data/chunks/chunk_a71f59bd_11.jsonl

{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 195, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n174\nFig. 4.21 Standard case for buckling.\nM\nM\nM\nM\nL\nFig. 4.22 Load cases less susceptible to buckling.\nW\nL\nM = WL/8\n(a)\nM\nβM\nβM\nM\nL\n(b)\nbuckling resistance moment, Mb, is obtained from\nthe following:\nFor class 1 plastic or cass 2 compact sections\nMb = pbSx\n(4.22)\nFor class 3 semi-compact sections, Mb, is given\nby\nMb = pbZx\n(4.23)\nor alternatively\nMb = pbSx,eff\n(4.24)\nFor class 4 slender sections\nMb = pbZx,eff\n(4.25)\nwhere\nSx\nplastic modulus about the major axis\nZx\nelastic modulus about the major axis\nSx,eff\neffective plastic modulus about the major\naxis (see clause 3.5.6)\nZx,eff\neffective section modulus about the major\naxis (see clause 3.6.2)\nThis value of the buckling moment assumes the\nbeam is acted on by a uniform, single curvature\nmoment (Fig. 4.21), which is the most severe\narrangement in terms of lateral stability. However,\nwhere the beam is acted on by variable moments\n(Fig. 4.22(a)) or unequal end moments (Fig.\n4.22(b)), the maximum moment about the major\naxis, Mx, must satisfy the following conditions:\nMx ≤ Mb/mLT and Mx ≤ Mcx\n(4.26)\nwhere\nmLT\nequivalent uniform moment factor for\nlateral torsional buckling read from Table\n18 of BS 5950, reproduced as Table 4.10.\nFor the destabilising loading condition\nmLT = 1\nMcx\nmoment capacity of the section about the\nmajor axis\nExample 4.8 Design of a laterally unrestrained beam – rigorous method\n(BS 5950)\nRepeat Example 4.7 using the more rigorous method.\nAs previously noted the 305 × 305 × 137 UC section is class 1 plastic has a design strength, py = 265 N/mm2. From\nsteel tables (Appendix B), ry = 78.2 mm, D = 320.5 mm and T = 21.7 mm. The slenderness ratio, λ, is\n    \nλ  \n  \n.\n  \n.\n=\n=\n=\nL\nr\nE\ny\n10000\n78 2\n127 9\n    \nλ\nx   \n.\n. /\n.   .\n=\n=\n127 9\n320 5 21 7\n8 7\nFrom Table 4.8, ν = 0.678. Since the section is class 1 plastic, β w = 1.0 and the equivalent slenderness ratio, λLT, is\nλLT = uνλ  βw  = 0.9 × 0.678 × 127.9 ×   1 0.  = 78\n\n[TABLES]\n[[{\"0\": \"Design in structural steelwork to BS 5950\", \"1\": \"\"}, {\"0\": \"MM\", \"1\": \"W\"}, {\"0\": \"L\", \"1\": \"L\"}, {\"0\": \"MM\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"M = WL/8\"}, {\"0\": \"Fig. 4.21 Standard case for buckling.\", \"1\": \"(a)\"}, {\"0\": \"buckling resistance moment, M b, is obtained from\", \"1\": \"MβM\"}, {\"0\": \"the following:\", \"1\": \"\"}, {\"0\": \"For class 1 plastic or cass 2 compact sections\", \"1\": \"L\"}, {\"0\": \"(4.22)M b = p bSx\", \"1\": \"\"}, {\"0\": \"For class 3 semi-compact sections, M b, is givenby\", \"1\": \"βM\"}, {\"0\": \"(4.23)M b = p bZx\", \"1\": \"M\"}, {\"0\": \"or alternatively(4.24)M b = p bSx,eff\", \"1\": \"(b)\"}, {\"0\": \"For class 4 slender sections\", \"1\": \"Fig. 4.22 Load cases less susceptible to buckling.\"}, {\"0\": \"(4.25)M b = p bZx,eff\", \"1\": \"\"}, {\"0\": \"where\", \"1\": \"(Fig.  4.22(a))  or  unequal  end  moments (Fig.\"}, {\"0\": \"plastic modulus about the major axisSx\", \"1\": \"4.22(b)),  the  maximum  moment  about  the  major\"}, {\"0\": \"elastic modulus about the major axisZ x\", \"1\": \"axis, M x, must satisfy the following conditions:\"}, {\"0\": \"effective plastic modulus about the majorSx,eff\", \"1\": \"(4.26)M x ≤ M b/m LT and M x ≤ Mcx\"}, {\"0\": \"axis (see clause 3.5.6)\", \"1\": \"where\"}, {\"0\": \"effective section modulus about the majorZ x,eff\", \"1\": \"equivalent uniform moment factor form LT\"}, {\"0\": \"axis (see clause 3.6.2)\", \"1\": \"lateral torsional buckling read from Table\"}, {\"0\": \"This value of the buckling moment assumes the\", \"1\": \"18 of BS 5950, reproduced as Table 4.10.\"}, {\"0\": \"beam  is  acted  on  by  a  uniform,  single  curvature\", \"1\": \"For the destabilising loading condition\"}, {\"0\": \"moment  (Fig.  4.21),  which is the  most  severe\", \"1\": \"m LT = 1\"}, {\"0\": \"arrangement in terms of lateral stability. However,\", \"1\": \"moment capacity of the section about theM cx\"}, {\"0\": \"where  the  beam  is  acted  on  by  variable  moments\", \"1\": \"major axis\"}, {\"0\": \"Example 4.8 Design of a laterally unrestrained beam – rigorous method\", \"1\": \"\"}, {\"0\": \"(BS 5950)\", \"1\": \"\"}, {\"0\": \"Repeat Example 4.7 using the more rigorous method.\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"As previously noted the 305 × 305 × 137 UC section is class 1 plastic has a design strength, py = 265 N/mm2. From\"}, {\"0\": \"steel tables (Appendix B), r y = 78.2 mm, D = 320.5 mm and T = 21.7 mm. The slenderness ratio, λ, is\", \"1\": \"\"}, {\"0\": \"L rE==λ\", \"1\": \"10000=127 9\"}, {\"0\": \"y\", \"1\": \"78 2\"}, {\"0\": \"λ x127 9=\", \"1\": \"=  8 7\"}, {\"0\": \"320 5 21 7\", \"1\": \"\"}, {\"0\": \"From Table 4.8, ν = 0.678. Since the section is class 1 plastic, β\", \"1\": \"w = 1.0 and the equivalent slenderness ratio, λLT, is\"}, {\"0\": \"βλLT = uνλ\", \"1\": \"= 78w  = 0.9 × 0.678 × 127.9 ×    1 0.\"}, {\"0\": \"174\", \"1\": \"\"}], [], []]\n\n[FORMULAS]\nry = 78; D = 320; ν = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0195.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 196, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n175\nSegments with end moments only\nβ\n1.0\n0.9\n0.8\n0.7\n0.6\n0.5\n0.4\n0.3\n0.2\n0.1\n0.0\n−0.1\n−0.2\n−0.3\n−0.4\n−0.5\n−0.6\n−0.7\n−0.8\n−0.9\n−1.0\nm LT\n1.00\n0.96\n0.92\n0.88\n0.84\n0.80\n0.76\n0.72\n0.68\n0.64\n0.60\n0.56\n0.52\n0.48\n0.46\n0.44\n0.44\n0.44\n0.44\n0.44\n0.44\nβ negative\nx\nM\nx Lateral restraint\nx\nβM\nL LT\nx\nM\nx\nβM\nL LT\nx\nx\nx\nx\nM\nL LT\nL LT\nβM\nβM\nx\nx\nx\nx\nM\nSpecific cases (no intermediate lateral restraints)\nL LT\nL LT\nx\nx\nx\nx\nm LT = 0.850\nm LT = 0.925\nx\nx\nx\nx\nL LT\nL LT\nm LT = 0.925\nm LT = 0.744\nβ positive\nTable 4.10\nEquivalent uniform moment factor mLT for lateral\ntorsional buckling (based on Table 18, BS 5950)\nDesign of steel beams and joists\nFrom Table 4.9, p b = 165.4 N/mm2, hence\nMb = pbSx = 165.4 × 2300 × 103\n= 380.4 × 106 N mm = 380.4 kN m\nmLT from Table 4.10 = 0.89 (approx.), by interpolation between 0.85 and 0.925\n    \nM\nm\nb\nLT\n  \n.\n.\n= 380 4\n0 89  = 427.4 kN m < Mcx (= pySx = 265 × 2300 × 10−3) = 609.5 kN m\nThis gives, in this case, a buckling moment approximately 10% greater than in Example 4.7. This may enable a lighter\nmember to be selected, but for rolled sections it may not be really worth the additional effort.\nExample 4.8 continued\n\n[TABLES]\n[[{\"0\": \"Table 4.10Equivalent uniform moment factor m LT for lateraltorsional buckling (based on Table 18, BS 5950)\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"Segments with end moments onlyβ positivexxLLTLLTx Lateral restraint\", \"1\": \"x MxβMM\", \"2\": \"\", \"3\": \"βM\", \"4\": \"\", \"5\": \"βmLT1.001.00.960.90.920.80.880.70.840.60.800.50.760.40.720.30.680.20.640.10.600.00.56−0.10.52−0.20.48−0.30.46−0.40.44−0.50.44−0.60.44−0.70.44−0.80.44−0.90.44−1.0\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"x xβ negativexxxxxxβMMβMM\", \"1\": \"\", \"2\": \"LLTLLTSpecific cases (no intermediate lateral restraints)\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"xxxx\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"LLTLLTmLT = 0.850mLT = 0.925xxxxLLTLLT\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"mLT = 0.925mLT = 0.7441759/3/09, 1:27 PM\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}], [], []]\n\n[FORMULAS]\n103\n= 380", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0196.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 197, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n176\nTable 4.11\nEffective length, LE, for cantilevers without intermediate constraint (based on Table 14,\nBS 5950)\nRestraint conditions at supports\nRestraint conditons at tip\nLoading conditions\nNormal\nDestabilising\nRestrained laterally, torsionally\n1) Free\n0.8L\n1.4L\nand against rotation on plan\n2) Lateral restraint to top flange\n0.7L\n1.4L\n3) Torsional restraint\n0.6L\n0.6L\n4) Lateral and torsional restraint\n0.5L\n0.5L\nL\n4.8.11.4 Cantilever beams\nFor cantilevers, the effective length is given in clear\ndiagrammatical form in Table 14 of BS 5950, part\nof which is reproduced as Table 4.11.\n4.8.11.5 Summary of design procedures\nThe two alternative methods for checking lateral\ntorsional buckling of beams can be summarised as\nfollows.\nConservative method\n1. Calculate the design shear force, Fv, and bending\nmoment, Mx, at critical points along the beam.\n2. Select and classify UB or UC section.\n3. Check shear and bending capacity of section. If\nunsatisfactory return to (2).\n4. Determine the effective length of the beam, LE,\nusing Table 4.6.\nExample 4.9 Checking for lateral instability in a cantilever steel beam\n(BS 5950)\nContinue Example 4.3 to determine whether the cantilever is laterally stable. Assume the load is destabilising.\nFrom Table 4.11, LE = 1.4L = 1.4 m\nFor 610 × 229 × 113 UB\n    \nλ  \n  \n.   \n.\n=\n=\n=\nL\nr\nE\ny\n1400\n48 8\n28 7\n    \nλ\nx   \n.\n. /\n.   .\n=\n=\n28 7\n607 317 3\n0 82 ⇒ ν = 0.99\n(Table 4.8)\nλLT = uνλ  βw  = 0.9 × 0.99 × 28.7 ×   1 0.  = 28.4\nFrom Table 4.9, pb = 265 N/mm2 = py, and so lateral torsional buckling is not a problem with this cantilever.\n\n[TABLES]\n[[{\"0\": \"Design in structural steelwork to BS 5950\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"Table 4.11\", \"1\": \"Effective length, LE, for cantilevers without intermediate constraint (based on Table 14,\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"BS 5950)\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"Restraint conditions at supports\", \"1\": \"Restraint conditons at tip\", \"2\": \"\", \"3\": \"Loading conditions\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"Normal\", \"3\": \"Destabilising\"}, {\"0\": \"Restrained laterally, torsionally\", \"1\": \"1) Free\", \"2\": \"0.8L\", \"3\": \"1.4L\"}, {\"0\": \"and against rotation on plan\", \"1\": \"2) Lateral restraint to top flange\", \"2\": \"0.7L\", \"3\": \"1.4L\"}, {\"0\": \"\", \"1\": \"3) Torsional restraint\", \"2\": \"0.6L\", \"3\": \"0.6L\"}, {\"0\": \"\", \"1\": \"4) Lateral and torsional restraint\", \"2\": \"0.5L\", \"3\": \"0.5L\"}], [{\"0\": \"From Table 4.9, p b = 265 N/mm2 = py, and so lateral torsional buckling is not a problem with this cantilever.\", \"1\": \"\"}, {\"0\": \"4.8.11.4 Cantilever beams\", \"1\": \"Conservative method\"}, {\"0\": \"For cantilevers, the effective length is given in clear\", \"1\": \"\"}, {\"0\": \"diagrammatical form in Table 14 of BS 5950, part\", \"1\": \"1. Calculate the design shear force, Fv, and bending\"}, {\"0\": \"of which is reproduced as Table 4.11.\", \"1\": \"moment, M x, at critical points along the beam.\"}, {\"0\": \"\", \"1\": \"2. Select and classify UB or UC section.