data/chunks/chunk_a71f59bd_12.jsonl

{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 215, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n194\nmx, my\nequivalent uniform moment factors\nfor major axis and minor axis\nbuckling respectively obtained from\nTable 26 of BS 5950, reproduced as\nTable 4.16\nmLT\nequivalent uniform moment factor\nfor lateral torsional buckling\nobtained from Table 18 of BS 5950,\nreproduced as Table 4.10\nMb\nbuckling resistance moment of the\ncased column = Sxpb ≤ 1.5Mb for the\nuncased section. To determine pb, ry\nshould be taken as the greater of ry\nof the uncased section or\n0.2(B + 100) mm (Fig. 4.30).\nMx, My\nmaximum moment about the major\nand minor axes respectively\nZx, Zy\nelastic modulus about the major and\nminor axes respectively.\nExample 4.13 continued\nFor λx = 25.7 and py = 265 N/mm2 compressive strength, pc = 257 N/mm2 (Table 4.13). For λy = 41.2 and py =\n265 N/mm2 compressive strength, pc = 228 N/mm2 (Table 4.14).\nHence, pc is equal to 228 N/mm2.\nCompression resistance\nAg = 15 000 mm2\nAc = dcbc = 425 × 425 = 180 625 mm2\npy = 265 N/mm2\n(since T = 18.7 mm)\npc = 228 N/mm2\nfcu = 20 N/mm2\nCompression resistance of encased column, Pc, is given by\n    \nP\nA\nf A\np\np\nc\ng\ncu\nc\ny\nc\n  \n  .\n=\n+\n\n\n\n\n0 45\n    \nPc\n15000\n  20  180625\n265\n  \n  .\n=\n+\n×\n×\n\n\n\n\n0 45\n228 = 4.81 × 106 N = 4810 kN\nwhich should not be greater than the short strut capacity, P cs, given by\n    \nP\nA\nf A\np\np\ncs\ng\ncu\nc\ny\ny\n  \n  .\n=\n+\n\n\n\n\n0 25\n15000\n  20  180625\n265\n \n  .\n=\n+\n×\n×\n\n\n\n\n0 25\n265 = 4.878 × 106 N = 4878 kN\nOK\nHence the compression resistance of the encased column is 4878 kN. Comparing this with the compression resistance\nof the uncased column (Example 4.10) shows that the load capacity of the column has been increased from 3300 kN\nto 4878 kN, which represents an increase of approximately 45%.\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"Design in structural steelwork to BS 5950\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"Example 4.13 continued\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"For  λ\", \"1\": \"x  =  25.7  and  py  =  265  N/mm2  compressive  strength,  pc  =  257  N/mm2  (Table 4.13).  For  λ\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"y  =  41.2  and  py  =\"}, {\"0\": \"\", \"1\": \"265 N/mm2 compressive strength, pc = 228 N/mm2 (Table 4.14).\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"Hence, pc is equal to 228 N/mm2.\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"Compression resistance\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"Ag\", \"12\": \"= 15 000 mm2\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"Ac\", \"12\": \"= dcbc = 425 × 425 = 180 625 mm2\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"py\", \"12\": \"= 265 N/mm2(since T = 18.7 mm)\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"pc\", \"12\": \"= 228 N/mm2\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"fcu\", \"12\": \"= 20 N/mm2\"}, {\"0\": \"\", \"1\": \"Compression resistance of encased column, Pc, is given by\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"P\", \"3\": \"c\", \"4\": \"=\", \"5\": \" \", \"6\": \"A\", \"7\": \"g\", \"8\": \"+\", \"9\": \"\", \"10\": \"0 45\", \"11\": \"fcu\", \"12\": \" Acpc\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"py\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Pc\", \"4\": \"=\", \"5\": \" \", \"6\": \"15000\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"+\", \"11\": \"0 45\", \"12\": \" ××  20  180625228 = 4.81 × 106 N = 4810 kN\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"265\"}, {\"0\": \"which should not be greater than the short strut capacity, P cs, given by\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"P\", \"1\": \"cs\", \"2\": \"=\", \"3\": \"\", \"4\": \" A\", \"5\": \"g\", \"6\": \"+\", \"7\": \"\", \"8\": \"0 25\", \"9\": \"\", \"10\": \"fcu\", \"11\": \" Ac\", \"12\": \"py\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"p\", \"10\": \"y\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"=\", \"3\": \"\", \"4\": \" \", \"5\": \"15000\", \"6\": \"\", \"7\": \"\", \"8\": \"+\", \"9\": \"\", \"10\": \"0 25\", \"11\": \"×\", \"12\": \" × 20  180625265 = 4.878 × 106 N = 4878 kNOK\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"265\"}, {\"0\": \"\", \"1\": \"Hence the compression resistance of the encased column is 4878 kN. Comparing this with the compression resistance\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"of the uncased column (Example 4.10) shows that the load capacity of the column has been increased from 3300 kN\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}, {\"0\": \"\", \"1\": \"to 4878 kN, which represents an increase of approximately 45%.\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\"}], [{\"0\": \"of the uncased column (Example 4.10) shows that the load capacity of the column has been increased from 3300 kN\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"to 4878 kN, which represents an increase of approximately 45%.\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"m x, m y\", \"1\": \"equivalent uniform moment factors\", \"2\": \"buckling resistance moment of theM b\"}, {\"0\": \"\", \"1\": \"for major axis and minor axis\", \"2\": \"cased column = Sxpb ≤ 1.5Mb for the\"}, {\"0\": \"\", \"1\": \"buckling respectively obtained from\", \"2\": \"uncased section. To determine pb, ry\"}, {\"0\": \"\", \"1\": \"Table 26 of BS 5950, reproduced as\", \"2\": \"should be taken as the greater of ry\"}, {\"0\": \"\", \"1\": \"Table 4.16\", \"2\": \"of the uncased section or\"}, {\"0\": \"m LT\", \"1\": \"equivalent uniform moment factor\", \"2\": \"0.2(B + 100) mm (Fig. 4.30).\"}, {\"0\": \"\", \"1\": \"for lateral torsional buckling\", \"2\": \"M x, M y maximum moment about the major\"}, {\"0\": \"\", \"1\": \"obtained from Table 18 of BS 5950,\", \"2\": \"and minor axes respectively\"}, {\"0\": \"\", \"1\": \"reproduced as Table 4.10\", \"2\": \"elastic modulus about the major andZ x, Z y\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"minor axes respectively.\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0215.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 216, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n195\nDesign of compression members\nPROPERTIES OF UC SECTION\nArea of UC section, Ag\n= 15 000 mm2\nRadius of gyration about x–x axis, rx = 136 mm\nRadius of gyration about y–y axis, ry = 77.5 mm\nElastic modulus about x–x axis, Zx\n= 1760 × 103 mm3\nElastic modulus about y–y axis, Zy\n= 587 × 103 mm3\nPlastic modulus about x–x axis, Sx\n= 1950 × 103 mm3\nDesign strength, py\n= 265 N/mm2\n(since T = 18.7 mm)\nEffective length, LE\n= 7 m\nLOCAL CAPACITY\nAxial load, Fc\n= 2000 kN\nApplied moment about x–x axis, Mx = 100 kN m\nApplied moment about y–y axis, My = 20 kN m\nShort strut capacity, Pcs, is given by\n    \nP\nA\nf A\np\np\ncs\ng\ncu\nc\ny\ny\n  \n  .\n=\n+\n\n\n\n\n0 25\n2\n15000\n  20  425\n265\n \n  .\n=\n+\n×\n×\n\n\n\n\n0 25\n265 = 4.878 × 106 N = 4878 kN\nMoment capacity of column about the x–x axis, Mcx, is given by\nMcx = pyZx = 265 × 1760 × 103 = 466.4 × 106 N mm = 466.4 kN m\n59\nB = 306.8\n59\n305 × 305 × 118 kg m–1 UC\n55\nD = 314.5\n55\nbc = 425\ndc = 425\nExample 4.14 Encased steel column resisting an axial load and bending\n(BS 5950)\nIn Example 4.11 it was found that a 305 × 305 × 118 kg/m UC column was incapable of resisting the design load and\nmoments below:\nDesign axial load\n= 2000 kN\nDesign moment about x–x axis = 100 kN m\nDesign moment about y–y axis = 20 kN m\nAssuming that the same column is now encased in concrete as show below, determine its suitability. The effective\nlength of the column about both axes is 7 m.\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\nry = 77; dc = 425", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0216.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 217, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n196\nExample 4.14 continued\nMoment capacity of column about the y–y axis, Mcy, is given by\nMcy = pyZy = 265 × 587 × 103 = 155.6 × 106 N mm = 155.6 kN m\n    \nF\nP\nM\nM\nM\nM\nc\ncs\nx\ncx\ny\ncy\n  \n  \n  \n  \n.   \n.