse spacing of link\nreinforcement should not exceed 1.5d. Cl. 9.3.2(1)\nalso notes that shear reinforcement should not be\nprovided in slabs less than 200 mm deep.\n8.9.2 REINFORCEMENT DETAILS FOR SOLID\nSLABS\nThis section outlines EC 2 requirements regarding\nthe detailing of slabs with respect to:\n1. reinforcement percentages\n2. spacing of reinforcement\n3. anchorage and curtailment of reinforcement\n4. crack control\n1. Reinforcement areas (Cl. 5.4.2.1.1, EC 2)\nIn EC 2, the maximum and minimum percentages\nof longitudinal steel permitted in beams and slabs\nare the same, namely:\n\n[TABLES]\n[[{\"0\": \"Eurocodes 2: Design of concrete structures\", \"1\": \"\"}, {\"0\": \"Example 8.9 continued\", \"1\": \"\"}, {\"0\": \"Substituting into\", \"1\": \"\"}, {\"0\": \"ωB 2M =\", \"1\": \"\"}, {\"0\": \"8\", \"1\": \"\"}, {\"0\": \"×(1 35 3 75=\", \"1\": \"2+× 1 5) 7qk\"}, {\"0\": \"295 6\", \"1\": \"8\"}, {\"0\": \"Hence\", \"1\": \"\"}, {\"0\": \"qk = 28.8 kNm−1\", \"1\": \"\"}, {\"0\": \"MAXIMUM POINT LOAD, QK\", \"1\": \"\"}, {\"0\": \"Factored permanent load, WP = (1.35gk) span= (1.35 × 3.75) 7 = 35.44 kN\", \"1\": \"\"}, {\"0\": \"Factored variable load, Wv = 1.5Q kSubstituting into\", \"1\": \"\"}, {\"0\": \"BWWP=+M\", \"1\": \"BV\"}, {\"0\": \"8\", \"1\": \"4\"}, {\"0\": \"×35 44 =295 6\", \"1\": \"×71 5 7Qk+\"}, {\"0\": \"8\", \"1\": \"4\"}, {\"0\": \"Hence Q k = 100.8 kN\", \"1\": \"\"}, {\"0\": \"8.9 Design of one-way solid slabs\", \"1\": \"of shear reinforcement in slabs can be calculated\"}, {\"0\": \"\", \"1\": \"as for beams (section 8.8.2). This should not be\"}, {\"0\": \"8.9.1 DEPTH, BENDING, SHEAR\", \"1\": \"less than the minimum value, ρw,min, given by\"}, {\"0\": \"EC 2 requires a slightly different approach to BS\", \"1\": \"ρfck )/fykw,min = (0.08\"}, {\"0\": \"8110 for designing one-way spanning solid slabs.\", \"1\": \"For members with vertical shear reinforcement, the\"}, {\"0\": \"To calculate the depth of the slab, the designer must\", \"1\": \"maximum longitudinal spacing of successive series\"}, {\"0\": \"first estimate the percentage of steel required in\", \"1\": \"of shear links, smax, should not exceed\"}, {\"0\": \"the slab for bending. Generally, most slabs will be\", \"1\": \"\"}, {\"0\": \"lightly reinforced, i.e. ρ < 0.5%. The percentage of\", \"1\": \"smax = 0.75d\"}, {\"0\": \"reinforcement can be used, together with the sup-\", \"1\": \"Moreover, the maximum transverse spacing of link\"}, {\"0\": \"port conditions, to calculate an appropriate span/\", \"1\": \"reinforcement should not exceed 1.5d. Cl. 9.3.2(1)\"}, {\"0\": \"depth ratio using equations 8.32 and 8.33. The\", \"1\": \"also notes that shear reinforcement should not be\"}, {\"0\": \"effective depth of the slab is obtained by multiplying\", \"1\": \"provided in slabs less than 200 mm deep.\"}, {\"0\": \"this ratio by the span (see Example 8.10). The actual\", \"1\": \"\"}, {\"0\": \"area of steel required in the slab for bending is\", \"1\": \"8.9.2 REINFORCEMENT DETAILS FOR SOLID\"}, {\"0\": \"calculated using the equations developed in section\", \"1\": \"SLABS\"}, {\"0\": \"8.8.1. Provided this agrees with the assumed value,\", \"1\": \"This section outlines EC 2 requirements regarding\"}, {\"0\": \"the calculated depth of the slab is acceptable.