\"}, {\"0\": \"4.8.11.5 Summary of design procedures\", \"1\": \"3. Check shear and bending capacity of section. If\"}, {\"0\": \"The  two  alternative  methods  for  checking  lateral\", \"1\": \"unsatisfactory return to (2).\"}, {\"0\": \"torsional buckling of beams can be summarised as\", \"1\": \"4. Determine the effective length of the beam, L E,\"}, {\"0\": \"follows.\", \"1\": \"using Table 4.6.\"}], [], []]\n\n[FORMULAS]\nLE = 1; ν = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0197.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 198, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n177\n5. Determine Sx, ry, D and T from steel tables\n(Appendix B).\n6. Calculate the slenderness value, λ = (β)0.5LE/ry.\n7. Determine the bending strength, pb, from Table\n4.7 using λ and D/T.\n8. Calculate the buckling resistance moment, Mb,\nvia equation 4.18 or 4.19.\n9. Check Mx ≤ Mb. If unsatisfactory return to (2).\nRigorous method\n1. Steps (1)–(4) as for conservative method.\n2. Determine Sx and ry from steel tables; deter-\nmine x and u from either steel tables or for UB\nand UCs assuming x = D/T and u = 0.9.\n3. Calculate slenderness ratio, λ = LE/ry.\n4. Determine ν from Table 4.8 using λ/x and\nN = 0.5.\n5. Calculate equivalent slenderness ratio, λLT =\nuνλ√βw.\n6. Determine pb from Table 4.9 using λLT.\n7. Calculate Mb via equations 4.22–4.25.\n8. Obtain the equivalent uniform moment factor,\nmLT, from Table 4.10.\n9. Check Mx ≤ Mb/mLT. If unsatisfactory return to\n(2).\n4.9 Design of compression\nmembers\n4.9.1 STRUTS\nSteel compression members, commonly referred to\nas stanchions, include struts and columns. A strut\nis a member subject to direct compression only.\nA column, on the other hand, refers to members\nsubject to a combination of compressive loading\nand bending. Although most columns in real struc-\ntures resist compressive loading and bending, the\nstrut is a convenient starting point.\nStruts (and columns) differ fundamentally in their\nbehaviour under axial load depending on whether\nthey are slender or stocky. Most real struts and\ncolumns can neither be regarded as slender nor\nstocky, but as something in between, but let us look\nat the behaviour of stocky and slender struts first.\nStocky struts will fail by crushing or squashing\nof the material. For stocky struts the ‘squash load’,\nPs is given by the simple formula\nPs = pyAg\n(4.27)\nwhere\npy\ndesign strength of steel\nAg\ngross cross sectional area of the section\nDesign of compression members\nStrength of strut pc (N mm−2)\n300\n250\n200\n150\n100\n50\n0\n50\n100 150 200 250 300 350\nSlenderness ratio λ\nEuler buckling stress p E\nTable 24(a)\nTable 24(b)\nTable 24(c)\nTable 24(d)\nYield stress p y = 275 N mm−2\nF\nF\nFig. 4.23\nSlender struts will fail by buckling. For elastic\nslender struts pinned at each end, the ‘Euler load’,\nat which a perfect strut buckles elastically is given by\n    \nP\nEI\nL\nEA r\nL\nEA\nE\n2\ng\ny\n2\n2\ng\n2\n=\n=\n=\n \n  \n  \nπ\nπ\nπ\nλ\n2\n2\n2\n(4.28)\nusing \n    \nr\nI\nA\n  \n=\nand\n    \nλ  \n.\n= L\nr\nIf the compressive strength, pc, which is given by\npc = Ps/Ag\n(for stocky struts)\n(4.29)\nand\npc = PE/Ag\n(for slender struts)\n(4.30)\nare plotted against λ (Fig. 4.23), the area above\nthe two dotted lines represents an impossible situ-\nation in respect of these struts. In this area, the\nstrut has either buckled or squashed. Struts which\nfall below the dotted lines are theoretically able to\nwithstand the applied load without either buckling\nor squashing. In reality, however, this tends not to\nbe the case because of a combination of manufac-\nturing and practical considerations. For example,\nstruts are never completely straight, or are subject\nto exactly concentric loading. During manufac-\nture, stresses are locked into steel members which\neffectively reduce their load-carrying capacities.\nAs a result of these factors, failure of a strut will\nnot be completely due to buckling or squashing,\nbut a combination, with partially plastic stresses\nappearing across the member section. These non-\nideal factors or imperfections are found in practice,\nand laboratory tests have confirmed that in fact the\nfailure line for real struts lies along a series of lines\nsuch as a,b,c and d in Fig. 4.23. These are derived\nfrom the Perry-Robertson equation which includes\nallowances for the various imperfections.\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"Design of compression members\"}, {\"0\": \"5. Determine  Sx,  ry,  D  and  T  from  steel  tables(Appendix B).\", \"1\": \"300Yield stress py = 275 N mm−2\"}, {\"0\": \"6. Calculate the slenderness value, λ = (β)0.5L E/ry.\", \"1\": \"F250\"}, {\"0\": \"7. Determine the bending strength, pb, from Table\", \"1\": \"Euler buckling stress pE200\"}, {\"0\": \"4.7 using λ and D/T.\", \"1\": \"Table 24(a)Table 24(b)\"}, {\"0\": \"8. Calculate the buckling resistance moment, M b,via equation 4.18 or 4.19.\", \"1\": \"Strength of strut pc (Nmm−2)150Table 24(c)\"}, {\"0\": \"9. Check M x ≤ M b. If unsatisfactory return to (2).\", \"1\": \"Table 24(d)100\"}, {\"0\": \"Rigorous method\", \"1\": \"50\"}, {\"0\": \"1. Steps (1)–(4) as for conservative method.\", \"1\": \"0F\"}, {\"0\": \"2. Determine  Sx  and  ry  from  steel  tables;  deter-mine x and u from either steel tables or for UB\", \"1\": \"50100150200250300350\"}, {\"0\": \"and UCs assuming x = D/T and u = 0.9.\", \"1\": \"Slenderness ratio λ\"}, {\"0\": \"3. Calculate slenderness ratio, λ = L E/ry.\", \"1\": \"Fig. 4.23\"}, {\"0\": \"4. Determine  ν from  Table  4.8  using  λ/x  and\", \"1\": \"Slender  struts  will  fail  by  buckling.  For  elastic\"}, {\"0\": \"N = 0.5.\", \"1\": \"slender struts pinned at each end, the ‘Euler load’,\"}, {\"0\": \"5. Calculate  equivalent  slenderness  ratio,  λLT  =\", \"1\": \"at which a perfect strut buckles elastically is given by\"}, {\"0\": \"uνλ√βw.\", \"1\": \"\"}, {\"0\": \"6. Determine p b from Table 4.9 using λLT.\", \"1\": \"y2222πππE A rEAEIgg===(4.28)P\"}, {\"0\": \"7. Calculate M b via equations 4.22–4.25.\", \"1\": \"E222λLL\"}, {\"0\": \"8. Obtain  the  equivalent  uniform  moment  factor,\", \"1\": \"\"}, {\"0\": \"m LT, from Table 4.10.\", \"1\": \"= L\"}, {\"0\": \"9. Check M x ≤ M b/m LT. If unsatisfactory return to(2).\", \"1\": \"I Ar  =λ .using andr\"}, {\"0\": \"\", \"1\": \"If the compressive strength, pc, which is given by\"}, {\"0\": \"4.9 Design of compression\", \"1\": \"(for stocky struts)(4.29)pc = Ps /Ag\"}, {\"0\": \"members\", \"1\": \"and\"}, {\"0\": \"\", \"1\": \"(for slender struts)(4.30)pc = PE/Ag\"}, {\"0\": \"4.9.1STRUTS\", \"1\": \"are  plotted  against  λ  (Fig.  4.23),  the  area  above\"}, {\"0\": \"Steel compression members, commonly referred to\", \"1\": \"the two dotted lines represents an impossible situ-\"}, {\"0\": \"as stanchions, include struts and columns. A strut\", \"1\": \"ation  in  respect  of  these  struts.  In  this  area,  the\"}, {\"0\": \"is  a  member  subject  to  direct  compression  only.\", \"1\": \"strut has either buckled or squashed. Struts which\"}, {\"0\": \"A  column,  on  the  other  hand,  refers  to  members\", \"1\": \"fall below the dotted lines are theoretically able to\"}, {\"0\": \"subject  to  a  combination  of  compressive  loading\", \"1\": \"withstand the applied load without either buckling\"}, {\"0\": \"and bending. Although most columns in real struc-\", \"1\": \"or squashing. In reality, however, this tends not to\"}, {\"0\": \"tures  resist  compressive  loading  and  bending,  the\", \"1\": \"be the case because of a combination of manufac-\"}, {\"0\": \"strut is a convenient starting point.\", \"1\": \"turing  and  practical  considerations.  For  example,\"}, {\"0\": \"Struts (and columns) differ fundamentally in their\", \"1\": \"struts are never completely straight, or are subject\"}, {\"0\": \"behaviour  under  axial  load  depending  on  whether\", \"1\": \"to  exactly  concentric loading.  During  manufac-\"}, {\"0\": \"they  are  slender  or  stocky.  Most  real  struts  and\", \"1\": \"ture, stresses are locked into steel members which\"}, {\"0\": \"columns  can  neither  be  regarded  as  slender  nor\", \"1\": \"effectively reduce their load-carrying  capacities.\"}, {\"0\": \"stocky, but as something in between, but let us look\", \"1\": \"As  a  result  of  these  factors,  failure  of  a  strut  will\"}, {\"0\": \"at the behaviour of stocky and slender struts first.\", \"1\": \"not  be  completely  due  to  buckling  or  squashing,\"}, {\"0\": \"Stocky  struts  will  fail  by  crushing  or  squashing\", \"1\": \"but  a  combination,  with  partially  plastic  stresses\"}, {\"0\": \"of the material. For stocky struts the ‘squash load’,\", \"1\": \"appearing across the member section. These non-\"}, {\"0\": \"Ps is given by the simple formula\", \"1\": \"ideal factors or imperfections are found in practice,\"}, {\"0\": \"(4.27)Ps = pyAg\", \"1\": \"and laboratory tests have confirmed that in fact the\"}, {\"0\": \"\", \"1\": \"failure line for real struts lies along a series of lines\"}, {\"0\": \"where\", \"1\": \"such as a,b,c and d in Fig. 4.23. These are derived\"}, {\"0\": \"design strength of steelpy\", \"1\": \"from the Perry-Robertson equation which includes\"}, {\"0\": \"gross cross sectional area of the sectionAg\", \"1\": \"allowances for the various imperfections.\"}], [], []]\n\n[FORMULAS]\nλ = LE/ry", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0198.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 199, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n178\nWhichever of the lines a–d is used depends\non the shape of section and the axis of buckling.\nTable 23 of BS 5950, part of which is reproduced\nas Table 4.12, specifies which of the lines is appro-\npriate for the shape of section, and Tables 24(a),\n(b), (c) and (d) enable values of pc to be read off\nappropriate to the section used. (Tables 24(b) and\n(c) of BS 5950 have been reproduced as Tables 4.13\nand 4.14 respectively.) Alternatively, Appendix C\nof BS 5950 gives the actual Perry-Robertson equa-\ntions which may be used in place of the tables if\nconsidered necessary.