\n+\n+\n=\n+\n+\n2000\n4878\n100\n466 4\n20\n155 6  = 0.41 + 0.21 + 0.13 = 0.75 < 1\nHence, the local capacity of the section is satisfactory.\nBUCKLING RESISTANCE\nRadii of gyration for cased section\nFor the cased section rx is the same as for the UC section = 136 mm\nFor the cased section ry = 0.2bc = 0.2 × 425 = 85 mm \u0001 0.2(B + 150) = 0.2(306.8 + 150) = 91.36 mm but not less\nthan that for the uncased section (= 77.5 mm)\nHence ry = 85 mm and rx = 136 mm.\nSlenderness ratio\n    \nλx\nE\nx\n=\n=\n=\n \n  \n  \n.\nL\nr\n7000\n136\n51 5\n    \nλy\nE\ny\n=\n=\n=\n \n  \n  \n.\nL\nr\n7000\n85\n82 4\nCompressive strength\nFor λx = 51.5 and py = 265 N/mm2 compressive strength, pc = 226 N/mm2 (Table 4.13). For λy = 82.4 and py =\n265 N/mm2 compressive strength, pc = 153 N/mm2 (Table 4.14).\nHence, pc is equal to 153 N/mm2.\nCompression resistance\nAg = 15 000 mm2\nAc = dcbc = 425 × 425 = 180 625 mm2\npy = 265 N/mm2\n(since T = 18.7 mm)\npc = 153 N/mm2\nfcu = 20 N/mm2\nCompression resistance of encased column, Pc, is given by\n    \nP\nA\nf A\np\np\nc\ng\ncu\nc\ny\nc\n  \n  .\n=\n+\n\n\n\n\n0 45\n15000\n  20  180625\n265\n \n  .\n=\n+\n×\n×\n\n\n\n\n0 45\n153 = 3.233 × 106 N = 3233 kN\nwhich is not greater than the short strut capacity, Pcs = 4878 kN (see above)\nOK\nBuckling resistance\nFor the uncased section,\n    \nλy\nE\ny\n  \n  \n.   \n=\n=\n=\nL\nr\n7000\n77 5\n90\n    \nλ y\nx   \n. /\n.   .\n=\n=\n90\n314 5 18 7\n5 4  ⇒ ν = 0.79\n(Table 4.8)\nλLT = uνλy  βw  = 0.851 × 0.79 × 90  1 = 61\nFrom Table 4.9, pb = 205 N/mm2\nMb = Sxpb = 1950 × 103 × 205 × 10−6 = 400 kN m\n\n[TABLES]\n[[{\"0\": \"Design in structural steelwork to BS 5950\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Example 4.14 continued\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Moment capacity of column about the y–y axis, Mcy, is given by\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"Mcy = pyZy = 265 × 587 × 103 = 155.6 × 106 N mm = 155.6 kN m\", \"2\": \"\"}, {\"0\": \"F PM MM M200010020cxy++=++\", \"1\": \"= 0.41 + 0.21 + 0.13 = 0.75 < 1\", \"2\": \"\"}, {\"0\": \"4878466 4155 6cscxcy\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Hence, the local capacity of the section is satisfactory.\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"BUCKLING RESISTANCE\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Radii of gyration for cased section\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"For the cased section rx is the same as for the UC section = 136 mm\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"For the cased section rythan that for the uncased section (= 77.5 mm)\", \"1\": \"= 0.2bc = 0.2 × 425 = 85 mm (cid:1) 0.2(B + 150) = 0.2(306.8 + 150) = 91.36 mm but not less\", \"2\": \"\"}, {\"0\": \"Hence r y = 85 mm and rx = 136 mm.\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Slenderness ratio\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"L rL r70007000EEλ===λ===     51 5     xy\", \"1\": \"82 4\", \"2\": \"\"}, {\"0\": \"13685xy\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Compressive strength\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"For  λ\", \"1\": \"x  =  51.5  and  py  =  265  N/mm2  compressive  strength,  pc  =  226  N/mm2  (Table  4.13).  For  λ\", \"2\": \"y  = 82.4  and  py  =\"}, {\"0\": \"265 N/mm2 compressive strength, pc = 153 N/mm2 (Table 4.14).\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Hence, pc is equal to 153 N/mm2.\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Compression resistance\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"= 15 000 mm2Ag\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Ac= dcbc = 425 × 425 = 180 625 mm2= 265 N/mm2(since T = 18.7 mm)\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"p y= 153 N/mm2\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"pc= 20 N/mm2\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"fcu\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"Compression resistance of encased column, P c, is given by\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"  0 45fAcuc=+P  A  pcgc\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"py\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"   ×  20  1806250 45=+ 15000\", \"1\": \"153 = 3.233 × 106 N = 3233 kN\", \"2\": \"\"}, {\"0\": \"265\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"which is not greater than the short strut capacity, Pcs = 4878 kN (see above)\", \"1\": \"OK\", \"2\": \"\"}, {\"0\": \"Buckling resistance\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"For the uncased section,\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"L r7000Eλ===      90y\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"77 5y\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"λ90y==    5 4  ⇒ ν = 0.79\", \"1\": \"(Table 4.8)\", \"2\": \"\"}, {\"0\": \"x314 5 18 7\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"βλLT = uνλw  = 0.851 × 0.79 × 90   1 = 61y\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"From Table 4.9, p b = 205 N/mm2\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"M b = Sx pb = 1950 × 103 × 205 × 10−6 = 400 kN m\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"196\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\n0.13 = 0; ν = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0217.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 218, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n197\nDesign of compression members\nFig. 4.32\n2c + T\n2c + t\nEffective bearing area\n4.9.7 DESIGN OF COLUMN BASEPLATES\nClause 4.13 gives guidance on the design of con-\ncentrically loaded column slab baseplates, which\ncovers most practical design situations. The plan\narea of the baseplate is calculated by assuming\na) the nominal bearing pressure between the\nbaseplate and support is uniform and\nb) the applied load acts over a portion of the\nbaseplate known as the effective area, the ex-\ntent of which for UB and UCs is as indicated\non Fig. 4.32.\nFor concrete foundations the bearing strength\nmay be taken as 0.6 times the characteristic cube\nstrength of the concrete base or the bedding mater-\nial (i.e. 0.6fcu), whichever is the lesser. The effective\narea of the baseplate, Abe, is then obtained from\n    \nAbe\naxial load\nbearing strength\n  \n=\n(4.42)\nIn determining the overall plan area of the plate\nallowance should be made for the presence of hold-\ning bolts.\nFor the cased section,\n    \nλy\nE\ny\n  \n  \n  \n.\n=\n=\n=\nL\nr\n7000\n85\n82 4\n    \nλy\nx   \n.\n. /\n.   .\n=\n=\n82 4\n314 5 18 7\n4 9 ⇒ ν = 0.82\n(Table 4.8)\nλLT = uνλy  βw  = 0.851 × 0.82 × 82.4  1 = 57.4\nFrom Table 4.9, pb = 213 N/mm2\nMb = Sxpb = 1950 × 103 × 213 × 10−6 = 415 kN m\nHence, Mb (for cased UC section = 415 kN m) < 1.5Mb (for uncased UC section = 1.5 × 400 = 600 kN m).\nChecking buckling resistance\n    \nF\nP\nm M\np Z\nm M\np Z\nc\nc\nx\nx\ny\nx\ny\ny\ny\ny\n  \n  \n+\n+\n  \n2000\n3233\n1\n100\n265\n1760\n10\n1\n20\n265\n587\n10\n3\n3\n  \n  \n  \n  \n  \n  \n  \n  \n+\n×\n×\n×\n+\n×\n×\n×\n−\n− = 0.62 + 0.21 + 0.13 = 0.96 < 1\nOK\n    \nF\nP\nm M\nM\nm M\np Z\nc\ncy\nLT\nx\nb\ny\ny\ny\ny\n  \n  \n+\n+\n  \n2000\n3233\n1\n100\n415\n1\n20\n265\n587\n10 3\n    \n  \n  \n  \n  \n+\n×\n+\n×\n×\n×\n− = 0.62 + 0.24 + 0.13 = 0.99 < 1\nOK\nHence, the section is now just adequate to resist the design axial load of 2000 kN and design moments about the\nx–x and y–y axes of 100 kN m and 20 kN m respectively.\nExample 4.14 continued\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"2c + t\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"2c + T\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}], [], []]\n\n[FORMULAS]\nν = 0; 0.13 = 0; 0.13 = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0218.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 219, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n198\nAREA OF BASEPLATE\nEffective area\n    \nA be\n2\naxial load\nbearing strength\nmm\n  \n  \n  \n.   \n  .\n  \n≥\n=\n×\n×\n=\n×\n3000\n10\n0 6\n30\n1 666\n10\n3\n5\nActual area\nAbe = (B + 2c)(D + 2c) − 2{(D − 2[T + c])([B + 2c] − [t + 2c])}\n1.666 × 105 = (306.8 + 2c)(314.5 + 2c) − 2{(314.5 − 2[18.7 + c])(306.8 − 11.9)} ⇒ c = 84.6 mm\nMinimum length of baseplate = D + 2c = 314.5 + 2 × 84.6 = 483.7 mm\nMinimum width of baseplate = B + 2c = 306.4 + 2 × 84.6 = 476 mm\nProvide 500 × 500 mm baseplate in grade S275 steel.