\", \"1\": \"the detailing of slabs with respect to:\"}, {\"0\": \"The designer will also need to check that the\", \"1\": \"\"}, {\"0\": \"slab will not fail in shear. The shear resistance of\", \"1\": \"1.reinforcement percentages\"}, {\"0\": \"the slab can be calculated as for beams (see section\", \"1\": \"2.spacing of reinforcement\"}, {\"0\": \"8.8.2). According to Cl. 6.2.1(3), where the design\", \"1\": \"3.anchorage and curtailment of reinforcement\"}, {\"0\": \"is less than the design shearshear force (VEd)\", \"1\": \"4.crack control\"}, {\"0\": \"resistance of the concrete alone (VRd,c) no shearreinforcement is necessary. The same requirement\", \"1\": \"1. Reinforcement areas (Cl. 5.4.2.1.1, EC 2)\"}, {\"0\": \"appears in BS 8110.\", \"1\": \"In EC 2, the maximum and minimum percentages\"}, {\"0\": \"Where VEd > VRd,c, all the shear force must be\", \"1\": \"of longitudinal steel permitted in beams and slabs\"}, {\"0\": \"resisted by the shear reinforcement. The actual area\", \"1\": \"are the same, namely:\"}, {\"0\": \"350\", \"1\": \"\"}], [], []]\n\n[FORMULAS]\n7 = 35; Wv = 1", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0371.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 372, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n351\n(a)\n(b)\nFace of support\n15%\n100%\nlbd\n0.15l, ≥ Ibd + d \nthe adjacent span as shown in Fig. 8.25 and be\nanchored. At an end support the moment to be\nresisted may be reduced to 15 per cent of the max-\nimum moment in the adjacent span.\nThe Concrete Centre publication ‘How to design\nconcrete structures using Eurocode 2’ recommends\nthat the simplified curtailment rules for continuous\nslabs shown in Fig. 8.26 should be used.\n \nA\nf\nf\nb d\nb d\ns\nctm\nyk\nt\nt\n1\n0 26\n0 0013\n .\n .\n≥\n≥\nAs1, As2 ≤ 0.04Ac\nwhere\nb\nbreadth of section\nd\neffective depth\nAc\ncross-sectional area of concrete (bh)\nfyk\ncharacteristic yield stress of reinforcement\nThe area of transverse or secondary reinforce-\nment should not generally be less than 20 per cent\nof the principal reinforcement, i.e.:\nAs (trans) ≥ 0.2 × As (main)\n2. Spacing of reinforcement\nThe clear distance between reinforcing bar should\nnot be less than the following:\n(a) maximum bar diameter\n(b) 20 mm\n(c) dg + 5 mm where dg is the maximum aggregate\nsize.\nThe maximum bar spacing in slabs, smax, for the\nmain reinforcement should not exceed the following:\nsmax = 3h ≤ 400 mm generally and 2h ≤ 250 mm\nin areas of maximum moment\nwhere h denotes the overall depth of the slab.\nThe maximum bar spacing in slabs, smax, for the\nsecondary reinforcement should not exceed the\nfollowing:\nsmax = 3.5h ≤ 450 mm generally and\n3h ≤ 400 mm in areas of maximum moment\n3. Anchorage and curtailment (Cl 9.2.1.3 & 9.3.1.2)\nFor detailing the main reinforcement in slabs,\nsimilar provisions to those outlined earlier for beams\napply.\nCl. 9.3.1.2 recommends that where end supports\nare simply supported, 50 per cent of the calculated\nspan reinforcement should continue up to the sup-\nport and be anchored to resist a force, FE, given by\nFE = |VEd|.a1/z + NEd\nwhere |VEd|, z and NEd are as defined for equation\n8.41 and a1 is equal to d if the slab is not rein-\nforced for shear or is given by equation 8.40 if\nshear reinforcement is provided.\nIf partial fixity exists along an edge of a slab\ntop reinforcement should be provided capable of\nresisting at least 25 per cent of the maximum\nmoment in the adjacent span. This reinforcement\nshould extend at least 0.2 times the length of\nDesign of one-way solid slabs\nFig. 8.26 Simplified curtailment rules for slabs (a) simple\nsupport (b) continuous member.