\nTable 4.13\nCompressive strength, pc (N/mm2) with λ < 110 for strut curve b\n(Table 24(b), BS 5950)\nλ\nSteel grade and design strength py (N/mm2)\nS275\nS355\nS460\n235\n245\n255\n265\n275\n315\n325\n335\n345\n355\n400\n410\n430\n440\n460\n15\n235\n245\n255\n265\n275\n315\n325\n335\n345\n355\n399\n409\n428\n438\n457\n20\n234\n243\n253\n263\n272\n310\n320\n330\n339\n349\n391\n401\n420\n429\n448\n25\n229\n239\n248\n258\n267\n304\n314\n323\n332\n342\n384\n393\n411\n421\n439\n30\n225\n234\n243\n253\n262\n298\n307\n316\n325\n335\n375\n384\n402\n411\n429\n35\n220\n229\n238\n247\n256\n291\n300\n309\n318\n327\n366\n374\n392\n400\n417\n40\n216\n224\n233\n241\n250\n284\n293\n301\n310\n318\n355\n364\n380\n388\n404\n42\n213\n222\n231\n239\n248\n281\n289\n298\n306\n314\n351\n359\n375\n383\n399\n44\n211\n220\n228\n237\n245\n278\n286\n294\n302\n310\n346\n354\n369\n377\n392\n46\n209\n218\n226\n234\n242\n275\n283\n291\n298\n306\n341\n349\n364\n371\n386\n48\n207\n215\n223\n231\n239\n271\n279\n287\n294\n302\n336\n343\n358\n365\n379\n50\n205\n213\n221\n229\n237\n267\n275\n283\n290\n298\n330\n337\n351\n358\n372\n52\n203\n210\n218\n226\n234\n264\n271\n278\n286\n293\n324\n331\n344\n351\n364\n54\n200\n208\n215\n223\n230\n260\n267\n274\n281\n288\n318\n325\n337\n344\n356\n56\n198\n205\n213\n220\n227\n256\n263\n269\n276\n283\n312\n318\n330\n336\n347\n58\n195\n202\n210\n217\n224\n252\n258\n265\n271\n278\n305\n311\n322\n328\n339\n60\n193\n200\n207\n214\n221\n247\n254\n260\n266\n272\n298\n304\n314\n320\n330\n62\n190\n197\n204\n210\n217\n243\n249\n255\n261\n266\n291\n296\n306\n311\n320\n64\n187\n194\n200\n207\n213\n238\n244\n249\n255\n261\n284\n289\n298\n302\n311\n66\n184\n191\n197\n203\n210\n233\n239\n244\n249\n255\n276\n281\n289\n294\n301\n68\n181\n188\n194\n200\n206\n228\n233\n239\n244\n249\n269\n273\n281\n285\n292\n70\n178\n185\n190\n196\n202\n223\n228\n233\n238\n242\n261\n265\n272\n276\n282\n72\n175\n181\n187\n193\n198\n218\n223\n227\n232\n236\n254\n257\n264\n267\n273\n74\n172\n178\n183\n189\n194\n213\n217\n222\n226\n230\n246\n249\n255\n258\n264\n76\n169\n175\n180\n185\n190\n208\n212\n216\n220\n223\n238\n241\n247\n250\n255\n78\n166\n171\n176\n181\n186\n203\n206\n210\n214\n217\n231\n234\n239\n241\n246\nTable 4.12\nStrut table selection (based on Table 23, BS 5950)\nType of section\nThicknessa\nAxis of buckling\nx–x\ny–y\nHot-finished structural hollow section\n24(a)\n24(a)\nRolled I-section\nUp to 40 mm\n24(a)\n24(b)\nRolled H-section\nUp to 40 mm\n24(b)b\n24(c)c\nOver 40 mm\n24(c)\n24(d)\nNotes. a For thicknesses between 40 and 50 mm the value of pc may be taken as the\naverage of the values for thicknesses up to 40 mm and over 40 mm.\nb Reproduced as Table 4.13.\nc Reproduced as Table 4.14.\n\n[TABLES]\n[[{\"0\": \"S27523524525526527515235245255265275\", \"1\": \"S355315325335345355315325335345355\", \"2\": \"S460400410430440460399409428438457\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"2023424325326327225229239248258267302252342432532623522022923824725640216224233241250422132222312392484421122022823724546209218226234242482072152232312395020521322122923752203210218226234542002082152232305619820521322022758195202210217224601932002072142216219019720421021764187194200207213661841911972032106818118819420020670178185190196202721751811871931987417217818318919476169175180185190781661711761811861789780415467193_C04178\", \"1\": \"310320330339349304314323332342298307316325335291300309318327284293301310318 2812892983063142782862943023102752832912983062712792872943022672752832902982642712782862932602672742812882562632692762832522582652712782472542602662722432492552612662382442492552612332392442492552282332392442492232282332382422182232272322362132172222262302082122162202232032062102142179/3/09, 1:29 PM\", \"2\": \"391401420429448384393411421439375384402411429366374392400417355364380388404351359375383399346354369377392341349364371386336343358365379330337351358372324331344351364318325337344356312318330336347305311322328339298304314320330291296306311320284289298302311276281289294301269273281285292261265272276282254257264267273246249255258264238241247250255231234239241246\"}], [], [], [{\"S355\": \"315 325 335 345 355\"}]]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0199.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 200, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n179\n80\n163\n168\n172\n177\n181\n197\n201\n204\n208\n211\n224\n226\n231\n233\n237\n82\n160\n164\n169\n173\n177\n192\n196\n199\n202\n205\n217\n219\n223\n225\n229\n84\n156\n161\n165\n169\n173\n187\n190\n193\n196\n199\n210\n212\n216\n218\n221\n86\n153\n157\n161\n165\n169\n182\n185\n188\n190\n193\n203\n205\n208\n210\n213\n88\n150\n154\n158\n161\n165\n177\n180\n182\n185\n187\n196\n198\n201\n203\n206\n90\n146\n150\n154\n157\n161\n172\n175\n177\n179\n181\n190\n192\n195\n196\n199\n92\n143\n147\n150\n153\n156\n167\n170\n172\n174\n176\n184\n185\n188\n189\n192\n94\n140\n143\n147\n150\n152\n162\n165\n167\n169\n171\n178\n179\n182\n183\n185\n96\n137\n140\n143\n146\n148\n158\n160\n162\n164\n165\n172\n173\n176\n177\n179\n98\n134\n137\n139\n142\n145\n153\n155\n157\n159\n160\n167\n168\n170\n171\n173\n100\n130\n133\n136\n138\n141\n149\n151\n152\n154\n155\n161\n162\n164\n165\n167\n102\n127\n130\n132\n135\n137\n145\n146\n148\n149\n151\n156\n157\n159\n160\n162\n104\n124\n127\n129\n131\n133\n141\n142\n144\n145\n146\n151\n152\n154\n155\n156\n106\n121\n124\n126\n128\n130\n137\n138\n139\n141\n142\n147\n148\n149\n150\n151\n108\n118\n121\n123\n125\n126\n133\n134\n135\n137\n138\n142\n143\n144\n145\n147\n110\n115\n118\n120\n121\n123\n129\n130\n131\n133\n134\n138\n139\n140\n141\n142\n112\n113\n115\n117\n118\n120\n125\n127\n128\n129\n130\n134\n134\n136\n136\n138\n114\n110\n112\n114\n115\n117\n122\n123\n124\n125\n126\n130\n130\n132\n132\n133\n116\n107\n109\n111\n112\n114\n119\n120\n121\n122\n122\n126\n126\n128\n128\n129\n118\n105\n106\n108\n109\n111\n115\n116\n117\n118\n119\n122\n123\n124\n124\n125\n120\n102\n104\n105\n107\n108\n112\n113\n114\n115\n116\n119\n119\n120\n121\n122\n122\n100\n101\n103\n104\n105\n109\n110\n111\n112\n112\n115\n116\n117\n117\n118\n124\n97\n99\n100\n101\n102\n106\n107\n108\n109\n109\n112\n112\n113\n114\n115\n126\n95\n96\n98\n99\n100\n103\n104\n105\n106\n106\n109\n109\n110\n111\n111\n128\n93\n94\n95\n96\n97\n101\n101\n102\n103\n103\n106\n106\n107\n107\n108\n130\n90\n92\n93\n94\n95\n98\n99\n99\n100\n101\n103\n103\n104\n105\n105\n135\n85\n86\n87\n88\n89\n92\n93\n93\n94\n94\n96\n97\n97\n98\n98\n140\n80\n81\n82\n83\n84\n86\n87\n87\n88\n88\n90\n90\n91\n91\n92\n145\n76\n77\n78\n78\n79\n81\n82\n82\n83\n83\n84\n85\n85\n86\n86\n150\n72\n72\n73\n74\n74\n76\n77\n77\n78\n78\n79\n80\n80\n80\n81\n155\n68\n69\n69\n70\n70\n72\n72\n73\n73\n73\n75\n75\n75\n76\n76\n160\n64\n65\n65\n66\n66\n68\n68\n69\n69\n69\n70\n71\n71\n71\n72\n165\n61\n62\n62\n62\n63\n64\n65\n65\n65\n65\n66\n67\n67\n67\n68\n170\n58\n58\n59\n59\n60\n61\n61\n61\n62\n62\n63\n63\n63\n64\n64\n175\n55\n55\n56\n56\n57\n58\n58\n58\n59\n59\n60\n60\n60\n60\n60\n180\n52\n53\n53\n53\n54\n55\n55\n55\n56\n56\n56\n57\n57\n57\n57\n185\n50\n50\n51\n51\n51\n52\n52\n53\n53\n53\n54\n54\n54\n54\n54\n190\n48\n48\n48\n48\n49\n50\n50\n50\n50\n50\n51\n51\n51\n51\n52\n195\n45\n46\n46\n46\n46\n47\n47\n48\n48\n48\n49\n49\n49\n49\n49\n200\n43\n44\n44\n44\n44\n45\n45\n45\n46\n46\n46\n46\n47\n47\n47\n210\n40\n40\n40\n40\n41\n41\n41\n41\n42\n42\n42\n42\n42\n43\n43\n220\n36\n37\n37\n37\n37\n38\n38\n38\n38\n38\n39\n39\n39\n39\n39\n230\n34\n34\n34\n34\n34\n35\n35\n35\n35\n35\n35\n36\n36\n36\n36\n240\n31\n31\n31\n31\n32\n32\n32\n32\n32\n32\n33\n33\n33\n33\n33\n250\n29\n29\n29\n29\n29\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n260\n27\n27\n27\n27\n27\n27\n28\n28\n28\n28\n28\n28\n28\n28\n28\n270\n25\n25\n25\n25\n25\n26\n26\n26\n26\n26\n26\n26\n26\n26\n26\n280\n23\n23\n23\n23\n24\n24\n24\n24\n24\n24\n24\n24\n24\n24\n24\n290\n22\n22\n22\n22\n22\n22\n22\n22\n22\n22\n23\n23\n23\n23\n23\n300\n20\n20\n21\n21\n21\n21\n21\n21\n21\n21\n21\n21\n21\n21\n21\n310\n19\n19\n19\n19\n19\n20\n20\n20\n20\n20\n20\n20\n20\n20\n–\n320\n18\n18\n18\n18\n18\n18\n18\n19\n19\n19\n19\n19\n19\n19\n19\n330\n17\n17\n17\n17\n17\n17\n17\n17\n17\n18\n18\n18\n18\n18\n18\n340\n16\n16\n16\n16\n16\n16\n16\n16\n17\n17\n17\n17\n17\n17\n17\n350\n15\n15\n15\n15\n15\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\nTable 4.13\n(cont’d)\nλ\nSteel grade and design strength py (N/mm2)\nS275\nS355\nS460\n235\n245\n255\n265\n275\n315\n325\n335\n345\n355\n400\n410\n430\n440\n460\n\n[TABLES]\n[[{\"0\": \"Table 4.13(cont’d )λ\", \"1\": \"Steel grade and design strength py (N/mm2)\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"S27523524525526527580163168172177181821601641691731778415616116516917386153157161165169881501541581611659014615015415716192143147150153156941401431471501529613714014314614898134137139142145100130133136138141102127130132135137104124127129131133106121124126128130108118121123125126110115118120121123112113115117118120114110112114115117116107109111112114118105106108109111120102104105107108122100101103104105124979910010110212695969899100128939495969713090929394951358586878889140808182838414576777878791507272737474155686969707016064656566661656162626263170585859596017555555656571805253535354185505051511904848484849195454646464620043444444442104040404041220363737373723034343434342403131313132250292929292926027272727272702525252525280232323232429022222222223002020212121310191919191932018181818183301717171717340161616161635015151515159780415467193_C04179\", \"1\": \"S35531532533534535519720120420821119219619920220518719019319619918218518819019317718018218518717217517717918116717017217417616216516716917115816016216416515315515715916014915115215415514514614814915114114214414514613713813914114213313413513713812913013113313412512712812913012212312412512611912012112212211511611711811911211311411511610911011111211210610710810910910310410510610610110110210310398999910010192939394948687878888818282838376777778787272737373686869696964656565656161616262585858595955555556565252535353 5050505050474748484845454546464141414242383838383835353535353232323232303030303027282828282626262626242424242422222222222121212121202020202018181919191717171718161616171716161616169/3/09, 1:29 PM\", \"2\": \"S46040041043044046022422623123323721721922322522921021221621822120320520821021319619820120320619019219519619918418518818919217817918218318517217317617717916716817017117316116216416516715615715916016215115215415515614714814915015114214314414514713813914014114213413413613613813013013213213312612612812812912212312412412511911912012112211511611711711811211211311411510910911011111110610610710710810310310410510596979798989090919192848585868679808080817575757676707171717266676767686363636464606060606056575757575454545454515151515249494949494646474747424242434339393939393536363636333333333330303030302828282828262626262624242424242323232323212121212120202020–1919191919181818181817171717171616161616179\"}], [], [], [{\"S355\": \"315 325 335 345 355\"}, {\"S355\": \"197 201 204 208 211\\n192 196 199 202 205\\n187 190 193 196 199\\n182 185 188 190 193\\n177 180 182 185 187\\n172 175 177 179 181\\n167 170 172 174 176\\n162 165 167 169 171\\n158 160 162 164 165\\n153 155 157 159 160\\n149 151 152 154 155\\n145 146 148 149 151\\n141 142 144 145 146\\n137 138 139 141 142\\n133 134 135 137 138\\n129 130 131 133 134\\n125 127 128 129 130\\n122 123 124 125 126\\n119 120 121 122 122\\n115 116 117 118 119\\n112 113 114 115 116\\n109 110 111 112 112\\n106 107 108 109 109\\n103 104 105 106 106\\n101 101 102 103 103\\n98 99 99 100 101\\n92 93 93 94 94\\n86 87 87 88 88\\n81 82 82 83 83\\n76 77 77 78 78\\n72 72 73 73 73\\n68 68 69 69 69\\n64 65 65 65 65\\n61 61 61 62 62\\n58 58 58 59 59\\n55 55 55 56 56\\n52 52 53 53 53\\n50 50 50 50 50\\n47 47 48 48 48\\n45 45 45 46 46\\n41 41 41 42 42\\n38 38 38 38 38\\n35 35 35 35 35\\n32 32 32 32 32\\n30 30 30 30 30\\n27 28 28 28 28\\n26 26 26 26 26\\n24 24 24 24 24\\n22 22 22 22 22\\n21 21 21 21 21\\n20 20 20 20 20\\n18 18 19 19 19\\n17 17 17 17 18\\n16 16 16 17 17\\n16 16 16 16 16\"}]]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0200.