\nBASEPLATE THICKNESS\nAssuming a baseplate thickness of less than 40 mm the design strength pyp = 265 N/mm2. The actual baseplate\nthickness, tp, is\ntp = c[3ω/pyp]0.5 = 84.6[3 × (0.6 × 30)/265]0.5 = 34.9 mm\nHence, a 500 mm × 500 mm × 35 mm thick baseplate in grade S275 steel should be suitable.\nThe required minimum baseplate thickness, tp,\nis given by\ntp = c[3ω/pyp]0.5\n(4.43)\nwhere\nc\nis the largest perpendicular distance from\nthe edge of the effective portion of the\nbaseplate to the face of the column\ncross-section (Fig. 4.32)\nω\npressure on the underside of the plate\nassuming a uniform distribution throughout\nthe effective portion, but ≤ 0.6fcu\npyp\ndesign strength of the baseplate which may\nbe taken from Table 4.3\nExample 4.15 Design of a steel column baseplate (BS 5950)\nDesign a baseplate for the axially loaded column shown below assuming it is supported on concrete of compression\ncharacteristic strength 30 N/mm2.\nAxial load = 3000 kN\nB = 306.8 mm\nD = 314.5 mm\nt p\n305 × 305 × 118 kg m–1 UC\nT = 18.7 mm\nb\nb\na\na\n\n[TABLES]\n[[{\"0\": \"aa\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\nc = 84; 84.6 = 483; .5 = 84; B = 306", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0219.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 220, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n199\nFloor systems for steel framed structures\n4.10 Floor systems for steel\nframed structures\nIn temporary steel framed structures such as car\nparks and Bailey bridges the floor deck can be\nformed from steel plates. In more permanent steel\nframed structures the floors generally comprise:\n• precast, prestressed concrete slabs\n• in-situ reinforced concrete slabs\n• composite metal deck floors.\nPrecast floors are normally manufactured using\nprestressed hollow core planks, which can easily\nspan up to 6– 8 m (Fig. 4.33(a)). The top surface\ncan be finished with a levelling screed or, if a com-\nposite floor is required, with an in-situ concrete\nstructural topping. This type of floor slab offers a\nnumber of advantages over other flooring systems\nincluding:\n• elimination of shuttering and propping\n• reduced floor depth by supporting the precast\nunits on shelf angles Fig. 4.33(b)\n• rapid construction since curing or strength devel-\nopment of the concrete is unnecessary.\nPrecast floors are generally designed to act non-\ncompositely with the supporting steel floor beams.\nNevertheless, over recent years, hollow core planks\nthat act compositely with the floor beams have been\ndeveloped and are slowly beginning to be speci-\nfied. A major drawback of precast concrete slabs,\nwhich restricts their use in many congested city\ncentre developments, is that the precast units are\nheavy and cranage may prove difficult. In such\nlocations, in-situ concrete slabs have invariably\nbeen found to be more practical.\nIn-situ reinforced concrete floor slabs can be\nformed using conventional removable shuttering\nand are normally designed to act compositely with\nthe steel floor beams. Composite action is achieved\nby welding steel studs to the top flange of the steel\nbeams and embedding the studs in the concrete\nwhen cast (Fig. 4.34). The studs prevent slippage\nand also enable shear stresses to be transferred be-\ntween the slab and supporting beams. This increases\nboth the strength and stiffnesses of the beams,\nthereby allowing significant reductions in construc-\ntion depth and weight of steel beams to be achieved\n(see Example 4.16). Composite construction not\nonly reduces frame loadings but also results in\nFig. 4.34 In-situ reinforced concrete slab.\nSteel studs\nwelded to top\nflange of steel\nbeam\nReinforced\nconcrete slab\nFig. 4.33 Precast concrete floor: (a) hollow core plank-section (b) precast concrete plank supported on shelf angles.\n(a)\n(b)\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"Floor systems for steel framed structures\"}, {\"0\": \"4.10 Floor systems for steel\", \"1\": \"Precast floors are generally designed to act non-\"}, {\"0\": \"framed structures\", \"1\": \"compositely with the supporting steel floor beams.\"}, {\"0\": \"\", \"1\": \"Nevertheless, over recent years, hollow core planks\"}, {\"0\": \"In  temporary  steel  framed  structures  such  as  car\", \"1\": \"that act compositely with the floor beams have been\"}, {\"0\": \"parks  and  Bailey  bridges  the  floor  deck  can  be\", \"1\": \"developed  and  are  slowly  beginning  to  be  speci-\"}, {\"0\": \"formed from steel plates. In more permanent steel\", \"1\": \"fied.  A  major  drawback  of  precast  concrete  slabs,\"}, {\"0\": \"framed structures the floors generally comprise:\", \"1\": \"which  restricts  their  use  in  many  congested  city\"}, {\"0\": \"• precast, prestressed concrete slabs\", \"1\": \"centre  developments,  is  that  the  precast  units  are\"}, {\"0\": \"•in-situ reinforced concrete slabs\", \"1\": \"heavy  and  cranage  may  prove  difficult.  In  such\"}, {\"0\": \"•composite metal deck floors.\", \"1\": \"locations, in-situ  concrete slabs  have invariably\"}, {\"0\": \"\", \"1\": \"been found to be more practical.\"}, {\"0\": \"Precast  floors  are  normally  manufactured  using\", \"1\": \"In-situ  reinforced  concrete  floor  slabs  can  be\"}, {\"0\": \"prestressed  hollow  core  planks,  which  can  easily\", \"1\": \"formed  using  conventional  removable  shuttering\"}, {\"0\": \"span up to 6– 8 m (Fig. 4.33(a)). The top surface\", \"1\": \"and are normally designed to act compositely with\"}, {\"0\": \"can be finished with a levelling screed or, if a com-\", \"1\": \"the steel floor beams. Composite action is achieved\"}, {\"0\": \"posite  floor  is  required,  with  an  in-situ  concrete\", \"1\": \"by welding steel studs to the top flange of the steel\"}, {\"0\": \"structural  topping.  This  type  of  floor  slab  offers  a\", \"1\": \"beams  and  embedding  the  studs  in  the  concrete\"}, {\"0\": \"number  of  advantages  over  other  flooring  systems\", \"1\": \"when  cast  (Fig.  4.34).  The  studs  prevent  slippage\"}, {\"0\": \"including:\", \"1\": \"and also enable shear stresses to be transferred be-\"}, {\"0\": \"\", \"1\": \"tween the slab and supporting beams. This increases\"}, {\"0\": \"•elimination of shuttering and propping\", \"1\": \"both  the  strength  and  stiffnesses  of  the  beams,\"}, {\"0\": \"•reduced  floor  depth  by  supporting  the  precast\", \"1\": \"thereby allowing significant reductions in construc-\"}, {\"0\": \"units on shelf angles Fig. 4.33( b)\", \"1\": \"tion depth and weight of steel beams to be achieved\"}, {\"0\": \"•rapid construction since curing or strength devel-\", \"1\": \"(see  Example  4.16 ).  Composite  construction  not\"}, {\"0\": \"opment of the concrete is unnecessary.\", \"1\": \"only  reduces  frame loadings  but  also  results in\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0220.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 221, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n200\nBeam A\nBeam B\n600\n300\n300\n250\n250\n300\n300\n8 m\nExample 4.16 Advantages of composite construction (BS 5950)\nTwo simply supported, solid steel beams 250 mm wide and 600 mm deep are required to span 8 m. Both beams are\nmanufactured using two smaller beams, each 250 mm wide and 300 mm deep, positioned one above the other. In\nbeam A the two smaller beams are not connected but act independently whereas in beam B they are fully joined and\nact together as a combined section.\n(a) Assuming the permissible strength of steel is 165 N/mm2, determine the maximum uniformly distributed load that\nBeam A and Beam B can support.\n(b) Calculate the mid-span deflections of both beams assuming that they are subjected to a uniformly distributed\nload of 140 kN/m.\n(A) LOAD CAPACITY\n(i) Beam A\nElastic modulus of single 250 × 300 mm beam, Zs, is\n    \nZ\nI\ny\nbd\nd\ns\ns\n3\n3\n/2\nmm\n=\n=\n=\n×\n=\n×\n   \n/\n  \n  \n  .