\n0.2l\n100%\n25%\nFig. 8.25\nSection A-A\nA\nA\n25%\n15%\n25%\n100%\n\n[TABLES]\n[[{\"0\": \"slabs shown in Fig. 8.26 should be used.\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}, {\"0\": \"25%\", \"1\": \"0.2l100%\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"(a)\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [], []]\n\n[FORMULAS]\nsmax = 3", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0372.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 373, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n352\n4. Crack widths (Cl. 7.3.3, EC 2)\nAccording to EC 2, where the overall depth of\nthe slab does not exceed 200 mm and the code\nprovisions with regard to reinforcement areas, spac-\ning of reinforcement, anchorage and curtailment\nof bars, etc., discussed above have been applied,\nno further measures specifically to control cracking\nare necessary.\nFurthermore, where at least the minimum rein-\nforcement area has been provided, the limitation\nof crack width to less than 0.3 mm for reinforced\nconcrete may generally be achieved by either limit-\ning the maximum bar spacing or maximum bar\ndiameter in accordance with the values given in\nTable 8.12.\nDETERMINE EFFECTIVE DEPTH OF SLAB AND AREA OF MAIN STEEL\nEstimate overall depth of slab\nAssume reinforcement ratio, ρ, is 0.35%. From equation 8.32, basic span/effective depth ratio for a simply supported\nslab is approximately 31.\nMinimum effective depth, \n \nd \n \n \n \n=\n=\n≈\nspan\nbasic ratio\nmm\n4650\n31\n150\nHence, take d = 155 mm\nMinimum cover, cmin = max{cmin,dur; cmin,b; 10 mm}\nFor Class XC1 exposure, cmin,dur = 15 mm (Table 8.10)\nAssume diameter of main steel, Φ = 10 mm, therefore cmin,b = 10 mm\nNominal cover to reinforcement, cnom = cmin + ∆cdev = 15 + 10 = 25 mm\nOverall depth of slab (h) = d + Φ/2 + cnom = 155 + 10/2 + 25 = 185 mm\nLoading\nPermanent\nSelf weight of slab (gk) = 0.185 × 25(kNm−3) = 4.625 kNm−2\nVariable\nTotal variable load (qk) = 4 kNm−2\nUltimate load\nFor 1m width of slab total ultimate load = (1.35gk + 1.5qk)span\n= (1.35 × 4.625 + 1.5 × 4)4.65 = 56.93 kN\nExample 8.10 Design of a one-way spanning floor (EC 2)\nDesign the floor shown in Fig. 8.27 for an imposed load of 4 kN mm−2, assuming the following material strengths:\nfck = 30 Nmm−2, fyk = 500 Nmm−2. The environmental conditions fall within exposure class XC1.\nFig. 8.27\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\n4.65 = 56", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0373.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 374, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n353\nExample 8.10 continued\nDesign of one-way solid slabs\nDesign moment\nDesign moment, \n \nM\nWl\nEd\nkNm\n \n \n.\n .\n \n. \n=\n=\n×\n8\n56 93\n4 65\n8\n33 1\nUltimate moment\nUltimate moment of resistance (MRd) = 0.167fckbd 2\n= 0.167 × 30 × 1000 × 1552 × 10−6 = 120 kNm\nSince MRd > MEd no compression reinforcement is required.\nMain reinforcement (As1)\n \nK\nM\nf bd\no\nck\n \n \n. \n \n \n .\n=\n=\n×\n×\n×\n=\n2\n6\n2\n33 1\n10\n30\n1000\n155\n0 046\nz = d(0.5 + ( .\n \n/ . )\n0 25\n3\n3 4\n−K o\n) ≤ 0.95d = 0.95 × 155 = 147.3 mm\n= 155(0.5 + ( .\n .\n/ . )\n0 25\n3\n0 046 3 4\n−\n×\n) = 155 × 0.957\n \nA\nM\nf z\ns\nyk\nmm\n1\n6\n2\n0 87\n33 1\n10\n0 87\n500\n147 3\n517\n \n.\n \n. \n.\n \n \n.\n \n \n=\n=\n×\n×\n×\n=\nHence from Table 3.22, H10 at 150 mm centres (As1 = 523 mm2 m−1) would be suitable.\nThe required reinforcement ratio, ρ, in this case is\n \nρ \n \n \n .\n .