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 201, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n180\nTable 4.14\nCompressive strength, pc (N/mm2) for strut curve c (Table 24(c), BS 5950)\nλ\nSteel grade and design strength py (N/mm2)\nS275\nS355\nS460\n235\n245\n255\n265\n275\n315\n325\n335\n345\n355\n400\n410\n430\n440\n460\n15\n235\n245\n255\n265\n275\n315\n325\n335\n345\n355\n398\n408\n427\n436\n455\n20\n233\n242\n252\n261\n271\n308\n317\n326\n336\n345\n387\n396\n414\n424\n442\n25\n226\n235\n245\n254\n263\n299\n308\n317\n326\n335\n375\n384\n402\n410\n428\n30\n220\n228\n237\n246\n255\n289\n298\n307\n315\n324\n363\n371\n388\n396\n413\n35\n213\n221\n230\n238\n247\n280\n288\n296\n305\n313\n349\n357\n374\n382\n397\n40\n206\n214\n222\n230\n238\n270\n278\n285\n293\n301\n335\n343\n358\n365\n380\n42\n203\n211\n219\n227\n235\n266\n273\n281\n288\n296\n329\n337\n351\n358\n373\n44\n200\n208\n216\n224\n231\n261\n269\n276\n284\n291\n323\n330\n344\n351\n365\n46\n197\n205\n213\n220\n228\n257\n264\n271\n279\n286\n317\n324\n337\n344\n357\n48\n195\n202\n209\n217\n224\n253\n260\n267\n274\n280\n311\n317\n330\n337\n349\n50\n192\n199\n206\n213\n220\n248\n255\n262\n268\n275\n304\n310\n323\n329\n341\n52\n189\n196\n203\n210\n217\n244\n250\n257\n263\n270\n297\n303\n315\n321\n333\n54\n186\n193\n199\n206\n213\n239\n245\n252\n258\n264\n291\n296\n308\n313\n324\n56\n183\n189\n196\n202\n209\n234\n240\n246\n252\n258\n284\n289\n300\n305\n315\n58\n179\n186\n192\n199\n205\n229\n235\n241\n247\n252\n277\n282\n292\n297\n306\n60\n176\n183\n189\n195\n201\n225\n230\n236\n241\n247\n270\n274\n284\n289\n298\n62\n173\n179\n185\n191\n197\n220\n225\n230\n236\n241\n262\n267\n276\n280\n289\n64\n170\n176\n182\n188\n193\n215\n220\n225\n230\n235\n255\n260\n268\n272\n280\n66\n167\n173\n178\n184\n189\n210\n215\n220\n224\n229\n248\n252\n260\n264\n271\n68\n164\n169\n175\n180\n185\n205\n210\n214\n219\n223\n241\n245\n252\n256\n262\n70\n161\n166\n171\n176\n181\n200\n204\n209\n213\n217\n234\n238\n244\n248\n254\n72\n157\n163\n168\n172\n177\n195\n199\n203\n207\n211\n227\n231\n237\n240\n246\n74\n154\n159\n164\n169\n173\n190\n194\n198\n202\n205\n220\n223\n229\n232\n238\n76\n151\n156\n160\n165\n169\n185\n189\n193\n196\n200\n214\n217\n222\n225\n230\n78\n148\n152\n157\n161\n165\n180\n184\n187\n191\n194\n207\n210\n215\n217\n222\n80\n145\n149\n153\n157\n161\n176\n179\n182\n185\n188\n201\n203\n208\n210\n215\n82\n142\n146\n150\n154\n157\n171\n174\n177\n180\n183\n195\n197\n201\n203\n207\n84\n139\n142\n146\n150\n154\n167\n169\n172\n175\n178\n189\n191\n195\n197\n201\n86\n135\n139\n143\n146\n150\n162\n165\n168\n170\n173\n183\n185\n189\n190\n194\n88\n132\n136\n139\n143\n146\n158\n160\n163\n165\n168\n177\n179\n183\n184\n187\n90\n129\n133\n136\n139\n142\n153\n156\n158\n161\n163\n172\n173\n177\n178\n181\n92\n126\n130\n133\n136\n139\n149\n152\n154\n156\n158\n166\n168\n171\n173\n175\n94\n124\n127\n130\n133\n135\n145\n147\n149\n151\n153\n161\n163\n166\n167\n170\n96\n121\n124\n127\n129\n132\n141\n143\n145\n147\n149\n156\n158\n160\n162\n164\n98\n118\n121\n123\n126\n129\n137\n139\n141\n143\n145\n151\n153\n155\n157\n159\n100\n115\n118\n120\n123\n125\n134\n135\n137\n139\n140\n147\n148\n151\n152\n154\n102\n113\n115\n118\n120\n122\n130\n132\n133\n135\n136\n143\n144\n146\n147\n149\n104\n110\n112\n115\n117\n119\n126\n128\n130\n131\n133\n138\n139\n142\n142\n144\n106\n107\n110\n112\n114\n116\n123\n125\n126\n127\n129\n134\n135\n137\n138\n140\n108\n105\n107\n109\n111\n113\n120\n121\n123\n124\n125\n130\n131\n133\n134\n136\n110\n102\n104\n106\n108\n110\n116\n118\n119\n120\n122\n126\n127\n129\n130\n132\n112\n100\n102\n104\n106\n107\n113\n115\n116\n117\n118\n123\n124\n125\n126\n128\n114\n98\n100\n101\n103\n105\n110\n112\n113\n114\n115\n119\n120\n122\n123\n124\n116\n95\n97\n99\n101\n102\n108\n109\n110\n111\n112\n116\n117\n118\n119\n120\n118\n93\n95\n97\n98\n100\n105\n106\n107\n108\n109\n113\n114\n115\n116\n117\n120\n91\n93\n94\n96\n97\n102\n103\n104\n105\n106\n110\n110\n112\n112\n113\n\n[TABLES]\n[[{\"0\": \"Table 4.14λ\", \"1\": \"Compressive strength, pc (N/mm2) for strut curve c (Table 24(c), BS 5950)Steel grade and design strength py (N/mm2)\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"S2752352452552652751523524525526527520233242252261271252262352452542633022022823724625535213221230238247402062142222302384220321121922723544200208216224231461972052132202284819520220921722450192199206213220521891962032102175418619319920621356183189196202209581791861921992056017618318919520162173179185191197641701761821881936616717317818418968164169175180185701611661711761817215716316817217774154159164169173761511561601651697814815215716116580145149153157161821421461501541578413914214615015486135139143146150881321361391431469012913313613914292126130133136139941241271301331359612112412712913298118121123126129100115118120123125102113115118120122104110112115117119106107110112114116108105107109111113110102104106108110112100102104106107114981001011031051169597991011021189395979810012091939496971809780415467193_C04180\", \"1\": \"S355315325335345355315325335345355308317326336345299308317326335289298307315324280288296305313270278285293301266273281288296261269276284291257264271279286253260267274280248255262268275244250257263270239245252258264234240246252258229235241247252225230236241247220225230236241215220225230235210215220224229205210214219223200204209213217195199203207211190194198202205185189193196200180184187191194176179182185188171174177180183167169172175178162165168170173158160163165168 1531561581611631491521541561581451471491511531411431451471491371391411431451341351371391401301321331351361261281301311331231251261271291201211231241251161181191201221131151161171181101121131141151081091101111121051061071081091021031041051069/3/09, 1:29 PM\", \"2\": \"S460400410430440460398408427436455387396414424442375384402410428363371388396413349357374382397335343358365380329337351358373323330344351365317324337344357311317330337349304310323329341297303315321333291296308313324284289300305315277282292297306270274284289298262267276280289255260268272280248252260264271241245252256262234238244248254227231237240246220223229232238214217222225230207210215217222201203208210215195197201203207189191195197201183185189190194177179183184187172173177178181166168171173175161163166167170156158160162164151153155157159147148151152154143144146147149138139142142144134135137138140130131133134136126127129130132123124125126128119120122123124116117118119120113114115116117110110112112113\"}], [], [], [{\"S355\": \"315 325 335 345 355\"}]]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0201.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 202, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n181\nTable 4.14\n(cont’d )\nλ\nSteel grade and design strength py (N/mm2)\nS275\nS355\nS460\n235\n245\n255\n265\n275\n315\n325\n335\n345\n355\n400\n410\n430\n440\n460\n122\n89\n90\n92\n93\n95\n99\n100\n101\n102\n103\n107\n107\n109\n109\n110\n124\n87\n88\n90\n91\n92\n97\n98\n99\n100\n100\n104\n104\n106\n106\n107\n126\n85\n86\n88\n89\n90\n94\n95\n96\n97\n98\n101\n102\n103\n103\n104\n128\n83\n84\n86\n87\n88\n92\n93\n94\n95\n95\n98\n99\n100\n100\n101\n130\n81\n82\n84\n85\n86\n90\n91\n91\n92\n93\n96\n96\n97\n98\n99\n135\n77\n78\n79\n80\n81\n84\n85\n86\n87\n87\n90\n90\n91\n92\n92\n140\n72\n74\n75\n76\n76\n79\n80\n81\n81\n82\n84\n85\n85\n86\n87\n145\n69\n70\n71\n71\n72\n75\n76\n76\n77\n77\n79\n80\n80\n81\n81\n150\n65\n66\n67\n68\n68\n71\n71\n72\n72\n73\n75\n75\n76\n76\n76\n155\n62\n63\n63\n64\n65\n67\n67\n68\n68\n69\n70\n71\n71\n72\n72\n160\n59\n59\n60\n61\n61\n63\n64\n64\n65\n65\n66\n67\n67\n67\n68\n165\n56\n56\n57\n58\n58\n60\n60\n61\n61\n61\n63\n63\n64\n64\n64\n170\n53\n54\n54\n55\n55\n57\n57\n58\n58\n58\n60\n60\n60\n60\n61\n175\n51\n51\n52\n52\n53\n54\n54\n55\n55\n55\n56\n57\n57\n57\n58\n180\n48\n49\n49\n50\n50\n51\n52\n52\n52\n53\n54\n54\n54\n54\n55\n185\n46\n46\n47\n47\n48\n49\n49\n50\n50\n50\n51\n51\n52\n52\n52\n190\n44\n44\n45\n45\n45\n47\n47\n47\n47\n48\n49\n49\n49\n49\n49\n195\n42\n42\n43\n43\n43\n45\n45\n45\n45\n45\n46\n46\n47\n47\n47\n200\n40\n41\n41\n41\n42\n43\n43\n43\n43\n43\n44\n44\n45\n45\n45\n210\n37\n37\n38\n38\n38\n39\n39\n39\n40\n40\n40\n40\n41\n41\n41\n220\n34\n34\n35\n35\n35\n36\n36\n36\n36\n36\n37\n37\n37\n37\n38\n230\n31\n32\n32\n32\n32\n33\n33\n33\n33\n34\n34\n34\n34\n34\n35\n240\n29\n29\n30\n30\n30\n30\n31\n31\n31\n31\n31\n31\n32\n32\n32\n250\n27\n27\n27\n28\n28\n28\n28\n28\n29\n29\n29\n29\n29\n29\n29\n260\n25\n25\n26\n26\n26\n26\n26\n26\n27\n27\n27\n27\n27\n27\n27\n270\n23\n24\n24\n24\n24\n24\n25\n25\n25\n25\n25\n25\n25\n25\n25\n280\n22\n22\n22\n22\n22\n23\n23\n23\n23\n23\n23\n24\n24\n24\n24\n290\n21\n21\n21\n21\n21\n21\n21\n22\n22\n22\n22\n22\n22\n22\n22\n300\n19\n19\n20\n20\n20\n20\n20\n20\n20\n20\n21\n21\n21\n21\n21\n310\n18\n18\n18\n19\n19\n19\n19\n19\n19\n19\n19\n19\n19\n19\n20\n320\n17\n17\n17\n17\n18\n18\n18\n18\n18\n18\n18\n18\n18\n18\n18\n330\n16\n16\n16\n16\n17\n17\n17\n17\n17\n17\n17\n17\n17\n17\n17\n340\n15\n15\n15\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n350\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\nDesign of compression members\n4.9.2 EFFECTIVE LENGTH\nAs mentioned in section 4.9.1, the compressive\nstrength of struts is primarily related to their slen-\nderness ratio. The slenderness ratio, λ, is given by\n    \nλ  \n= L\nr\nE\n(4.31)\nwhere\nLE\neffective length of the member\nr\nradius of gyration obtained from steel tables.\nThe concept of effective length was discussed in\nsection 4.8.11.1, in the context of lateral torsional\nbuckling, and is similarly applicable to the design of\nstruts and columns. The effective length is simply\na function of the actual length of the member and\nthe restraint at the member ends.\nThe formulae in Appendix C of BS 5950 and the\ngraph in Fig. 4.23 relate to standard restraint con-\nditions in which each end is pinned. In reality each\nend of the strut may be free, pinned, partially fixed,\n\n[TABLES]\n[[{\"0\": \"Table 4.14(cont’d )λ\", \"1\": \"Steel grade and design strength py (N/mm2)\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"S275235245255265275122899092939512487889091921268586888990128838486878813081828485861357778798081140727475767614569707171721506566676868155626363646516059596061611655656575858170535454555517551515252531804849495050185464647474819044444545451954242434343200404141414221037373838382203434353535230313232323224029293030302502727272828260252526262627023242424242802222222222290212121212130019192020203101818181932017171717183301616161617340151515161635015151515154.9.2EFFECTIVE LENGTHAs  mentioned in section  4.9.1,\", \"1\": \"S355315325335345355991001011021039798991001009495969798929394959590919192938485868787798081818275767677777171727273676768686963646465656060616161575758585854545555555152525253494950505047474747484545454545434343434339393940403636363636333333333430313131312828282929262626272724252525252323232323212122222220202020201919191919 1818181818171717171716161616161515151515the  compressive\", \"2\": \"S4604004104304404601071071091091101041041061061071011021031031049899100100101969697989990909192928485858687798080818175757676767071717272666767676863636464646060606061565757575854545454555151525252494949494946464747474444454545404041414137373737383434343435313132323229292929292727272727252525252523242424242222222222212121212119191919201818181818171717171716161616161515151515The concept of effective length was discussed insection  4.8.11.1,  in  the  context  of  lateral  torsional\"}], [], [], [{\"S355\": \"315 325 335 345 355\"}, {\"S355\": \"99 100 101 102 103\\n97 98 99 100 100\\n94 95 96 97 98\\n92 93 94 95 95\\n90 91 91 92 93\\n84 85 86 87 87\\n79 80 81 81 82\\n75 76 76 77 77\\n71 71 72 72 73\\n67 67 68 68 69\\n63 64 64 65 65\\n60 60 61 61 61\\n57 57 58 58 58\\n54 54 55 55 55\\n51 52 52 52 53\\n49 49 50 50 50\\n47 47 47 47 48\\n45 45 45 45 45\\n43 43 43 43 43\\n39 39 39 40 40\\n36 36 36 36 36\\n33 33 33 33 34\\n30 31 31 31 31\\n28 28 28 29 29\\n26 26 26 27 27\\n24 25 25 25 25\\n23 23 23 23 23\\n21 21 22 22 22\\n20 20 20 20 20\\n19 19 19 19 19\\n18 18 18 18 18\\n17 17 17 17 17\\n16 16 16 16 16\\n15 15 15 15 15\"}]]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0202.