\n  \n12\n250\n300\n6\n3 75\n10\n2\n6\nCombined elastic modulus of two 250 × 300 mm beams acting separately, Zc = 2Zs = 7.5 × 106 mm3\nMoment capacity of combined section, M, is\nM = σZc = 165 × 7.5 × 106 = 1.2375 × 109 N mm\nHence, load carrying capacity of Beam A, ωA, is\n      \nM  \n  \n(   \n)\n=\n=\n×\nω\nω\nb2\n3 2\n8\n8\n10\n8\nA\n = 1.2375 × 109 N mm\n⇒ ωA = 154.7 N/mm = 154.7 kN/m\n(ii) Beam B\nIn beam B the two smaller sections act together and behave like a beam 250 mm wide and 600 mm deep. The elastic\nmodulus of the combined section, Z, is\n    \nZ\nI\ny\nbd\nd\n    \n/\n  \n  \n  \n  \n=\n=\n=\n×\n=\n×\n3\n3\n/2\nmm\n12\n250\n600\n6\n15\n10\n2\n6\nBending strength of beam, M = σZ = 165 × 15 × 106 = 2.475 × 106 N mm\nHence, load carrying capacity of Beam B, ωB, is\n      \nM  \n  \n(   \n)   .\n  \n=\n=\n×\n=\n×\nω\nω\nb2\n3 2\n9\n8\n8\n10\n8\n2 475\n10\nB\nN mm\n⇒ ωB = 309.4 N/mm = 309.4 kN/m\n\n[TABLES]\n[[{\"0\": \"\"}, {\"0\": \"\"}], [{\"0\": \"250250\"}, {\"0\": \"\"}, {\"0\": \"\"}], [], []]\n\n[FORMULAS]\n106 = 1; ωA = 154; ωB = 309", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0221.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 222, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n201\nFloor systems for steel framed structures\nsmaller and hence cheaper foundations. One draw-\nback of this form of construction is that shuttering\nis needed and the slab propped until the concrete\ndevelops adequate strength.\nA development of this approach, which can elimi-\nnate the need for tensile steel reinforcement and\npropping of the slab during construction, is to use\nprofiled metal decking as permanent shuttering\n(Fig. 4.35). Three common types of metal decking\nused in composite slab construction are shown\nin Fig. 4.36. The metal decking is light and easy\nto work, which makes for simple and rapid con-\nstruction. This system is most efficient for slab\nspans of between 3–4 m, and beam spans of up to\naround 12 m. Where longer spans and/or higher\nFig. 4.35 Composite metal deck floor.\nloads are to be supported, the steel beams may be\nsubstituted with cellular beams or stub-girders (Fig.\n4.37). In structures where there is a need to reduce\nthe depth of floor construction, for example tall\nbuildings, the Slimflor system developed by British\nSteel can be used. Floor spans are limited to 7.5 m\nusing this system, which utilises stiff steel beams,\nfabricated from universal column sections welded\nto a steel plate, and deep metal decking which\nrests on the bottom flange of the beam (Fig. 4.38).\nNowadays, almost all composite floors in steel\nframed buildings are formed using profiled metal\ndecking and the purpose of the following sections\nis to discuss the design of (a) composite slabs and\n(b) composite beams.\n4.10.1 COMPOSITE SLABS\nComposite slabs (i.e. metal decking plus concrete)\nare normally designed to BS 5950: Part 4. Although\nexplicit procedures are given in the standard, these\ntend to be overly conservative when compared with\nthe results of full-scale tests. Therefore, designers\nmostly rely on load/span tables produced by metal\ndeck manufacturers in order to determine the thick-\nness of slab and mesh reinforcement required for a\ngiven floor arrangement, fire rating, method of\nconstruction, etc. Table 4.17 shows an example of\na typical load/span table available from one sup-\nplier of metal decking. Concrete grades in the range\n30–40 N/mm2 are common. Slab depths may vary\nbetween 100 and 200 mm. Example 4.20 illustrates\nthe use of this table.\n4.10.2 COMPOSITE BEAMS\nOnce the composite slab has been designed, design\nof the primary and secondary composite beams (i.e.\nsteel beams plus slab) can begin. This is normally\nExample 4.16 continued\n(B) DEFLECTION\nMid-span deflection of beam A (for a notional load of ω = 140 kN/m), δA, is\n      \nδ\nω\nA\nmm\n=\n=\n×\n×\n×\n×\n×\n×\n×\n=\n \n  \n  \n  (   \n)\n  \n  \n ( .\n  \n)  \n.  \n5\n384\n5\n140\n8\n10\n384\n205\n10\n2 5 625\n10\n32 3\n4\n3 4\n3\n8\nb\nEI\nMid-span deflection of beam B, δB, is\n      \nδ\nω\nB\nmm\n=\n=\n×\n×\n×\n×\n×\n×\n×\n=\n \n  \n  \n  (   \n)\n  \n  \n .   \n   \n5\n384\n5\n140\n8\n10\n384\n205\n10\n4 5\n10\n8\n4\n3 4\n3\n9\nb\nEI\nHence, it can be seen that the load capacity has been doubled and the stiffness quadrupled by connecting the two\nbeams, a fact which will generally be found to hold for other composite sections.\n\n[TABLES]\n[[{\"0\": \"beams, a fact which will generally be found to hold for other composite sections.\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"loads are to be supported, the steel beams may be\"}, {\"0\": \"\", \"1\": \"substituted with cellular beams or stub-girders (Fig.\"}, {\"0\": \"\", \"1\": \"4.37 ). In structures where there is a need to reduce\"}, {\"0\": \"\", \"1\": \"the  depth  of  floor  construction,  for  example  tall\"}, {\"0\": \"\", \"1\": \"buildings, the Slimflor system developed by British\"}, {\"0\": \"\", \"1\": \"Steel can be used. Floor spans are limited to 7.5 m\"}, {\"0\": \"\", \"1\": \"using  this  system,  which  utilises  stiff  steel  beams,\"}, {\"0\": \"\", \"1\": \"fabricated  from  universal  column  sections  welded\"}, {\"0\": \"\", \"1\": \"to  a  steel  plate,  and  deep  metal  decking  which\"}, {\"0\": \"\", \"1\": \"rests on the bottom flange of the beam (Fig. 4.38).\"}, {\"0\": \"\", \"1\": \"Nowadays,  almost  all  composite  floors  in  steel\"}, {\"0\": \"\", \"1\": \"framed  buildings  are  formed  using  profiled  metal\"}, {\"0\": \"\", \"1\": \"decking  and  the  purpose  of  the  following  sections\"}, {\"0\": \"\", \"1\": \"is to discuss the design of (a) composite slabs and\"}, {\"0\": \"\", \"1\": \"(b) composite beams.\"}, {\"0\": \"\", \"1\": \"4.10.1 COMPOSITE SLABS\"}, {\"0\": \"\", \"1\": \"Composite slabs (i.e. metal decking plus concrete)\"}, {\"0\": \"\", \"1\": \"are normally designed to BS 5950: Part 4. Although\"}, {\"0\": \"Fig. 4.35 Composite metal deck floor.\", \"1\": \"explicit procedures are given in the standard, these\"}, {\"0\": \"\", \"1\": \"tend to be overly conservative when compared with\"}, {\"0\": \"\", \"1\": \"the  results  of  full-scale  tests.  Therefore,  designers\"}, {\"0\": \"smaller and hence cheaper foundations. One draw-\", \"1\": \"mostly rely on load/span tables produced by metal\"}, {\"0\": \"back of this form of construction is that shuttering\", \"1\": \"deck manufacturers in order to determine the thick-\"}, {\"0\": \"is needed and the slab propped until the concrete\", \"1\": \"ness of slab and mesh reinforcement required for a\"}, {\"0\": \"develops adequate strength.\", \"1\": \"given  floor  arrangement,  fire  rating,  method  of\"}, {\"0\": \"A development of this approach, which can elimi-\", \"1\": \"construction, etc. Table 4.17 shows an example of\"}, {\"0\": \"nate  the  need  for  tensile  steel  reinforcement  and\", \"1\": \"a  typical  load/span  table  available  from  one  sup-\"}, {\"0\": \"propping of the slab during construction, is to use\", \"1\": \"plier of metal decking. Concrete grades in the range\"}, {\"0\": \"profiled  metal  decking  as  permanent  shuttering\", \"1\": \"30–40 N/mm2 are common. Slab depths may vary\"}, {\"0\": \"(Fig. 4.35 ). Three common types of metal decking\", \"1\": \"between 100 and 200 mm. Example 4.20 illustrates\"}, {\"0\": \"used in  composite  slab  construction  are  shown\", \"1\": \"the use of this table.\"}, {\"0\": \"in  Fig.  4.36.  The  metal  decking  is  light  and  easy\", \"1\": \"\"}, {\"0\": \"to  work,  which  makes  for  simple  and  rapid  con-\", \"1\": \"4.10.2 COMPOSITE BEAMS\"}, {\"0\": \"struction.  This  system is  most  efficient  for  slab\", \"1\": \"Once the composite slab has been designed, design\"}, {\"0\": \"spans of between 3–4 m, and beam spans of up to\", \"1\": \"of the primary and secondary composite beams (i.e.\"}, {\"0\": \"around  12  m.  Where  longer  spans  and/or  higher\", \"1\": \"steel beams plus slab) can begin. This is normally\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0222.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 223, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n202\n112\n26\n136\n15\n55\n70\nCover width 900 mm\n(a)\n40\n112.5\n152.5\n137.5\nCover width 610 mm\nCover width 600 mm\nTrough shear-\nbond clip\n56\n210\n87.5\n21\n38\n(b)\n(c)\n51\n15\nFig. 4.36 Steel decking: (a) re-entrant, (b) trapezoidal (c) deep deck.