\n%\n=\n=\n×\n=\n=\nA\nbd\ns1\n517\n1000\n155\n0 00334\n0 334\nCheck estimated effective depth of slab\nDesign service stress, \n \nσ\nν\ns\nyk\nreq\npro\nNmm\n \n \n \n \n \n. \n≈\n×\n×\n=\n×\n×\n=\n−\n5\n8\n5\n8\n500\n517\n523\n308 9\n2\nf\nA\nA\nModification factor = \n \n310\n310\n308 9\n1 0\nσs\n \n.\n .\n=\n=\nBasic span/effective depth ratio is given by\n \nB\n \nd\nK\nf\nf\n \n .\n .\n \n/\n=\n+\n\n\n\n+\n−\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n11\n1 5\n3 2\n1\n3 2\nck\no\nck\no\nρ\nρ\nρ\nρ\n \n=\n+\n×\n×\n\n\n\n+\n×\n×\n×\n−\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n=\n−\n−\n−\n−\n .\n .\n \n.\n \n . \n \n.\n \n \n \n.\n/\n1 0 11\n1 5 30\n30\n10\n3 34\n10\n3 2\n30\n30\n10\n3 34\n10\n1\n33 4\n3\n3\n3\n3\n3 2\nModified span/effective depth ratio = basic ratio × modification factor\n= 26.4 × 1.0 = 33.4\nActual span/effective depth ratio \n \n=\n=\n \n \n4650\n155\n30 < allowable OK\nTherefore, use a slab with an effective depth = 155 mm, overall depth = 185 mm and main steel = H10 at 150 mm\ncentres.\nCheck bar spacing and steel area\nMaximum bar spacing for main steel < 3h = 3 × 185 = 555 mm ≤ 250 mm (in areas of maximum moment) OK\nMaximum area of steel, As,max = 0.04Ac = 0.04 × 185 × 103 = 7400 mm2 m−1 OK\n\n[TABLES]\n[[{\"0\": \"Design of one-way solid slabs\"}, {\"0\": \"Example 8.10 continued\"}, {\"0\": \"Design moment×\"}, {\"0\": \"Wl56 93 4 65==Design moment, M 33 1 kNmEd\"}, {\"0\": \"88\"}, {\"0\": \"Ultimate moment\"}, {\"0\": \"Ultimate moment of resistance (MRd) = 0.167fckbd 2\"}, {\"0\": \"= 0.167 × 30 × 1000 × 1552 × 10−6 = 120 kNm\"}, {\"0\": \"Since MRd > MEd no compression reinforcement is required.\"}, {\"0\": \"Main reinforcement (As1)\"}, {\"0\": \"6×33 110M ===\"}, {\"0\": \"0 046K o22××301000155fbd ck\"}, {\"0\": \"−(0 25 3/3 4)) ≤ 0.95d = 0.95 × 155 = 147.3 mmK oz = d(0.5 +\"}, {\"0\": \"−×) = 155 × 0.9570 25 3 0 046 3 4)= 155(0.5 + (\"}, {\"0\": \"6×33 110 2===\"}, {\"0\": \"M f 517 Amms1××0 870 87500147 3 zyk\"}, {\"0\": \"Hence from Table 3.22, H10 at 150 mm centres (A s1 = 523 mm2 m−1) would be suitable.The required reinforcement ratio, ρ, in this case is\"}, {\"0\": \"A b5171====0 003340 334ρ %\"}, {\"0\": \"ds×1000155\"}, {\"0\": \"Check estimated effective depth of slab\"}, {\"0\": \"5 8A A5 8517req−2σ≈××=××=500308 9 f NmmDesign service stress, syk\"}, {\"0\": \"ν523pro 310310\"}, {\"0\": \"==1 0Modification factor = σ\"}, {\"0\": \"308 9s\"}, {\"0\": \"Basic span/effective depth ratio is given by\"}, {\"0\": \"3 2B\"}, {\"0\": \" ρ ρ ρ ρ oo+−=+111 53 21 K f f ckck\"}, {\"0\": \"d\"}, {\"0\": \"3 2−−3330 1030 10\"}, {\"0\": \" × × × × =+×−=+ 1 0 11 1 530 3 2 30 1 33 4−−33\"}, {\"0\": \"3 34 103 34 10\"}, {\"0\": \"Modified span/effective depth ratio = basic ratio × modification factor\"}, {\"0\": \"= 26.4 × 1.0 = 33.4\"}, {\"0\": \"4650==Actual span/effective depth ratio 30 < allowable OK\"}, {\"0\": \"155 Therefore, use a slab with an effective depth = 155 mm, overall depth = 185 mm and main steel = H10 at 150 mm\"}, {\"0\": \"centres.\"}, {\"0\": \"Check bar spacing and steel area\"}, {\"0\": \"Maximum bar spacing for main steel < 3h = 3 × 185 = 555 mm ≤ 250 mm (in areas of maximum moment) OK\"}, {\"0\": \"Maximum area of steel, A s,max = 0.04A c = 0.04 × 185 × 103 = 7400 mm2 m−1 OK\"}], [], []]\n\n[FORMULAS]\n155 = 147; 1.0 = 33; max = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0374.