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 203, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n182\nL E = (0.5) 0.7L\nL E = 0.85L\nL E = (0.7) 0.85L\nL E = (1.0) 1.0L\nL E = (1.0) 1.2L\nL E = 1.5L\nL E = (2.0) 2.0L\n(1)\n(2)\n(3)\n(4)\n(5)\n(6)\n(7)\nFig. 4.24\nTable 4.15\nNominal effective length, LE, for a compression member (Table 22, BS 5950)\na) non-sway mode\nRestraint (in the plane under consideration) by other parts of the structure\nLE\nEffectively held in position at both ends\nEffectively restrained in direction at both ends (1)\n0.7L\nPartially restrained in direction at both ends (2)\n0.85L\nRestrained in direction at one end (3)\n0.85L\nNot restrained in direction at either end (4)\n1.0L\nb) sway mode\nOne end\nOther end\nEffectively held in position and restrained\nNot held\nEffectively restrained in direction (5)\n1.2L\nin direction\nin position\nPartially restrained in direction (6)\n1.5L\nNot restrained in direction (7)\n2.0L\nthat the design effective lengths are greater than\nthe theoretical values where one or both ends of the\nmember are partially or wholly restrained. This is\nbecause, in practice, it is difficult if not impossible\nto guarantee that some rotation of the member will\nnot take place. Furthermore, the effective lengths are\nalways less than the actual length of the compres-\nsion member except when the structure is unbraced.\nor fully fixed (rotationally). Also, whether or not the\ntop of the strut is allowed to move laterally with\nrespect to the bottom end is important i.e. whether\nthe structure is braced or unbraced. Figure 4.24\nsummarises these restraints, and Table 22 of BS\n5950, reproduced above as Table 4.15 stipulates\nconservative assumptions of effective length LE from\nwhich the slenderness λ can be calculated. Note\n\n[TABLES]\n[[{\"0\": \"Table 4.15Nominal effective length, L E, for a compression member (Table 22, BS 5950)\", \"1\": \"\"}, {\"0\": \"a) non-sway mode\", \"1\": \"\"}, {\"0\": \"Restraint (in the plane under consideration) by other parts of the structure\", \"1\": \"L E\"}, {\"0\": \"Effectively held in position at both endsEffectively restrained in direction at both ends (1)\", \"1\": \"0.7L\"}, {\"0\": \"Partially restrained in direction at both ends (2)\", \"1\": \"0.85L\"}, {\"0\": \"Restrained in direction at one end (3)\", \"1\": \"0.85L\"}, {\"0\": \"Not restrained in direction at either end (4)\", \"1\": \"1.0L\"}, {\"0\": \"b) sway mode\", \"1\": \"\"}, {\"0\": \"One endOther end\", \"1\": \"\"}, {\"0\": \"Effectively held in position and restrainedNot heldEffectively restrained in direction (5)\", \"1\": \"1.2L\"}, {\"0\": \"in directionin positionPartially restrained in direction (6)\", \"1\": \"1.5L\"}, {\"0\": \"Not restrained in direction (7)\", \"1\": \"2.0L\"}], [{\"0\": \"in directionin position\", \"1\": \"Partially restrained in direction (6)1.5L\"}, {\"0\": \"\", \"1\": \"Not restrained in direction (7)2.0L\"}, {\"0\": \"or fully fixed (rotationally). Also, whether or not the\", \"1\": \"that  the  design  effective  lengths  are  greater  than\"}, {\"0\": \"top  of  the  strut  is  allowed  to  move  laterally  with\", \"1\": \"the theoretical values where one or both ends of the\"}, {\"0\": \"respect to the bottom end is important i.e. whether\", \"1\": \"member  are  partially  or  wholly  restrained.  This  is\"}, {\"0\": \"the  structure  is  braced  or  unbraced.  Figure  4.24\", \"1\": \"because, in practice, it is difficult if not impossible\"}, {\"0\": \"summarises  these  restraints,  and  Table  22  of  BS\", \"1\": \"to guarantee that some rotation of the member will\"}, {\"0\": \"5950,  reproduced  above  as  Table  4.15  stipulates\", \"1\": \"not take place. Furthermore, the effective lengths are\"}, {\"0\": \"conservative assumptions of effective length L E from\", \"1\": \"always less than the actual length of the compres-\"}, {\"0\": \"which  the  slenderness  λ  can  be  calculated.  Note\", \"1\": \"sion member except when the structure is unbraced.\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0203.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 204, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n183\nExample 4.10 Design of an axially loaded column (BS 5950)\nA proposed 5 metre long internal column in a ‘rigid’ jointed steel structure is to be loaded concentrically with 1000 kN\ndead and 1000 kN imposed load (Fig. 4.25 ). Assuming that fixity at the top and bottom of the column gives effective\nrotational restraints, design column sections assuming the structure will be (a) braced and (b) unbraced.\nDesign of compression members\nFig. 4.25\n1000 kN dead load\n1000 kN imposed load\nFloors\nColumns\nBRACED COLUMN\nDesign axial loading\nFactored loading, Fc = (1.4 × 1000) + (1.6 × 1000) = 3000 kN\nEffective length\nFor the braced case the column is assumed to be effectively held in position at both ends, and restrained in direction\nat both ends. It will buckle about the weak (y–y) axis. From Table 4.15 therefore, the effective length, LE, is\nLE = 0.7L = 0.7 × 5 = 3.5 m\nSection selection\nThis column design can only really be done by trial and error.\nInitial trial. Try 254 × 254 × 107 UC:\npy = 265 N/mm2\nry = 65.7 mm\nAg = 13 700 mm2\nb/T = 6.3\nd/t = 15.4\nλ = LE/ry = 3500/65.7 = 53\nFrom Table 4.12, use Table 24(c) of BS 5950 (i.e. Table 4.14 ), from which pc = 208 N/mm2. UC section is not slender\nsince b/T < 15ε = 15 × (275/265)0.5 = 15.28 and d/t < 40ε = 40.74 (Table 4.4). From clause 4.7.4 of BS 5950,\ncompression resistance of column, Pc, is\nPc = Agpc = 13700 × 208/103 = 2850 kN < 3000 kN\nNot OK\nSecond trial. Try 305 × 305 × 118 UC:\npy = 265 N/mm2\nry = 77.5 mm\nAg = 15 000 mm2\nb/T = 8.20\nd/t = 20.7\nλ = LE/ry = 3500/77.5 = 45\nThen from Table 24(c) of BS 5950 (i.e. Table 4.14), pc = 222 N/mm2. UC section is not slender then\nPc = Agpc = 15000 × 222/103 = 3330 kN > 3000 kN\nOK\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"1000 kN dead load1000 kN imposed load\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"Floors\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}], [], []]\n\n[FORMULAS]\n5 = 3; ry = 65; b/T = 6; .7 = 53; .5 = 15; ry = 77; b/T = 8; .5 = 45", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0204.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 205, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n184\n4.9.3 COLUMNS WITH BENDING MOMENTS\nAs noted earlier, most columns in steel structures\nare subject to both axial load and bending. Ac-\ncording to clause 4.8.3.1 of BS 5950, such members\nshould be checked (for yielding or local buckling)\nat the points of greatest bending moment and axial\nload, which usually occur at the member ends. In\naddition, the buckling resistance of the member as\na whole should be checked.\n4.9.3.1 Cross-section capacity check\nThe purpose of this check is to ensure that\nnowhere across the section does the steel stress\nexceed yield. Generally, except for class 4 slender\ncross-sections, clause 4.8.3.2 states that the follow-\ning relationship should be satisfied:\n    \nF\nA p\nM\nM\nM\nM\nc\ng\ny\nx\ncx\ny\ncy\n  \n  \n  \n+\n+\n≤1\n(4.32)\nwhere\nFc\naxial compression load\nAg\ngross cross-sectional area of section\npy\ndesign strength of steel\nMx\napplied major axis moment\nMcx\nmoment capacity about the major axis in\nthe absence of axial load\nMy\napplied minor axis moment\nMcy\nmoment capacity about the minor axis in\nthe absence of axial load\nWhen the section is slender the expression in\nclause 4.8.3.2 (c) should be used. Note that\nparagraph (b) of this clause gives an alternative\nexpression for calculating the (local) capacity of\ncompression members of class 1 plastic or class 2\ncompact cross-section, which yields a more exact\nestimate of member strength.\n4.9.3.2 Buckling resistance check\nBuckling due to imposed axial load, lateral tor-\nsional buckling due to imposed moment, or a com-\nbination of buckling and lateral torsional buckling\nare additional possible modes of failure in most\npractical columns in steel structures.\nClause 4.8.3.3.1 of BS 5950 gives a simplified\napproach for calculating the buckling resistance of\ncolumns which involves checking that the follow-\ning relationships are both satisfied:\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+\n≤1\n(4.33)\n    \nF\nP\nm\nM\nM\nm M\np Z\nc\ncy\nLT\nLT\nb\ny\ny\ny\ny\n  \n  \n  \n+\n+\n≤1\n(4.34)\nwhere\nPc\nsmaller of Pcx and Pcy\nPcx\ncompression resistance of member\nconsidering buckling about the major axis\nPcy\ncompression resistance of member\nconsidering buckling about the minor axis\nmx\nequivalent uniform moment factor for\nmajor axis flexural buckling obtained\nfrom Table 26 of BS 5950, reproduced\nas Table 4.16\nmy\nequivalent uniform moment factor for minor\naxis flexural buckling obtained from Table\n26 of BS 5950, reproduced as Table 4.16\nmLT\nequivalent uniform moment factor for lateral\ntorsional buckling obtained from Table 18\nof BS 5950, reproduced as Table 4.10\nMLT\nmaximum major axis moment in the\nsegment length Lx governing Pcx\nMb\nbuckling resistance moment\nZx\nelastic modulus about the major axis\nZy\nelastic modulus about the minor axis and\nthe other symbols are as above.\nUNBRACED COLUMN\nFor the unbraced case, LE = 1.2L = 6.0 metres from Table 4.15, and the most economic member would appear to be\n305 × 305 × 158 UC:\npy = 265 N/mm2\nry = 78.9 mm\nAg = 20 100 mm2\nb/T = 6.21\nd/t = 15.7\nλ = LE/ry = 6000/78.9 = 76\nThen from Table 4.14, pc = 165 N/mm2. Section is not slender\nPc = Agpc = 20 100 × 165/103 = 3317 kN > 3000 kN\nOK\nHence, it can immediately be seen that for a given axial load, a bigger steel section will be required if the column is\nunbraced.