\nPrimary\ncomposite\nbeam\nSecondary composite beams\ncarried out in accordance with the recommenda-\ntions in Part 3: Section 3.1 of BS 5950, hereafter\nreferred to as BS 5950–3.1, and involves the follow-\ning steps:\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0223.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 224, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n203\nConcrete slab\nVoids in web reduce self-weight and\nallow easy service distribution\n(a)\nStud connectors\nVoids\nSteel decking\nBottom chord\nStub girders\nSecondary beam\n(b)\nFloor systems for steel framed structures\nFig. 4.37 Long span structures: (a) cellular beams (b) stub-girder.\nFig. 4.38 Slimflor system.\n1. Determine the effective breadth of the concrete\nslab.\n2. Calculate the moment capacity of the section.\n3. Evaluate the shear capacity of the section.\n4. Design the shear connectors.\n5. Assess the longitudinal shear capacity of the\nsection.\n6. Check deflection.\nEach of these steps is explained below by refer-\nence to simply supported secondary beams of class\n1 plastic UB section supporting a solid slab.\n4.10.2.1 Effective breadth of concrete slab, Be\nAccording to BS 5950–3.1, the effective breadth of\nconcrete slab, Be, acting compositely with simply\nsupported beams of length L should be taken as\nthe lesser of L/4 and the sum of the effective\nbreadths, be, of the portions of flange each side of\nthe centreline of the steel beam (Fig. 4.39).\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}], [{\"0\": \"Stud connectorsVoidsSteel decking\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}, {\"0\": \"Bottom chordStub girdersSecondary beam\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\", \"15\": \"\", \"16\": \"\", \"17\": \"\", \"18\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0224.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 225, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n204\nTable 4.17\nTypical load/span table for design of unpropped double span slab and deck made of\nnormal weight grade 35 concrete (PMF, CF70, Corus).\nMaximum span (m)\nDeck thickness (mm)\nSlab\n0.9\n1.2\nFire\nDepth\nMesh\nTotal applied load (kN/m2)\nRating\n(mm)\n3.50\n5.00\n10.0\n3.50\n5.00\n10.0\n1 hr\n125\nA142\n3.2\n3.2\n2.8\n4.0\n3.7\n3.0\n11/2hr\n135\nA193\n3.1\n3.1\n2.7\n3.9\n3.5\n2.8\n2 hr\n150\nA193\n2.9\n2.9\n2.5\n3.5\n3.2\n2.6\n200\nA393\n2.5\n2.5\n2.5\n3.5\n3.5\n3.5\n250\nA393\n2.1\n2.1\n2.1\n3.2\n3.2\n3.2\nb e\nb e\nBe is the lesser of:\n(a) beam span/4\n(b) 2be\nFig. 4.39 Effective breadth.\nFig. 4.40 Plastic neutral axis positions.\nCase 1\nCase 2\nCase 3\nd) The ultimate moment capacity of the com-\nposite section is independent of the method of\nconstruction i.e. propped or unpropped.\nThe moment capacity of a composite section de-\npends upon where the plastic neutral axis falls within\nthe section. Three outcomes are possible, namely:\n1. plastic neutral axis occurs within the concrete\nflange;\n2. plastic neutral axis occurs within the steel flange;\n3. plastic neutral axis occurs within the web\n(Fig. 4.40).\nOnly the first two cases will be discussed here.\n(i) Case 1: Rc > Rs. Figure 4.41 shows the stress\ndistribution in a typical composite beam section\nwhen the plastic neutral axis lies within the con-\ncrete slab.\nSince there is no resultant axial force on the\nsection, the force in the concrete, R c′, must equal\nthe force in the steel beam, Rs. Hence\n4.10.2.2 Moment capacity\nIn the analysis of a composite section to determine\nits moment capacity the following assumptions can\nbe made:\na) The stress block for concrete in compression at\nultimate conditions is rectangular with a design\nstress of 0.45fcu\nb) The stress block for steel in both tension and\ncompression at ultimate conditions is rectangu-\nlar with a design stress equal to py\nc) The tensile strength of the concrete is zero\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0225.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 226, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n205\nFig. 4.41 Stress distribution when plastic neutral axis lies within concrete flange.\nFloor systems for steel framed structures\nD s\nD\nA\nR s\np y\nBe\ny p\nR′c\n0.45f cu\nplastic neutral\naxis\nR c′ = Rs\n(4.44)\nwhere\nR c′ = design stress in concrete × area of\nconcrete in compression\n= (0.45fcu)(Beyp)\n(4.45)\nRs = steel design strength × area of steel section\n= pyA\nThe maximum allowable force in the concrete\nflange, R c, is given by\nRc = design stress in concrete ×\narea of concrete flange\n= (0.45fcu)(BeDs)\n(4.46)\nEliminating 0.45fcu from equations 4.45 and 4.46\ngives\nR′c = \n    \nR y\nD\nc\np\ns\n(4.47)\nCombining equations 4.44 and 4.47 and rearrang-\ning obtains the following expression for depth of\nthe plastic neutral axis, yp\n    \ny\nR\nR D\nD\np\ns\nc\ns\ns\n \n  \n \n=\n≤\n(4.48)\nTaking moments about the top of the concrete\nflange and substituting for yp, the moment capacity\nof the section, Mc, is given by\n    \nM\nR\nD\nD\nR y\nc\ns\ns\nc\np\n2\n2\n=\n+\n\n\n\n−\n′\n \n \n  \n    \n=\n+\n\n\n\n−\n′\n \n \n  \nR\nD\nD\nR R\nR D\ns\ns\nc\ns\nc\ns\n2\n2\n    \n=\n+\n\n\n\n−\n \n \n  \nR\nD\nD\nR D\nR\ns\ns\ns\n2\ns\nc\n2\n2\n(4.49)\nwhere\nD = depth of steel section\nDs = depth of concrete flange\nNote that the equation for Mc does not involve\nyp. Nonetheless it should be remembered that this\nequation may only be used to calculate Mc pro-\nvided that yp < Ds. This condition can be checked\neither via equation 4.48 or, since Rc′ = Rs (equation\n4.44) and Rc > R c′, by checking that Rc > Rs. Clearly\nif Rc < Rs, it follows that the plastic neutral axis\noccurs within the steel beam.\n(ii) Case 2: Rc < Rs. Figure 4.42a shows the stress\ndistribution in the section when the plastic neutral\naxis lies within the steel flange.\nBy equating horizontal forces, the depth of\nplastic neutral axis below the top of the steel flange,\ny, is obtained as follows\nRc + py(By) = Rs − py(By)\n    \n⇒\n=\n−\n   \n \ny\nR\nR\nBp\ns\nc\ny\n2\nResistance of the steel flange, Rf = py(BT ) ⇒ py\n= \n    \nR\nBT\nf\nSubstituting into the above expression for y gives\n    \ny\nR\nR\nR T\nT\n  \n \n \n=\n−\n≤\ns\nc\nf/\n \n2\n(4.50)\nThe expression for moment capacity is derived\nusing the equivalent stress distribution shown in\nFig. 4.42(b). Taking moments about the top of the\nsteel flange, the moment capacity of the section is\ngiven by\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"0.45fcu\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0226.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 227, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n206\nD s\nD\nR s − p yBy\np y\nBe\nR c\nT\nB\nR s\ny\np y\nD/2\n2p y\nR c\n(a)\n(b)\nFig. 4.42 Stress distribution when plastic neutral axis lies within steel flange.\nFig. 4.43 Slab thickness and depth of metal decking.\nDs\nSection X−X\nX\nX\nD p\nOmit from\nconsideration\nProfiled\nmetal\ndecking\n    \nM\nR D\nR D\np By y\nc\ns\nc\ns\ny\n2\n2\n  \n  \n  (\n)\n=\n+\n−2\n2\n    \n=\n+\n−\n \n  \n  \nR D\nR D\np By\ns\nc\ns\ny\n2\n2\n2\nSubstituting for py and y and simplifying gives\n    \nM\nR D\nR D\nR\nR\nR\nT\nc\ns\nc\ns\ns\nc\nf\n2\n2\n=\n+\n−\n \n  \n  (\n – \n)2\n4\n(4.51)\n(iii) Moment capacity of composite beam\nincorporating metal decking. The moment\ncapacity of composite beams incorporating profiled\nmetal decking is given by the following:\nCase 1: Plastic neutral axis is in the concrete flange\n    \nM\nR\nD\nD\nR\nR\nD\nD\nc\ns\ns\ns\nc\ns\np\n2\n2\n  \n  \n  \n  \n=\n+\n−\n−\n\n\n\n\n\n\n\n\n\n\n\n\n(4.52)\nCase 2: Plastic neutral axis is in the steel flange\n    \nM\nR D\nR\nD\nD\nR\nR\nR\nT\nc\ns\nc\ns\np\ns\nc\nf\n2\n2\n4\n  \n  \n  \n  (\n  \n)\n=\n+\n+\n\n\n\n−\n−\n2\n(4.53)\nThese expressions are derived in the same way\nas for beams incorporating solid slabs but further\nassume that (a) the ribs of the metal decking run\nperpendicular to the beams and (b) the concrete\nwithin the depth of the ribs is ignored (Fig. 4.43).\nThe stress distributions used to derive equations\n4.52 and 4.53 are shown in Fig. 4.44. Note that\nthe symbols in these equations are as previously\ndefined except for Rc, which is given by\nRc = 0.45fcuBe(Ds − Dp)\n(4.54)\nwhere Dp is the overall depth of the profiled metal\ndecking (Fig. 4.43).