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 375, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nEurocodes 2: Design of concrete structures\n354\nExample 8.10 continued\nMinimum reinforcement area, \n \nA\nf\nf\nb d\nb d\ns\nctm\nyk\nt\nt\n,min .\n .\n=\n≥\n0 26\n0 0013\n \n=\n×\n×\n=\n−\n .\n. \n \n \n \n/\n0 260 3\n30\n500\n10\n155\n233\n2 3\n3\n2\n1\nmm m\n≥ 0.0013bd = 0.0013 × 103 × 155 = 202 mm2 m−1 > As1\nOK\nSECONDARY REINFORCEMENT\nProvide H8 at 300 mm centres (As (tras) = 168 mm2 m−1)\nAs (trans) ≥ 0.2As (main) = 0.2 × 523 = 105 mm2 m−1 OK\nMaximum bar spacing < 3.5h = 3.5 × 185 = 648 mm ≤ 400 mm (in areas of maximum moment) OK\nSHEAR REINFORCEMENT\nVEd\nW\nRA\nRB\nVEd\n4.5 m\nUltimate load (W) = 51.7 kN\nDesign shear force (VEd)\n \nV\nW\nEd\nkN\n=\n=\n=\n \n \n.\n \n. \n2\n56 93\n2\n28 5\nShear resistance of concrete alone, VRd,c\nfck = 30 N mm−2\nCRd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2\n \nk\nd\n \n \n . .\n=\n+\n=\n+\n=\n≤\n1\n200\n1\n200\n155\n2 1\n2 0\nAssuming 50 per cent of mid-span steel does not extend to the supports As1 = 260 mm2 m−1. Hence\n \nρ1\n3\n260\n10\n155\n0 00168\n0 02\n \n \n \n .\n .\n=\n=\n×\n=\n<\nA\nb d\ns1\nw\nOK\nσcp = 0\nνmin = 0.035k3/2f ck\n1/2 = 0.035 × 23/2 × 301/2 = 0.542\nVRd,c = [CRd,ck(100ρ1fck)1/3 + k1σcp]bwd\n= [0.12 × 2(100 × 0.00168 × 30)1/3]103 × 155\n= 63780 N ≥ (νmin + k1σcp)bwd = 0.542 × 103 × 155 = 84010 N\nSince VRd,c (= 84 kN) > VEd (28.5 kN) no shear reinforcement is required which is normally the case for slabs.\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"Eurocodes 2: Design of concrete structures\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"Example 8.10 continued\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"Minimum reinforcement area,\", \"1\": \"A\", \"2\": \",min s\", \"3\": \"fctm=≥ 0 26b d 0 0013b d\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"fyk\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"2 3×0 3 30−31=×=\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"0 2610 155 233 mm m500\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"≥ 0.0013bd = 0.0013 × 103 × 155 = 202 mm2 m−1 > A s1\", \"4\": \"OK\"}, {\"0\": \"\", \"1\": \"SECONDARY REINFORCEMENT\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"Provide H8 at 300 mm centres (A s (tras) = 168 mm2 m−1)\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"A s (trans) ≥ 0.2A s (main) = 0.2 × 523 = 105 mm2 m−1 OK\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"Maximum bar spacing < 3.5h = 3.5 × 185 = 648 mm ≤ 400 mm (in areas of maximum moment) OK\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"SHEAR REINFORCEMENT\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"W\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"4.5 mRARB\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"VEd\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"VEd\", \"4\": \"\"}, {\"0\": \"Ultimate load (W ) = 51.7 kN\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"Design shear force (VEd)\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"W56 93===V 28 5 kNEd\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"22\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"Shear resistance of concrete alone, VRd,c\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"fck = 30 N mm−2\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"C Rd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"200200=+=+=≤k 1 1 2 1 2 0\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"d155\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"Assuming 50 