\nExample 4.10 continued\n\n[TABLES]\n[[{\"0\": \"Design in structural steelwork to BS 5950\"}, {\"0\": \"Example 4.10 continued\"}, {\"0\": \"UNBRACED COLUMN\"}, {\"0\": \"For the unbraced case, L E = 1.2L = 6.0 metres from Table 4.15, and the most economic member would appear to be305 × 305 × 158 UC:\"}, {\"0\": \"b/T = 6.21d/t = 15.7py = 265 N/mm2ry = 78.9 mmAg = 20 100 mm2\"}, {\"0\": \"λ = L E/r y = 6000/78.9 = 76\"}, {\"0\": \"Then from Table 4.14, pc = 165 N/mm2. Section is not slender\"}, {\"0\": \"OKPc = Agpc = 20 100 × 165/103 = 3317 kN > 3000 kN\"}, {\"0\": \"Hence, it can immediately be seen that for a given axial load, a bigger steel section will be required if the column is\"}, {\"0\": \"unbraced.\"}, {\"0\": \"4.9.3.2 Buckling resistance check4.9.3 COLUMNS WITH BENDING MOMENTS\"}, {\"0\": \"Buckling  due  to  imposed  axial  load,  lateral  tor-As  noted  earlier,  most  columns  in  steel  structures\"}, {\"0\": \"sional buckling due to imposed moment, or a com-are  subject  to  both  axial  load  and  bending.  Ac-\"}, {\"0\": \"bination of buckling and lateral torsional bucklingcording to clause 4.8.3.1 of BS 5950, such members\"}, {\"0\": \"are  additional  possible  modes  of  failure  in  mostshould  be  checked  (for  yielding  or  local  buckling)\"}, {\"0\": \"practical columns in steel structures.at the points of greatest bending moment and axial\"}, {\"0\": \"Clause  4.8.3.3.1  of  BS  5950  gives  a  simplifiedload, which usually occur at the member ends. In\"}, {\"0\": \"approach for calculating the buckling resistance ofaddition, the buckling resistance of the member as\"}, {\"0\": \"columns  which  involves  checking  that  the  follow-a whole should be checked.\"}, {\"0\": \"ing relationships are both satisfied:\"}, {\"0\": \"F P4.9.3.1 Cross-section capacity checkm Mm Myycxx++     ≤ 1(4.33)\"}, {\"0\": \"The  purpose  of this  check is to  ensure thatp Zp Z\"}, {\"0\": \"cyxyy    nowhere  across  the  section  does  the  steel  stress\"}, {\"0\": \"F Pm MmMexceed  yield.  Generally,  except  for  class  4  slenderyycLTLT++     ≤ 1(4.34)\"}, {\"0\": \"cross-sections, clause 4.8.3.2 states that the follow-Mp Zcybyy\"}, {\"0\": \"ing relationship should be satisfied:where\"}, {\"0\": \"M MM MPcsmaller of Pcx and PcyFycx++     ≤ 1(4.32)\"}, {\"0\": \"PcxA pgycxcy    considering buckling about the major axis\"}, {\"0\": \"compression resistance of memberPcywhere\"}, {\"0\": \"considering buckling about the minor axis axial compression loadFc\"}, {\"0\": \"equivalent uniform moment factor form xgross cross-sectional area of sectionAg\"}, {\"0\": \"major axis flexural buckling obtaineddesign strength of steelpy\"}, {\"0\": \"from Table 26 of BS 5950, reproducedapplied major axis momentM x\"}, {\"0\": \"as Table 4.16moment capacity about the major axis inMcx\"}, {\"0\": \"equivalent uniform moment factor for minorm ythe absence of axial load\"}, {\"0\": \"axis flexural buckling obtained from Tableapplied minor axis momentM y\"}, {\"0\": \"26 of BS 5950, reproduced as Table 4.16moment capacity about the minor axis inMcy\"}, {\"0\": \"equivalent uniform moment factor for lateralm LTthe absence of axial load\"}, {\"0\": \"torsional buckling obtained from Table 18\"}, {\"0\": \"When  the  section  is  slender  the  expression  inof BS 5950, reproduced as Table 4.10\"}, {\"0\": \"clause  4.8.3.2 (c) should  be  used.  Note thatmaximum major axis moment in theM LT\"}, {\"0\": \"paragraph  (b)  of  this  clause  gives  an  alternativesegment length L x governing Pcx\"}, {\"0\": \"expression  for  calculating  the  (local)  capacity  ofbuckling resistance momentM b\"}, {\"0\": \"compression  members  of  class  1  plastic  or  class  2elastic modulus about the major axisZ x\"}, {\"0\": \"compact  cross-section,  which  yields  a  more  exactelastic modulus about the minor axis andZ y\"}, {\"0\": \"estimate of member strength.the other symbols are as above.\"}, {\"0\": \"184\"}], [], []]\n\n[FORMULAS]\nLE = 1; ry = 78; b/T = 6; .9 = 76", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0205.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 206, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n185\nTable 4.16\nEquivalent uniform moment factor m for flexural buckling\n(Table 26, BS 5950)\n1.00\n0.96\n0.92\n0.88\n0.84\n0.80\n0.76\n0.72\n0.68\n0.64\n0.60\n0.58\n0.56\n0.54\n0.52\n0.50\n0.48\n0.46\n0.44\n0.42\n0.40\nSegments with end moments only\nβ positive\nx\nx\nM\nL\nβM\nx\nx\nx\nx βM\nM\nL\nSegments between intermediate lateral restraints\nSpecific cases\nx\nx\nL\nx\nx\nL\nx\nx\nL\nx\nx\nL\nm = 0.90\nm = 0.95\nm = 0.95\nm = 0.80\nβ\nm\n1.0\n0.9\n0.8\n0.7\n0.6\n0.5\n0.4\n0.3\n0.2\n0.1\n0.0\n−0.1\n−0.2\n−0.3\n−0.4\n−0.5\n−0.6\n−0.7\n−0.8\n−0.9\n−1.0\nβ negative\nDesign of compression members\nExample 4.11 Column resisting an axial load and bending (BS 5950)\nSelect a suitable column section in grade S275 steel to support a factored axial concentric load of 2000 kN and\nfactored bending moments of 100 kN m about the major axis, and 20 kN m about the minor axis (Fig. 4.26), applied\nat both ends of the column. The column is 10 m long and is fully fixed against rotation at top and bottom, and the\nfloors it supports are braced against sway.\nINITIAL SECTION SELECTION\n305 × 305 × 118 UC:\npy = 265 N/mm2, plastic\nSx = 1950 cm3\nZx = 1760 cm3\nAg = 150 cm2\nSy = 892 cm3\nZy = 587 cm3\nt = 11.9 mm\nx = D/T = 314.5/18.7\nu = 0.9\nd = 246.6 mm\n= 16.8\nry = 7.75 cm\nNote: In this case the classification procedure is slightly different in respect of web classification. From Table 11 of\nBS 5950 r1 = Fc/dtpy but −1 < r1 ≤ 1\n= 2 × 106/246.6 × 11.9 × 265 = 2.87\nTherefore, r1 = 1. Limiting d/t = 80ε/1 + r1 = \n  \n80\n1\n1  \n+\nε = 40(276/265)1/2 = 40.75\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"Design of compression members\", \"5\": \"\"}, {\"0\": \"Table 4.16(Table 26, BS 5950)β positivexx\", \"1\": \"\", \"2\": \"Equivalent uniform moment factor m for flexural bucklingSegments with end moments onlyβ negative\", \"3\": \"\", \"4\": \"βm1.001.0\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"0.960.9\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"MLβM\", \"2\": \"MxxLxxβM\", \"3\": \"\", \"4\": \"0.920.80.880.70.840.60.800.50.760.40.720.30.680.20.640.10.600.00.58−0.10.56−0.20.54−0.30.52−0.40.50−0.50.48−0.60.46−0.70.44−0.80.42−0.9−1.0\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"0.40Segments between intermediate lateral restraints\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}, {\"0\": \"\", \"1\": \"Specific cases\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}, {\"0\": \"xL = 0.90m\", \"1\": \"xxxxxxxLLLm = 0.95m = 0.95m = 0.80\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\"}], [], []]\n\n[FORMULAS]\nt = 11; 265 = 2; r1 = 1", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0206.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 207, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n186\nExample 4.11 continued\n2000 kN\n20 kN m\n100 kN m\n10 m\nFig. 4.26\nActual d/t = 20.7, hence web is plastic.\nb/T = 8.20 < 9ε = 9(275/265)1/2 = 9.17, hence flange is plastic.\nMcx = pySx = 265 × 1 950 × 10−3 = 516.75 kN m\nMcy = pySy = 265 × 892 × 10−3 = 236.38 kN m\nCROSS-SECTION CAPACITY CHECK\nSubstituting into \n    \nF\nA p\nM\nM\nM\nM\nc\ng\ny\nx\ncx\ny\ncy\n  \n  \n+\n+\n gives\n  \n2000  10\n150  10\n3\n2\n×\n×\n×\n+\n+\n=\n+\n+\n=\n<\n  \n  \n.\n  \n.\n  .\n  .\n  .\n  .\n  \n265\n100\n516 75\n20\n236 38\n0 503\n0 193\n0 085\n0 781\n1\nOK\nBUCKLING RESISTANCE CHECK\nIn-plane buckling\nFrom Table 4.15, effective length LE = 0.7L = 7 m\nλ = LE/ry = 7000/77.5 = 90\nFrom Table 4.12, for buckling about the x–x axis and y–y axis use, respectively, Table 24(b) of BS 5950 (Table 4.13)\nfrom which pcx = 157 N/mm2, and Table 24(c) of BS 5950 (Table 4.14) from which pcy = 139 N/mm2. Then\nPcx = Agpcx = 150 × 102 × 157 × 10−3 = 2355 kN\nPcy = Agpcy = 150 × 102 × 139 × 10−3 = 2085 kN = Pc\nRatio of end moments about both x–x and y–y axes, β = 1. Hence from Table 4.16, mx = my = 1\nSubstituting into \n    \nF\nP\nm M\np Z\nm M\np Z\nc\nc\nx\nx\ny\nx\ny\ny\ny\ny\ngives\n  \n  \n \n+\n+\n  \n2000\n2085\n10\n10\n3\n3\n  \n  \n  \n  \n+\n×\n×\n×\n+\n×\n×\n×\n−\n−\n1  100\n265  1760\n1  20\n265  587\n = 0.96 + 0.21 + 0.13 = 1.30 > 1\nNot OK\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}], [], []]\n\n[FORMULAS]\n.5 = 90; β = 1; 0.13 = 1", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0207.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 208, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n187\nDesign of compression members\nExample 4.11 continued\nLateral torsional buckling\n    \n(\n)\n  ( . )\n.   \n.\n.\nβw\nE\ny\n0 5\n0 5\n1 0\n7000\n77 5\n90\nL\nr\n=\n=\nand\nD/T = 16.8\nFrom Table 4.7, pb = 196 N/mm2\nMb = pbSx = 196 × 1950 × 10−3 = 382.2 kN m\nRatio of end moments about both major axes, β = 1. Hence from Table 4.10, mLT = 1\nSubstituting into \n    \nF\nP\nm M\nM\nm M\np Z\nc\ncy\nLT\nLT\nb\ny\ny\ny\ny\ngives\n  \n  \n \n+\n+\n  \n2000\n2085\n382 2\n10 3\n  \n.\n  \n \n+\n×\n+\n×\n×\n×\n−\n1  100\n1  20\n265  587\n = 0.96 + 0.26 + 0.13 = 1.35 > 1\nNot OK\nHence, a bigger section should be selected.\nSECOND SECTION SELECTION\nTry 356 × 368 × 177 UC:\npy = 265 N/mm2, plastic\nSx = 3460 cm3\nZx = 3100 cm3\nAg = 226 cm2\nSy = 1670 cm3\nZy = 1100 cm3\nry = 9.52 cm3\nx = D/T = 368.3/23.8\nu = 0.9\n= 15.5\nMcx = pySx = 265 × 3460 × 10−3 = 916.9 kN m\nMcy = pySy = 265 × 1670 × 10−3 = 442.55 kN m\nCROSS-SECTION CAPACITY CHECK\nAgain using \n    \nF\nA p\nM\nM\nM\nM\nc\ng\ny\nx\ncx\ny\ncy\ngives\n  \n  \n \n+\n+\n  \n2000  10\n226  10\n3\n2\n×\n×\n×\n+\n+\n \n  \n.   \n.\n265\n100\n916 9\n20\n442 55  = 0.33 + 0.11 + 0.05 = 0.49 < 1\nOK\nBUCKLING RESISTANCE CHECK\nIn-plane buckling\nFrom Table 4.15, effective length LE = 0.7L = 7 m\nλ = LE/ry = 7000/95.2 = 73.5\nFrom Table 4.12, for buckling about the x–x axis and y–y axis use, respectively, Table 24(b) of BS 5950 (Table 4.13)\nfrom which pcx = 190 N/mm2, and Table 24(c) of BS 5950 (Table 4.14) from which pcy = 169 N/mm2. Then\nPcx = Agpcx = 226 × 102 × 190 × 10−3 = 4294 kN\nPcy = Agpcy = 226 × 102 × 169 × 10−3 = 3819.4 kN = Pc\nRatio of end moments about both x–x and y–y axes, β = 1. Hence from Table 4.16, mx = my = 1\nSubstituting into \n    \nF\nP\nm M\np Z\nm M\np Z\nc\nc\nx\nx\ny\nx\ny\ny\ny\ny\ngives\n  \n  \n \n+\n+\n  \n2000\n3819 4\n10\n10\n3\n3\n.   \n  \n  \n  \n+\n×\n×\n×\n+\n×\n×\n×\n−\n−\n1  100\n265  3100\n1  20\n265  1100\n = 0.52 + 0.12 + 0.07 = 0.71 < 1\nOK\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"Design of compression members\"}, {\"0\": \"\", \"1\": \"Example 4.11 continued\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"Lateral torsional buckling\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"β()w\", \"3\": \"L rE0 5=  (\", \"4\": \"1 0\", \"5\": \")\", \"6\": \"70000 5\", \"7\": \"=andD/T  =  16.8  90\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"y\", \"4\": \"\", \"5\": \"\", \"6\": \"77 5\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"From Table 4.7, p b = 196 N/mm2\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"M b = pbSx = 196 × 1950 × 10−3 = 382.2 kN m\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"Ratio of end moments about both major axes, β = 1. Hence from Table 4.10, m LT = 1\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"Substituting into\", \"1\": \"F Pc\", \"2\": \"m Mm MLT++\", \"3\": \"y gives\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"cy\", \"2\": \"Mp Zby\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"2000\", \"2\": \"×1  100++\", \"3\": \"×1\", \"4\": \"20\", \"5\": \"\", \"6\": \"\", \"7\": \"= 0.96 + 0.26 + 0.13 = 1.35 > 1\", \"8\": \"Not OK\"}, {\"0\": \"\", \"1\": \"2085\", \"2\": \"382 2\", \"3\": \"×265\", \"4\": \"587\", \"5\": \"×\", \"6\": \"−10 3\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"Hence, a bigger section should be selected.