\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Rc\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"2pyD/2\", \"4\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"Rc\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"pyRs\", \"3\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0227.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 228, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n207\nD s\nD\nR s\np y\nB e\ny p\nR′c\n0.45f cu\nPlastic\nneutral\naxes\n(a)\np y\n(b)\nR s\ny\nD/2\nDs − Dp\n0.45f cu\nR c\n2p y\nFig. 4.44 Stress distributions in composite beams incorporating profiled metal decking: (a) plastic neutral axis is within the\nconcrete flange (b) plastic neutral axis is within the steel flange.\nFloor systems for steel framed structures\nShear studs are available in a range of diameters\nand lengths as indicated in Table 4.18. The 19 mm\ndiameter by 100 mm high stud is by far the most\ncommon in buildings. In slabs comprising profiled\nmetal decking and concrete, the heights of the studs\nshould be at least 35 mm greater than the overall\ndepth of the decking. Also, the centre-to-centre\ndistance between studs along the beam should lie\nbetween 5φ and 600 mm or 4Ds if smaller, where\nφ is the shank diameter and Ds the depth of the\nconcrete slab. Some of the other code recommen-\ndations governing the minimum size, transverse\nspacing and edge distances of shear connectors are\nshown in Fig. 4.45.\n(ii) Design procedure. The shear strength of\nheaded studs can be determined using standard\npush-out specimens consisting of a short section\nof beam and slab connected by two or four studs.\nTable 4.18 gives the characteristic resistances, Qk,\nof headed studs embedded in a solid slab of\nnormal weight concrete. For positive (i.e. sagging)\nmoments, the design strength, Qp, should be\ntaken as\nQp = 0.8Qk\n(4.56)\nA limit of 80% of the static capacity of the shear\nconnector is deemed necessary for design to ensure\nthat full composite action is achieved between the\nslab and the beam.\nThe capacity of headed studs in composite slabs\nwith the ribs running perpendicular to the beam\nshould be taken as their capacity in a solid slab\n4.10.2.3 Shear capacity\nAccording to BS 5950–3.1, the steel beam should\nbe capable of resisting the whole of the vertical\nshear force, Fv. As discussed in 4.8.5, the shear\ncapacity, Pv, of a rolled I-section is given by\nPv = 0.6pytD\nLike BS 5950: Part 1, BS 5950–3.1 recommends\nthat where the co-existent shear force exceeds 0.5Pv\nthe moment capacity of the section, Mcv, should be\nreduced in accordance with the following:\nMcv = Mc − (Mc − Mf)(2Fv/Pv − 1)2\n(4.55)\nwhere\nMf\nis the plastic moment capacity of that part\nof the section remaining after deduction of\nthe shear area Av defined in Part 1 of\nBS 5950\nPv\nis the lesser of the shear capacity and the\nshear buckling resistance, both determined\nfrom Part 1 of BS 5950\n4.10.2.4 Shear connectors\n(i) Headed studs. For the steel beams and slab\nto act compositely and also to prevent separation\nof the two elements under load, they must be struc-\nturally tied. This is normally achieved by providing\nshear connectors in the form of headed studs as\nshown in Fig. 4.45. The shear studs are usually\nwelded to the steel beams through the metal\ndecking.\n\n[TABLES]\n[[{\"0\": \"Floor systems for steel framed structures\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"0.45fcu\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Rc\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"2pyD/2\"}], [{\"0\": \"yp\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0228.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 229, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n208\nFig. 4.45 Geometrical requirements for placing of studs (CIRIA Report 99).\n3ø h\n50 mm\nø\n50 mm\n20 mm\nD p\nD s\n35 mm\n15 mm\n0.4ø\n1.5ø\n4ø\n25 mm or 1.25ø\nTable 4.18\nCharacteristic resistance, Qk, of headed studs in normal\nweight concrete (Table 5, BS 5950–3.1)\nShank diameter\nHeight\nCharacteristic strength (N/mm2)\n(mm)\n(mm)\n25\n30\n35\n40\n25\n100\n146\n154\n161\n168 kN\n22\n100\n119\n126\n132\n139 kN\n19\n100\n95\n100\n104\n109 kN\n19\n75\n82\n87\n91\n96 kN\n16\n75\n70\n74\n78\n82 kN\n13\n65\n44\n47\n49\n52 kN\nFor full shear connection, the total number of\nstuds, Np, required over half the span of a simply\nsupported beam in order to develop the positive\nmoment capacity of the section can be determined\nusing the following expression:\nNp = Fc/Qp\n(4.57)\nwhere\nFc = Apy (if plastic neutral axis lies in the\nconcrete flange)\nFc = 0.45fcuBeDs (if plastic neutral axis lies in the\nsteel beam)\nQp = design strength of shear studs = 0.8Q k\n(in solid slab) and kQ k (in a composite slab\nformed using ribbed profile sheeting aligned\nperpendicular to the beam)\nmultiplied by a reduction factor, k, given by the\nfollowing expressions:\nfor one stud per rib\nk = 0.85(br/Dp){(h/Dp) − 1} ≤ 1\nfor two studs per rib\nk = 0.6(br/Dp){(h/Dp) − 1} ≤ 0.8\nfor three or more studs per rib\nk = 0.5(br/Dp){(h/Dp) − 1} ≤ 0.6\nwhere\nbr\nbreadth of the concrete rib\nDp\noverall depth of the profiled steel sheet\nh\noverall height of the stud but not more than\n2Dp or Dp + 75 mm, although studs of\ngreater height may be used\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"ø\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}], [{\"0\": \"Design in structural steelwork to BS 5950\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"50 mm3øh\", \"1\": \"25 mm or 1.25ø\", \"2\": \"1.5ø0.4ø4ø\", \"3\": \"ø20 mm\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"15 mm\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"35 mmDp\", \"13\": \"\", \"14\": \"Ds\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"50 mm\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\", \"9\": \"\", \"10\": \"\", \"11\": \"\", \"12\": \"\", \"13\": \"\", \"14\": \"\"}], [], []]\n\n[FORMULAS]\nNp = Fc/Qp; Fc = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0229.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 230, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n209\nFloor systems for steel framed structures\nIn composite floors made with metal decking it\nmay not always be possible to provide full shear con-\nnection between the beam and the slab because\nthe top flange of the beam may be too narrow\nand/or the spacing of the ribs may be too great to\naccommodate the total number of studs required. In\nthis case, the reader should refer to BS 5950–3.1:\nAppendix B, which gives alternative expressions for\nmoment capacity of composite sections with partial\nshear connection.\n(A) PLASTIC NEUTRAL AXIS OF COMPOSITE SECTION\nResistance of the concrete flange, Rc, is\nRc = (0.45fcu)BeDs = (0.45 × 35)1500 × 130 × 10−3 = 3071.25 kN\nFrom Table 4.3, since T(= 16 mm) ≤ 16 mm and steel grade is S275, design strength of beam, py = 275 N/mm2.\nResistance of steel beam, Rs, is\nRs = Apy = 95 × 102 × 275 × 10−3 = 2612.5 kN\nSince Rc > Rs, the plastic neutral axis falls within the concrete slab. Confirm this by calculating yp:\n    \ny\nAp\nf\nB\np\ny\ncu\ne\n(0.45\n  \n)\n  \n.   \n.\n  \n  \n=\n=\n×\n×\n×\n2612 5\n10\n0 45\n35\n1500\n3\n = 110.6 mm < Ds\nOK\n(B) MOMENT CAPACITY\nSince yp < Ds use equation 4.49 to calculate the moment capacity of the section, Mc. Hence\n    \nM\nAp\nD\nD\nR\nR\nD\nc\ny\ns\ns\nc\ns\n2\n2\nkN m\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−\n2612 5\n10 130\n412 8\n2\n2612 5\n3071 25\n130\n2\n10\n734 4\n3\n6\nFig. 4.46\nExample 4.17 Moment capacity of a composite beam (BS 5950)\nDetermine the moment capacity of the section shown in Fig. 4.46 assuming the UB is of grade S275 steel and the\ncharacteristic strength of the concrete is 35 N/mm2.\n1500 mm\n130 mm\n406 × 178 × 74 UB\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"Floor systems for steel framed structures\"}, {\"0\": \"In  composite  floors  made  with  metal  decking  it\", \"1\": \"accommodate the total number of studs required. In\"}, {\"0\": \"may not always be possible to provide full shear con-\", \"1\": \"this case, the reader should refer to BS 5950 –3.1:\"}, {\"0\": \"nection  between  the  beam  and  the  slab  because\", \"1\": \"Appendix B, which gives alternative expressions for\"}, {\"0\": \"the  top  flange  of  the  beam  may  be  too  narrow\", \"1\": \"moment capacity of composite sections with partial\"}, {\"0\": \"and/or the spacing of the ribs may be too great to\", \"1\": \"shear connection.