per cent of mid-span steel does not extend to the supports A s1 = 260 mm2 m−1. Hence\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"=ρ 1\", \"2\": \"A bds1=\", \"3\": \"260=<OK 0 00168 0 02\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"w\", \"3\": \"3×10 155\", \"4\": \"\"}, {\"0\": \"σ\", \"1\": \"cp = 0\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"ν\", \"1\": \"min = 0.035k3/2f ck\", \"2\": \"\", \"3\": \"1/2 = 0.035 × 23/2 × 301/2 = 0.542\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"VRd,c = [CRd,ck(100ρ\", \"2\": \"\", \"3\": \"σ1fck)1/3 + k1cp]bwd\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"= [0.12 × 2(100 × 0.00168 × 30)1/3]103 × 155\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"= 63780 N ≥ (ν\", \"3\": \"σmin + k1cp)bwd = 0.542 × 103 × 155 = 84010 N\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"Since VRd,c (= 84 kN) > VEd (28.5 kN) no shear reinforcement is required which is normally the case for slabs.\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"354\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}], [], [], [{\"\": \"\", \"None\": \"\"}, {\"\": \"\", \"None\": \"\"}]]\n\n[FORMULAS]\nc = 0; .18/γc = 0; .18/1.5 = 0; νmin = 0; 301/2 = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0375.png"} |
| {"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition ( PDFDrive ).pdf", "page_num": 376, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n355\n16-H08-300\nFE = VEd(d/z) = 28.5/0.9 = 31.7 kN\nAs,req = (31.7 × 103)/(500/1.15) = 73 mm2\nlbd = (36 × 10) × 73/260 = 101 mm measured from front face of support to end of bar\n ≥ 0.3lb,min = 0.3 × (36 × 10) = 108 mm\n30mm\n08\n01-H10-150\n155mm\n08\n01\nCurtail 50% of bars x mm in from centre\nline of both supports where\nx = 0.15l – al(= d)\n = 0.15 × 4650 – 155 = 542 mm\n01\n01\n16-H08-300\n01-H10-150\n01-H10-150\n542\n108\n150\nExample 8.10 continued\nREINFORCEMENT DETAILS\nThe sketches below show the main reinforcement requirements for the slab.\nReinforcement areas and bar spacing\nMinimum reinforcement areas and bar spacing rules were checked above and found to be satisfactory.\nCrack width\nSince the overall depth of the slab does not exceed 200 mm and the rest of the code provision have been met, no\nfurther measures specifically to control cracking are necessary.\nfck = 25 N mm–2\nfyk = 500 Nmm–2\nρc = 25 kNm–3\nDesign of one-way solid slabs\nExample 8.11 Analysis of one-way spanning floor (EC 2)\nCalculate the maximum uniformly distributed variable load that the floor shown below can carry.\n\n[TABLES]\n[[{\"0\": \"16-H08-300\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"0101\", \"2\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"155mm30mm\", \"4\": \"\"}, {\"0\": \"0808\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"54201\", \"1\": \"108\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"150\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"ρ\", \"3\": \"fck = 25 N mm–2fyk = 500 Nmm–2c = 25 kNm–3\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [], [], [{\"\": \"\", \"None\": \"\"}, {\"\": \"108\\n1\", \"None\": \"\"}, {\"\": \"\", \"None\": \"50\"}], [{\"\": \"\", \"None\": \"\"}, {\"\": \"\", \"None\": \"\"}, {\"\": \"\", \"None\": \"\"}, {\"\": \"\", \"None\": \"\"}, {\"\": \"\", \"None\": \"\"}, {\"\": \"\", \"None\": \"\"}]]\n\n[FORMULAS]\n.5/0.9 = 31; min = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0376.png"} |
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