\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"SECOND SECTION SELECTION\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"Try 356 × 368 × 177 UC:\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"py\", \"3\": \"= 265 N/mm2, plastic\", \"4\": \"\", \"5\": \"\", \"6\": \"Sx\", \"7\": \"= 3460 cm3= 3100 cm3Z x\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"= 226 cm2Ag\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"Sy\", \"7\": \"= 1670 cm3= 1100 cm3Zy\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"= 9.52 cm3ry\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"x = D/T = 368.3/23.8\", \"7\": \"u = 0.9\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"= 15.5\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Mcx = pySx = 265 × 3460 × 10−3 = 916.9 kN m\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Mcy = pySy = 265 × 1670 × 10−3 = 442.55 kN m\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"CROSS-SECTION CAPACITY CHECK\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"Fc Again using\", \"1\": \"+\", \"2\": \"M MM Mxy+   gives\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"A py\", \"1\": \"\", \"2\": \"cxcy\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"2×32000  10\", \"3\": \"100+\", \"4\": \"+\", \"5\": \"\", \"6\": \"20\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"××226  10 265\", \"3\": \"916 9\", \"4\": \"\", \"5\": \"\", \"6\": \"442 55\", \"7\": \"= 0.33 + 0.11 + 0.05 = 0.49 < 1\", \"8\": \"OK\"}, {\"0\": \"\", \"1\": \"BUCKLING RESISTANCE CHECK\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"In-plane buckling\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"From Table 4.15, effective length L E = 0.7L = 7 m\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"λ = L E/ry = 7000/95.2 = 73.5\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"From Table 4.12, for buckling about the x–x axis and y–y axis use, respectively, Table 24( b) of BS 5950 (Table 4.13)\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"from which pcx = 190 N/mm2, and Table 24(c) of BS 5950 (Table 4.14 ) from which pcy = 169 N/mm2. Then\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Pcx = Ag pcx = 226 × 102 × 190 × 10−3 = 4294 kN\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"Pcy = Ag pcy = 226 × 102 × 169 × 10−3 = 3819.4 kN = Pc\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"Ratio of end moments about both x–x and y–y axes, β = 1. Hence from Table 4.16, m x = m y = 1\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"Substituting into\", \"1\": \"F Pc\", \"2\": \"m Mm Mxy++\", \"3\": \"gives\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}, {\"0\": \"\", \"1\": \"c\", \"2\": \"p Zp Zxy\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\", \"7\": \"\", \"8\": \"\"}], [], []]\n\n[FORMULAS]\nβ = 1; mLT = 1; 0.13 = 1; ry = 9; .9\n= 15; 0.05 = 0; .2 = 73; β = 1; 0.07 = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0208.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 209, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n188\nFig. 4.27 Load eccentricity for columns in simple construction.\nBeam R\nBeam S\nSection L–L\n(a)\nx\nex\nD/2 100\n100 t/2\nx\nFx\ny\nFy\nLoad from\nbeam S\nD\nLoad from\nbeam R\nA\nL\nB\nL\nLoading at column edge\ndue to beam on cap plate\n(c)\n(b)\nExample 4.11 continued\nLateral torsional buckling\n    \n(\n)\n  ( . )\n.   \n.\n.\n.\nβw\nE\ny\n0 5\n0 5\n1 0\n7000\n95 2\n73 5\nL\nr\n=\n=\nand\nD/T = 15.5\nFrom Table 4.7, pb = 220 N/mm2\nMb = pbSx = 220 × 3460 × 10−3 = 761.2 kN m\nRatio of end moments about both major axes, β = 1. Hence from Table 4.10, mLT = 1\nSubstituting into \n    \nF\nP\nm M\nM\nm M\np Z\nc\ncy\nLT\nLT\nb\ny\ny\ny\ny\ngives\n  \n  \n   \n+\n+\n≤1\n  \n2000\n3819 4\n761 2\n10 3\n.   \n.\n  \n  \n+\n×\n+\n×\n×\n×\n−\n1  100\n1  20\n265  1100\n = 0.52 + 0.13 + 0.07 = 0.72 < 1\nOK\nHence a 356 × 368 × 177 UC section is satisfactory.\nClause 4.8.3.3.2 of BS 5950 gives a more exact\napproach, but as in practice most designers tend to\nuse the simplified approach, the more exact method\nis not discussed here.\n4.9.4 COLUMN DESIGN IN ‘SIMPLE’\nCONSTRUCTION\nAt first sight it would appear that columns in\nso-called ‘simple construction’ are not subject to\nmoments, as the beams are all joined at connec-\ntions which allow no moment to develop. In fact,\nin most cases there is a bending moment due to\nthe eccentricity of the shear load from the beam.\nThis is summarised in Clause 4.7.7 of BS 5950\nand illustrated in Fig. 4.27. Note that where a\nbeam sits on a column cap plate, for example at A\n(Fig. 4.27(c)), it can be assumed that the reaction\nfrom the beam acts at the face of the column. How-\never, where the beam is connected to a column\nby means of a ‘simple’ connection, e.g. using web\ncleats, the reaction from the beam can be assumed\nto act 100 mm from the column (web or flange)\nface as illustrated in Fig. 4.27(b).\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"Beam S\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"Section L–L\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\", \"5\": \"\", \"6\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\", \"4\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"D\", \"4\": \"\"}], [{\"0\": \"LB\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"Loading at column edgedue to beam on cap plateL\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\nβ = 1; mLT = 1; 0.07 = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0209.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 210, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n189\nwhere\nFc\naxial compressive load\nPc\n= Agpc – for all classes except class 4 (clause\n4.7.4 of BS 5950) – in which Ag is the gross\ncross-sectional area of the section and pc is\nthe compressive strength, see section 4.9.1\nMx\nnominal major axis moment\nMy\nnominal minor axis moment\nMbs\nbuckling resistance moment for ‘simple’\ncolumns\npy\ndesign strength of steel\nZy\nelastic modulus about the minor axis.\nNote that this expression is similar to that used\nfor checking the buckling resistance of columns in\ncontinuous structures (equation 4.34), but with all\nequivalent moment factors taken as 1.0. For I- and\nH-sections Mbs = Mb determined as discussed in\nsection 4.8.11.3 but using the equivalent slender-\nness of the column, λLT, given by\nλLT = 0.5L /ry\n(4.36)\nwhere\nL\nis the distance between levels at which the\ncolumn is laterally restrained in both directions\nry\nis the radius of gyration about the minor axis.\nFig. 4.29\nDesign of compression members\nUltimate\nreaction from\nBeam B – 75 kN\nUltimate\nreaction from\nBeam A – 200 kN\nPlan on cap\nL = 7 m\nF\nUltimate load due\nto self-weight 5 kN\nL E = 0.85L\nEffectively held in position\nbut not restrained in\ndirection, i.e. pinned\nEffectively held in position\nand direction, i.e. fixed\nFig. 4.28 Column supporting a roof truss.\nF\nRoof truss\nColumn\nCap plate\nExample 4.12 Design of a steel column in ‘simple’ construction (BS 5950)\nSelect a suitable column section in S275 steel to support the ultimate loads from beams A and B shown in Fig. 4.29.\nAssume the column is 7 m long and is effectively held in position at both ends but only restrained in direction at the\nbottom.\nWhen a roof truss is supported on a column cap\nplate (Fig. 4.28), and the connection is unable to\ndevelop significant moments, it can be assumed\nthat the load from the truss is transmitted concen-\ntrically to the column.\nColumns in simple construction will not need\nto be checked for local capacity but it will still\nbe necessary to check for buckling, which involves\nsatisfying the following relationship:\n    \nF\nP\nM\nM\nM\np Z\nc\nc\nx\nbs\ny\ny\ny\n 1\n  \n  \n \n+\n+\n≤\n(4.35)\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\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"F\", \"4\": \"\"}], [{\"0\": \"\", \"1\": \"\", \"2\": \"to self-weight 5 kN\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0210.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 211, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n190\nSECTION SELECTION\nThis can only really be done by trial and error. Therefore, try a: 203 × 203 × 52 UC: Sx = 568 cm3, plastic.\nDESIGN LOADING AND MOMENTS\nUltimate reaction from beam A, RA = 200 kN; ultimate reaction from beam B, RB = 75 kN; assume self-weight of\ncolumn = 5 kN. Ultimate axial load, F, is\nF = RA + RB + self-weight of column\n= 200 + 75 + 5 = 280 kN\nLoad eccentricity for beam A,\nex = D/2 + 100 = 206.2/2 + 100 = 203.1 mm\nLoad eccentricity for beam B,\ney = t/2 + 100 = 8/2 + 100 = 104 mm\nMoment due to beam A,\nMx = RAex = 200 × 103 × 203.1 = 40.62 × 106 N mm\nMoment due to beam B,\nMy = RBey = 75 × 103 × 104 = 7.8 × 106 N mm\nEFFECTIVE LENGTH\nFrom Table 4.15, effective length coefficient = 0.85. Hence, effective length is\nLE = 0.85L = 0.85 × 7000 = 5950 mm\nBENDING STRENGTH\nFrom Table 4.12, relevant compressive strength values for buckling about the x–x axis are obtained from Table 24(b)\n(Table 4.13) and from Table 24(c) (Table 4.14) for bending about the y–y axis.\nλx = LE/rx = 5950/89 = 66.8\nFrom Table 4.13, pc = 208 N/mm2.\nλy = LE/ry = 5950/51.6 = 115.3\nFrom Table 4.14, pc = 103 N/mm2. Hence critical compressive strength of column is 103 N/mm2.\nBUCKLING RESISTANCE\nλLT = 0.5L/ry = 0.5 × 7000/51.6 = 67.8\nFrom Table 4.9, p b = 193 N/mm2. Buckling resistance moment capacity of column, Mbs, is given by\nMbs = Mb = pbSx = 193 × 568 × 103 = 109.6 × 106 N mm\nHence for stability,\n    \nF\nP\nM\nM\nM\np Z\nc\nc\nx\nbs\ny\ny\ny\n 1\n  \n  \n \n+\n+\n≤\n  \n280  10\n66.4  10\n40.6  10\n109.6  10\n7.8  10\n275  174  10\n3\n2\n6\n6\n6\n3\n×\n×\n×\n+\n×\n×\n+\n×\n×\n×\n \n  \n  \n103\n = 0.41 + 0.37 + 0.16 = 0.94 < 1\nTherefore, the 203 × 203 × 52 UC section is suitable.\nExample 4.12 continued\n\n[TABLES]\n[[{\"0\": \"Design in structural steelwork to BS 5950\"}, {\"0\": \"Example 4.12 continued\"}, {\"0\": \"SECTION SELECTION\"}, {\"0\": \"This can only really be done by trial and error. Therefore, try a: 203 × 203 × 52 UC: Sx = 568 cm3, plastic.