\"}, {\"0\": \"Example 4.17 Moment capacity of a composite beam (BS 5950)\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"Determine the moment capacity of the section shown in Fig. 4.46 assuming the UB is of grade S275 steel and the\"}, {\"0\": \"characteristic strength of the concrete is 35 N/mm2.\", \"1\": \"\"}], [], []]\n\n[FORMULAS]\n3 = 3071", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0230.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 231, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n210\nExample 4.18 Moment capacity of a composite beam (BS 5950)\nRepeat Example 4.17 assuming the beam is made of grade S355 steel. Also, design the shear connectors assuming the\nbeam is 6 m long and that full composite action is to be provided.\nPLASTIC NEUTRAL AXIS OF COMPOSITE SECTION\nAs before, the resistance of the concrete flange, Rc, is\nRc = (0.45fcu)Ac = (0.45 × 35)1500 × 130 × 10−3 = 3071.25 kN\nFrom Table 4.3, since T (= 16 mm) ≤ 16 mm and steel grade is S355, design strength of beam, py = 355 N/mm2.\nResistance of steel beam, R s, is\nRs = Apy = 95 × 102 × 355 × 10−3 = 3372.5 kN\nSince Rc < Rs, the plastic neutral axis will lie within the steel beam. Confirm this by calculating y:\n    \ny\nR\nR\nBp\n  \n  \n  \n.   \n  \n.\n  \n  \n.   \n  .  \n=\n−\n=\n×\n−\n×\n×\n×\n=\ns\nc\ny\n2\nmm\n3372 5\n10\n3071 25\n10\n2\n179 7\n355\n2 4\n3\n3\nOK\nMOMENT CAPACITY\nSince y < T use equation 4.51 to calculate moment capacity of the section, Mc.\nResistance of steel flange, Rf = BTpy = 179.7 × 16 × 355 = 1.02 × 106 N\nMoment capacity of composite section, Mc, is\n    \nM\nR D\nR D\nR\nR\nR\nT\nc\ns\nc\ns\ns\nc\nf\n2\n2\n  \n  \n  (\n  \n)\n=\n+\n−\n−\n2\n4\n  \n=\n×\n+\n×\n−\n−\n×\n×\n \n.   \n.   \n.\n  \n  ((\n.   \n.\n)  \n)\n.\n  \n3372 5\n10 412 8\n2\n3071 25\n10 130\n2\n3372 5\n3071 25\n10\n1 02\n10\n16\n4\n3\n3\n3 2\n6\n= 895.4 × 106 N mm = 895.4 kN m\nSHEAR STUDS\nFrom Table 4.18, characteristic resistance, Qk, of headed studs 19 mm diameter × 100 mm high embedded in grade\n35 concrete is 104 kN.\nDesign strength of studs under positive moment, Qp, is given by\nQp = 0.8Qk = 0.8 × 104 = 83.2 kN\nNumber of studs required \n    \n=\n=\n≥\n \n  \n.\n.\n \nR\nQ\nc\np\n 36.9\n3071 25\n83 2\nProvide 38 studs, evenly arranged in pairs, in each half span of beam as shown.\n225\nX\nX\n3 m\n19 pairs of studs @ 150 centres\nCentre line of beam\nSection X–X\n75\n\n[TABLES]\n[[{\"0\": \"Design in structural steelwork to BS 5950\"}, {\"0\": \"Example 4.18 Moment capacity of a composite beam (BS 5950)\"}, {\"0\": \"Repeat Example 4.17 assuming the beam is made of grade S355 steel. Also, design the shear connectors assuming the\"}, {\"0\": \"beam is 6 m long and that full composite action is to be provided.\"}, {\"0\": \"PLASTIC NEUTRAL AXIS OF COMPOSITE SECTION\"}, {\"0\": \"As before, the resistance of the concrete flange, Rc, is\"}, {\"0\": \"R c = (0.45fcu)A c = (0.45 × 35)1500 × 130 × 10−3 = 3071.25 kN\"}, {\"0\": \"From  Table  4.3,  since  T  (=  16  mm)  ≤  16  mm  and  steel  grade  is  S355,  design  strength  of  beam,  py  =  355  N/mm2.\"}, {\"0\": \"Resistance of steel beam, R s, is\"}, {\"0\": \"R s = Apy = 95 × 102 × 355 × 10−3 = 3372.5 kN\"}, {\"0\": \"Since Rc < R s, the plastic neutral axis will lie within the steel beam. Confirm this by calculating y:\"}, {\"0\": \"33−×−×3372 5103071 2510R  R      sc===\"}, {\"0\": \"2 4mmy       OK××22179 7355Bp    y\"}, {\"0\": \"MOMENT CAPACITY\"}, {\"0\": \"Since y < T use equation 4.51 to calculate moment capacity of the section, Mc.\"}, {\"0\": \"Resistance of steel flange, R f = BTpy = 179.7 × 16 × 355 = 1.02 × 106 N\"}, {\"0\": \"Moment capacity of composite section, Mc, is\"}, {\"0\": \"2−(  )DDRRTssc=+−\"}, {\"0\": \"MRRcsc224Rf\"}, {\"0\": \"32−×412 8130((3372 5  3071 25)  10)1633=×+×−\"}, {\"0\": \"3372 5  10  3071 25  10  6×221 02  104\"}, {\"0\": \"= 895.4 × 106 N mm = 895.4 kN m\"}, {\"0\": \"SHEAR STUDS\"}, {\"0\": \"From Table 4.18, characteristic resistance, Q k, of headed studs 19 mm diameter × 100 mm high embedded in grade35 concrete is 104 kN.\"}, {\"0\": \"Design strength of studs under positive moment, Q p, is given by\"}, {\"0\": \"Q p = 0.8Q k = 0.8 × 104 = 83.2 kN\"}, {\"0\": \"R Q3071 25c==≥Number of studs required      36.9\"}, {\"0\": \"83 2p\"}, {\"0\": \"Provide 38 studs, evenly arranged in pairs, in each half span of beam as shown.\"}], [], []]\n\n[FORMULAS]\n3 = 3071; 355 = 1; 104 = 83", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0231.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 232, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n211\n4.10.2.5 Longitudinal shear capacity\nA solid concrete slab, which is continuous over sup-\nports, will need to be reinforced with top and bottom\nsteel to resist the sagging and hogging moments due\nto the applied loading. Over beam supports, this steel\nwill also be effective in transferring longitudinal forces\nfrom the shear connectors to the slab without splitt-\ning the concrete. Design involves checking that the\napplied longitudinal shear force per unit length, ν, does\nnot exceed the shear resistance of the concrete, νr.\nThe total longitudinal shear force per unit length,\nν, is obtained using the following\nν = NQp/s\n(4.58)\nwhere\nN\nnumber of shear connectors in a group\ns\nlongitudinal spacing centre-to-centre of\ngroups of shear connectors\nQp\ndesign strength of shear connectors\nIn a solid slab, the concrete shear resistance, νr, is\nobtained using the following:\nνr = 0.7Asvfy + 0.03ηAcvfcu ≤ 0.8ηAcv    fcu\n(4.59)\nIn slabs with profiled steel sheeting, νr, is given by\nνr = 0.7Asv fy + 0.03ηAcvfcu + νp\n≤ 0.8ηAcv    fcu  + νp\n(4.60)\nwhere\nfcu\ncharacteristic strength of the concrete\n≤ 40 N/mm2\nη\n1.0 for normal weight concrete\nAcv\nmean cross-sectional area per unit length of\nthe beam of the concrete surface under\nconsideration\nAsv\ncross-sectional area, per unit length of the\nbeam, of the combined top and bottom\nreinforcement crossing the shear surface\n(Fig. 4.47)\nνp\ncontribution of the profiled steel decking.\nAssuming the decking is continuous across\nthe top flange of the steel beam and that the\nribs are perpendicular to the span of the\nbeam, νp is given by\nνp = tppyp\n(4.61)\nin which\ntp\nthickness of the steel decking\npyp\ndesign strength of the steel decking obtained\neither from Part 4 of BS 5950 or\nmanufacturer’s literature\n4.10.2.6 Deflection\nThe deflection experienced by composite beams\nwill vary depending on the method of construction.\nThus where steel beams are unpropped during\nconstruction, the total deflection, δT, will be the\nsum of the dead load deflection, δD, due to the self\nweight of the slab and beam, based on the proper-\nties of the steel beam alone, plus the imposed load\ndeflection, δI, based on the properties of the com-\nposite section:\nδT = δD + δI\nFor propped construction the total deflection is\ncalculated assuming that the composite section sup-\nports both dead and imposed loads.\nThe mid-span deflection of a simply supported\nbeam of length L subjected to a uniformly distrib-\nuted load, ω, is given by\nFig. 4.47\n(a) Solid slab\na\na\nb\nb\nA b\nA t\ne\ne\nA t\nSheeting\n(b) Composite slab with the sheeting\nspanning perpendicular to the beam\nSurface\nAsv\na–a\nb–b\ne–e\nA b + A t\n2A b\nA t\nFloor systems for steel framed structures\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}], [], []]\n\n[FORMULAS]\nνr = 0; δT = δD + δI", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0232.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 233, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n212\n    \nδ\nω\n  \n= 5 L\nEI\n4\n384\nwhere\nE\nelastic modulus = 205 kN/mm2 for steel\nI\nsecond moment of area of the section\nDeflections of simply supported composite beams\nshould be calculated using the gross value of the\nsecond moment of area of the uncracked section,\nIg, determined using a modular ratio approach. The\nactual value of modular ratio, αe, depends on the\nproportions of the loading which are considered to\nbe long term and short term. Imposed loads on\nfloors should be assumed to be 2/3 short term and\n1/3 long term. On this basis, appropriate values of\nmodular ratio for normal weight and lightweight\nconcrete are 10 and 15, respectively. For com-\nposite beams with solid slabs Ig is given by\n    \nI\nI\nB D\nAB D D\nD\nA\nB D\ng\ns\ne\ns\n3\ne\ne\ns\ns\ne\ne\ns\n=\n+\n+\n+\n+\n \n  \n  \n(\n  \n)\n(\n  \n)\n12\n4\n2\nα\nα\n(4.62)\nwhere Is is the second moment of area of the steel\nsection.\nEquation 4.62 is derived assuming that the con-\ncrete flange is uncracked and unreinforced (see\nAppendix D). In slabs with profiled steel sheeting\nthe concrete within the depth of the ribs may con-\nservatively be omitted and Ig is then given by:\n    \nI\nI\nB D\nD\ng\ns\ne\ns\np\ne\n  \n  \n(\n  \n)\n=\n+\n−\n3\n12α\n    \n+\n−\n+\n+\n+\n−\n \n(\n  \n)(\n  \n  \n)\n(\n  \n[\n  \n])\nAB D\nD\nD\nD\nD\nA\nB D\nD\ne\ns\np\ns\np\ne\ne\ns\np\n2\n4\nα\n(4.63)\nThe deflections under unfactored imposed loads\nshould not exceed the limits recommended in\nBS 5590, as summarised in Table 4.5.\nExample 4.19 Design of a composite floor (BS 5950)\nSteel UBs at 3.5 m centres with 9 m simple span are to support a 150 mm deep concrete slab of characteristic\nstrength 30 N/mm2 (Fig. 4.48). If the imposed load is 4 kN/m2 and the weight of the partitions is 1 kN/m2\na) select a suitable UB section in grade S355 steel\nb) check the shear capacity\nc) determine the number and arrangement of 19 mm diameter × 100 mm long headed stud connectors required\nd) assuming the slab is reinforced in both faces with H8@150 centres (A = 335 mm2/m), check the longitudinal\nshear capacity of the concrete\ne) calculate the imposed load deflection of the beam.\nAssume that weight of the finishes, and ceiling and service loads are 1.2 kN/m2 and 1 kN/m2 respectively. The density\nof normal weight reinforced concrete, ρc, can be taken as 24 kN/m3.\nBEAM SELECTION\nDesign moment\nBeam span, L = 9 m\nSlab span, b = 3.5 m\nDesign load per beam, ω = 1.4(Dsρc + finishes + ceiling/services)b + 1.6(qk + partition loading)b\n= 1.4(0.15 × 24 + 1.2 + 1)3.5 + 1.6(4 + 1)3.5 = 56.4 kN/m\nDesign moment, \n    \nM\nL\n  \n  \n.   \n  \n \n=\n=\n×\n=\nω 2\n8\nkN m\n56 4\n9\n8\n571\n2\nFig. 4.48\n3.5 m\n3.5 m\n3.5 m\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\nb = 3; ω = 1; b\n= 1; 3.5 = 56", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0233.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 234, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n213\nFloor systems for steel framed structures\nEffective width of concrete slab\nBe is the lesser of beam span/4 (= 9000/4 = 2250 mm) and beam spacing (= 3500 mm)\n∴ Be = 2250 mm\nMoment capacity\nUsing trial and error, try 356 × 171 × 45 UB in grade S355 steel\nResistance of the concrete flange, Rc, is\nRc = (0.45fcu)BeDs = (0.45 × 30) 2250 × 150 × 10−3 = 4556.3 kN\nFrom Table 4.3, since T (= 9.7 mm) < 16 mm and steel grade is S355, design strength of beam, py = 355 N/mm2.\nResistance of steel beam, Rs, is\nRs = Apy = 57 × 102 × 355 × 10−3 = 2023.5 kN\nSince Rc > Rs, the plastic neutral axis will lie in the concrete slab. Confirm this by substituting into the following\nexpression for yp\n    \n⇒\n=\n=\n×\n×\n×\n=\n \n(0.45\nmm < \np\ny\ncu\ne\ns\ny\nAp\nf\nB\nD\n  \n)\n  \n.   \n.\n  \n  \n  \n.  \n2023 5\n10\n0 45\n30\n2250\n66 6\n3\nOK\nHence, moment capacity of composite section, Mc, is\n    \nM\nAp\nD\nD\nR\nR\nD\nc\ny\ns\ns\nc\ns\n2\nkN m\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−\n2\n2023 5\n10 150\n352\n2\n2023 5\n4556 3\n150\n2\n10\n592 3\n3\n6\nMc > M + Msw = 571 + (45 × 9.8 × 10−3)92/8 = 575.5 kN m\nOK\nSHEAR CAPACITY\nShear force, Fv = \n    \n1\n2\n1\n2\nωL  \n=\n × 56.4 × 9 = 253.8 kN\nShear resistance, Pv = 0.6pytD = 0.6 × 355 × 6.9 × 352 × 10−3 = 517 kN > Fv\nOK\nAt mid-span, Fv (= 0) < 0.5Pv (= 258 kN) and therefore the moment capacity of the section calculated above is valid.\nSHEAR CONNECTORS\nFrom Table 4.18, characteristic resistance, Qk, of headed studs 19 mm diameter × 100 mm high is 100 kN.\nDesign strength of shear connectors, Qp, is\nQp = 0.8Qk = 0.8 × 100 = 80 kN\nLongitudinal force that needs to be transferred, Fc, is 2023.5 kN\nNumber of studs required \n    \n=\n=\n≥\n \n  \n.  \nR\nQ\nc\np\n 25.3\n2023 5\n80\nProvide 26 studs, evenly arranged in pairs, in each half span of beam.\nSpacing \n  \n= 4500\n12  = 375 mm, say 350 mm centres\nExample 4.19 continued\n125\n12 × 350 = 4200 (26 studs)\n175\n4500 mm\nCentre line\nof beam\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"Floor systems for steel framed structures\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Example 4.19 continued\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Effective width of concrete slab\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"Be is the lesser of beam span/4 (= 9000/4 = 2250 mm) and beam spacing (= 3500 mm)\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"∴ Be = 2250 mm\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Moment capacity\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Using trial and error, try 356 × 171 × 45 UB in grade S355 steel\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Resistance of the concrete flange, Rc, is\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"R c = (0.45fcu)BeDs = (0.45 × 30) 2250 × 150 × 10−3 = 4556.3 kN\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"From  Table 4.3,  since  T  (=  9.7  mm)  <  16  mm  and  steel  grade  is  S355,  design  strength  of  beam,  py  =  355  N/mm2.\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Resistance of steel beam, R s, is\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"R s = Apy = 57 × 102 × 355 × 10−3 = 2023.5 kN\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"Since  Rc  >  R s,  the  plastic  neutral  axis  will  lie  in  the  concrete  slab.  Confirm  this  by  substituting  into  the  following\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"expression for yp\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"3×Ap2023 5  10y⇒===\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"y      66 6 p××fB0 45  30  2250(0.45)cue\", \"1\": \"mm < Ds\", \"2\": \"OK\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Hence, moment capacity of composite section, Mc, is\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \" R Rs 2  . .3522023 5DDs3=+−=×+−        2023 5  10150     MApDcys\", \"1\": \"150×\", \"2\": \" 10\", \"3\": \"−6\", \"4\": \"=\", \"5\": \"592 3\", \"6\": \"kN m\"}, {\"0\": \"224556 3c\", \"1\": \"2\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"OKMc > M + M sw = 571 + (45 × 9.8 × 10−3)92/8 = 575.5 kN m\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"SHEAR CAPACITY\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"1 21 2ωL  = × 56.4 × 9 = 253.8 kNShear force, Fv =\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Shear resistance, Pv = 0.6pyt D = 0.6 × 355 × 6.9 × 352 × 10−3 = 517 kN > Fv\", \"1\": \"OK\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"At mid-span, Fv (= 0) < 0.5Pv (= 258 kN) and therefore the moment capacity of the section calculated above is valid.\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"SHEAR CONNECTORS\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"From Table 4.18, characteristic resistance, Q k, of headed studs 19 mm diameter × 100 mm high is 100 kN.\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Design strength of shear connectors, Q p, is\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Q p = 0.8Q k = 0.8 × 100 = 80 kN\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Longitudinal force that needs to be transferred, Fc, is 2023.5 kN\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"R Q2023 5c==≥Number of studs required\", \"1\": \"25.3\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"80p\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Provide 26 studs, evenly arranged in pairs, in each half span of beam.\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"4500= Spacing  = 375 mm, say 350 mm centres\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"12\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}], [], []]\n\n[FORMULAS]\n3 = 4556; 92/8 = 575; 9 = 253; Pv = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0234.png"}