\"}, {\"0\": \"DESIGN LOADING AND MOMENTS\"}, {\"0\": \"Ultimate  reaction  from  beam  A,  RA  =  200  kN;  ultimate  reaction  from  beam  B,  R B  =  75  kN;  assume  self-weight  ofcolumn = 5 kN. Ultimate axial load, F, is\"}, {\"0\": \"F = RA + RB + self-weight of column\"}, {\"0\": \"= 200 + 75 + 5 = 280 kN\"}, {\"0\": \"Load eccentricity for beam A,\"}, {\"0\": \"ex = D/2 + 100 = 206.2/2 + 100 = 203.1 mm\"}, {\"0\": \"Load eccentricity for beam B,\"}, {\"0\": \"e y = t /2 + 100 = 8/2 + 100 = 104 mm\"}, {\"0\": \"Moment due to beam A,\"}, {\"0\": \"M x = RAe x = 200 × 103 × 203.1 = 40.62 × 106 N mm\"}, {\"0\": \"Moment due to beam B,\"}, {\"0\": \"M y = RBey = 75 × 103 × 104 = 7.8 × 106 N mm\"}, {\"0\": \"EFFECTIVE LENGTH\"}, {\"0\": \"From Table 4.15, effective length coefficient = 0.85. Hence, effective length is\"}, {\"0\": \"L E = 0.85L = 0.85 × 7000 = 5950 mm\"}, {\"0\": \"BENDING STRENGTH\"}, {\"0\": \"From Table 4.12, relevant compressive strength values for buckling about the x–x axis are obtained from Table 24( b)\"}, {\"0\": \"(Table 4.13) and from Table 24(c) (Table 4.14 ) for bending about the y–y axis.\"}, {\"0\": \"λx = LE/rx = 5950/89 = 66.8\"}, {\"0\": \"From Table 4.13, pc = 208 N/mm2.\"}, {\"0\": \"λy = LE/ry = 5950/51.6 = 115.3\"}, {\"0\": \"From Table 4.14, pc = 103 N/mm2. Hence critical compressive strength of column is 103 N/mm2.\"}, {\"0\": \"BUCKLING RESISTANCE\"}, {\"0\": \"λLT = 0.5L /ry = 0.5 × 7000/51.6 = 67.8\"}, {\"0\": \"From Table 4.9, p b = 193 N/mm2. Buckling resistance moment capacity of column, M bs, is given by\"}, {\"0\": \"M bs = M b = p bSx = 193 × 568 × 103 = 109.6 × 106 N mm\"}, {\"0\": \"Hence for stability,\"}, {\"0\": \"F PM MMcxy++≤      1\"}, {\"0\": \"p Zcbsy\"}, {\"0\": \"366××280  1040.6  107.8  10++\"}, {\"0\": \"× ×     = 0.41 + 0.37 + 0.16 = 0.94 < 1263××××66.4  10109.6  10275  174  10 103\"}, {\"0\": \"Therefore, the 203 × 203 × 52 UC section is suitable.\"}, {\"0\": \"190\"}], [], []]\n\n[FORMULAS]\n100 = 203; .1 = 40; cient = 0; .6 = 115; 7000/51.6 = 67; 103\n = 0; 0.16 = 0", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0211.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 212, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n191\n4.9.5 SUMMARY OF DESIGN PROCEDURES FOR\nCOMPRESSION MEMBERS\nAxially loaded members\n1. Determine ultimate axial load Fc.\n2. Select trial section and check it is non-slender.\n3. Determine rx, ry and Ag from steel tables.\n4. Determine effective lengths, LEX and LEY, using\nTable 4.15.\n5. Calculate slenderness ratios, λEX (= LEX/rx) and\nλEY (= LEY/ry).\n6. Select suitable strut curves from Table 4.12.\n7. Determine compressive strength, pc, using Table\n4.13, 4.14 or similar.\n8. Calculate compression resistance of member,\nPc = Agpc.\n9. Check Fc ≤ Pc. If unsatisfactory return to 2.\nMembers subject to axial load and bending\n1. Determine ultimate axial load, Fc, and bend-\ning moments, Mx and My.\n2. Select and classify trial section.\n3. Calculate moment capacities of section, Mcx\nand Mcy. If either Mx > Mcx or My > Mcy return\nto 2.\n4. Check cross-section capacity of section via\nequation 4.32. If unsatisfactory return to 2.\n5. Determine effective lengths, LEX and LEY, using\nTable 4.15.\n6. Calculate slenderness ratios, λEX (= LEX/rx) and\nλEY (= LEY/ry).\n7. Select suitable strut curves from Table 4.12.\n8. Determine the major and minor axes com-\npressive strengths, pcx and pcy, using Table 4.13,\n4.14 or similar.\n9. Calculate compressive resistances, Pcx (= pcxAg)\nand Pcy (= pcy Ag).\n10. Evaluate buckling resistance of section, Mb.\n11. Determine equivalent uniform moment factors\nfor flexural buckling, mx and my, using Table\n4.16.\n12. Check buckling resistance of member using\nequations (4.33) and (4.34). If unsatisfactory\nreturn to 2.\nCompression members in simple construction\n1. Determine ultimate axial load, Fc, and bend-\ning moments, Mx and My.\n2. Select and classify trial section.\n3. Determine effective lengths, LEX and LEY, using\nTable 4.15.\n4. Calculate slenderness ratios, λEX (= LEX/rx) and\nλEY (= LEY/ry).\nDesign of compression members\n6. Select suitable strut curves from Table 4.12.\n7. Determine compressive strength, pc, using Table\n4.13, 4.14 or similar.\n8. Calculate compression resistance, Pc = Agpc. If\nPc < Fc return to 2.\n9. Calculate effective slenderness ratio, λLT =\n0.5L/ry.\n10. Calculate buckling resistance of section, Mbs =\nMb = pbSx.\n11. Check buckling resistance of member using\nequation (4.35). If unsatisfactory return to (2).\n4.9.6 DESIGN OF CASED COLUMNS\nAs discussed in section 4.5, steel columns are some-\ntimes cased in concrete for fire protection. How-\never, the concrete also increases the strength of the\nsection, a fact which can be used to advantage in\ndesign provided that the conditions stated in Clause\n4.14.1 of BS 5950 are met. Some of these condi-\ntions are illustrated in Fig. 4.30.\nBS 5950 gives guidance on the design of UC\nsections encased in concrete for the following load-\ning conditions which are discussed below.\n(i) axially loaded columns\n(ii) columns subject to axial load and bending.\nLongitudinal\nbars\nB\n50 min\n75 max\n50 min\n75 max\nD\ndc\nbc\nCharacteristic\nstrength of concrete \u0003 20 N/mm2\nReinforcement\nReinforcement: steel fabric type D98\n(BS 4483) or \u0003 5 mm diameter longitudinal\nbars and links at a maximum spacing of 200 mm\n=\n=\n=\n=\nFig. 4.30 Cased UC section.\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"bc\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\", \"3\": \"\"}], [], []]\n\n[FORMULAS]\nPc = Agpc; Pc = Agpc; Mb = pbSx", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0212.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 213, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\nDesign in structural steelwork to BS 5950\n192\nselect selection\ncalculate effective length*1, L E\ncalculate radii of gyration of cased section*2, r y and r x\ncalculate compression resistance of column*3, Pc\nFig. 4.31 Design procedure for axially loaded cased columns.\nNotes to Fig. 4.31\n1 The effective length, LE, should not exceed the\nleast of:\n(i)\n40bc\n(ii)\n    \n100b\nd\nc\n2\nc\n(iii) 250r\nwhere bc and dc are as indicated in Fig. 4.30\nand r is the minimum radius of gyration of the\nuncased UC section i.e. ry.\n2 The radius of gyration of the cased section\nabout the y–y axis, ry, is taken as 0.2bc but not\nmore than 0.2(B + 150) mm and not less than\nthat of the steel section alone.\nThe radius of gyration of the cased section\nabout the x–x axis, rx, is taken as that of the\nsteel section alone.\n3 The compression resistance of the cased section,\nPc, 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(4.37)\nHowever, this should not be greater than the\nshort strut capacity of the section, Pcs, which is\ngiven by:\n    \nP\nA\nf A\np\np\ncs\ng\ncu\nc\ny\ny\n  \n  .\n=\n+\n\n\n\n\n0 25\n(4.38)\nwhere\nAc\nis the gross sectional area of the concrete\nbut neglecting any casing in excess of\n75 mm from the overall dimensions of\nthe UC section or any applied finish\nAg\nis the gross sectional area of the UC\nsection\nfcu\nis the characteristic strength of the\nconcrete which should not be greater than\n40 N/mm2\npc\nis the compressive strength of the UC\nsection determined as discussed for\nuncased columns (section 4.9.1), but using\nrx and ry for the cased section (see note 2)\nand taking py ≤ 355 N/mm2\npy\ndesign strength of the UC section which\nshould not exceed 355 N/mm2.\n4.9.6.1 Axially loaded columns\nThe design procedure for this case is shown in\nFig. 4.31.\n4.9.6.2 Cased columns subject to axial load\nand moment\nThe design procedure here is similar to that when\nthe column is axially loaded but also involves check-\ning the member’s cross-section capacity and buck-\nling resistance using the following relationships:\n1. Cross-section capacity check\n    \nF\nP\nM\nM\nM\nM\nc\ncs\nx\ncx\ny\ncy\n  \n  \n  \n+\n+\n≤1\n(4.39)\nwhere\nFc\naxial compression load\nPcs\nshort strut capacity (equation 4.38)\nMx\napplied moment about major axis\nMcx\nmajor axis moment capacity of steel\nsection\nMy\napplied moment about minor axis\nMcy\nminor axis moment capacity of steel\nsection\n2. Buckling resistance check\nMajor axis \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+\n≤1\n(4.40)\nMinor axis \n    \nF\nP\nm\nM\nM\nm M\np Z\nc\ncy\nLT\nx\nb\ny\ny\ny\ny\n  \n  \n  \n+\n+\n≤1 (4.41)\nwhere\nFc\nmaximum compressive axial force\nPc\nsmaller of Pcx and Pcy (equation 4.37)\nPcx\ncompression resistance of member\nconsidering buckling about the major\naxis\nPcy\ncompression resistance of member\nconsidering buckling about the\nminor axis\n\n[TABLES]\n[[{\"0\": \"\", \"1\": \"\"}, {\"0\": \"Design in structural steelwork to BS 5950\", \"1\": \"\"}, {\"0\": \"select selection\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"\"}, {\"0\": \"calculate effective length*1, LE\", \"1\": \"\"}, {\"0\": \"\", \"1\": \"calculate radii of gyration of cased section*2, ry and rx\"}, {\"0\": \"\", \"1\": \"\"}], [], []]", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0213.png"}
{"pdf_file": "Design of Structural Elements_ Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes, 3rd Edition   ( PDFDrive ).pdf", "page_num": 214, "prompt": "Analyze this page with diagrams, tables, formulas, calculations, and explanations.", "response": "[NATIVE TEXT]\n193\nDesign of compression members\nExample 4.13 Encased steel column resisting an axial load (BS 5950)\nCalculate the compression resistance of a 305 × 305 × 118 kg/m UC column if it is encased in concrete of com-\npressive strength 20 N/mm2 in the manner shown below. Assume that the effective length of the column about both\naxes is 3.5 m.\n59\nB = 306.8\n59\n305 × 305 × 118 kg m–1 UC\n55\nD = 314.5\n55\nbc = 425\ndc = 425\nPROPERTIES OF UC SECTION\nArea of UC section (Ag) = 15000 mm2\n(Appendix B)\nRadius of gyration (rx)\n= 136 mm\nRadius of gyration (ry)\n= 77.5 mm\nDesign strength (py)\n= 265 N/mm2\n(since T = 18.7 mm)\nEffective length (LE)\n= 3.5 m\nEFFECTIVE LENGTH\nCheck that the effective length of column (= 3500 mm) does not exceed the least of:\n(i)\n40bc = 40 × 425 = 17 000 mm\n(ii)\n    \n100\n100\n425\n425\n42 500\n2\nb\nd\nc\n2\nc\nmm\n  \n  \n  \n \n=\n×\n=\n(iii) 250ry = 250 × 77.5 = 19 375 mm\nOK\nRADII OF GYRATION FOR THE CASED SECTION\nFor the cased section rx is the same as for UC section = 136 mm\nFor the cased section ry = 0.2bc = 0.2 × 425 = 85 mm \u0004 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\nCOMPRESSION RESISTANCE\nSlenderness ratio\n    \nλ x\nE\nx\n  \n  \n  \n.\n=\n=\n=\nL\nr\n3500\n136\n25 7\n    \nλ y\nE\ny\n  \n  \n  \n.\n=\n=\n=\nL\nr\n3500\n85\n41 2\nCompressive strength\nFrom Table 4.12, relevant compressive strength values for buckling about the x–x axis are obtained from Table 24(b)\nof BS 5950 (Table 4.13) and from Table 24(c) of BS 5950 (Table 4.14) for bending about the y–y axis.\n\n[TABLES]\n[[{\"0\": \"B = 306.8\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}, {\"0\": \"\", \"1\": \"\", \"2\": \"\"}], [], []]\n\n[FORMULAS]\ndc = 425", "page_type": "mixed", "image": "Design of Structural Elements_ Concr_a6902759_p0214.png"}