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ÿÿð™ÿð»ÿÿðàð�ÿÿð™ÿð»ÿÿð�ÿÿð™ÿð°ÿÿð�ÿÿð™ÿð°ÿðîð�ÿÿð™ð ÿÿðîð�ÿÿð™ð ÿÿàÿð�ÿÿð™ð°ÿÿÿÿð�ÿÿð™ð°ÿÿÿÿð�ÿÿð™ð°ÿÿðàðÿ ÿ� ÿðàÿ � � ðàÿ � � ðîÿð™™ÿ °°ðîÿð™™ÿ °°ðîÿð™™™�ÿ»°ÿîÿÿ™™ÿÿð»ÿÿîÿÿ™™ÿÿð»ÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ1CreateButton("fim","&Fim","CloseWindow(`main')")Zÿmain?main(;SISTRUT - Suporte ao Usuário €OŸm##¶O,‡lÔÔŸŸÀÀÀZÿsecondarysecond;SISTRUT - Suporte ao Usuário €OŸs##¶O,‡lddppŸŸÀÀÀn e ÿÿÿÿ ^-ÿÿÿÿj1&ÿÿÿÿÿÿÿÿj4€CD - Verificação de pilares pela NB1 - 78 (2)‚< ìF \€|€€†"€ÈBack()€†"€‰€‚ÿ VERIFICAÇÃO DE PILARES SEGUNDO A NB1 - 78 ( 2 ) 'j$ €€°€‚ÿ¾’ìÑ, &€%€°€€€‚‚ÿSegue abaixo uma explicação mais detalhada referente à verificação de pilares executada pelo módulo Pilares de Concreto do sistema SISTRUT :1- *€ €°€†"€‚ÿÃÑK d€‡€°€ƒƒƒƒ‚ƒƒƒƒƒ‚‚ƒ‚ƒƒƒƒ‚‚ƒƒƒ‚ƒƒƒƒ‚‚‚‚ƒƒ‚‚ƒƒ‚ÿ Cargas : G = 117.42g = 1.30Q = 38.58q = 1.40Coeficientes de Materiais:concreto = 1.40aço = 1.15Fluência:n = 1.25 = 2.0De acordo com a Norma Brasileira, além dos esforços inicialmente aplicados, devemos acrescentar excentricidades adicionais para verificação da armadura proposta assim temos:carga normal adotadaNk = ( 117.42 x 1.30 + 38.58 x 1.40 ) / 1.40 = 147.6 tmomentos iniciais:Mkx inicial = 0 tm"á2A P€Ã€°€ƒƒƒƒ‚‚ƒ‚‚ƒ‚‚€€‚€‚‚‚ƒ‚ƒ‚ÿMky inicial = 0 tmcomprimento de flambagem:Lxe = Lye = 10 mexcentricidades acidentais:h / 30 ou 2.0 cm (máximo)Solicitando uma verificação da coluna acima proposta pelo módulo Pilares de Concreto do sistema SISTRUT, temos o seguinte resultado: Verificacao de Pilares segundo a Norma Brasileira - NB1/78 Forca Normal Nk(t) *FIM* 147.6 Momento inicial Mxk(tm) *0* Momento inicial Myk(tm) *0* Compr. flambagem Lxe(m) 10Compr. flambagem Lye(m) 10 Ú?3 4€µ€°€ƒƒ‚‚‚‚‚‚‚‚‚‚‚‚‚ÿ Lambda x = 43 Lambda y = 87 Situacao - Compressao Centrada caso 1/2 Mx= 3.94(a)= 3.94 tm 1) ok My= 0.00(a)= 0.00 tm Mtotal= 3.94 tm incl 0 graus Esforcos resistentes do Pilar Mx= 28.94(1a)- 5.81(2a)= 23.13 tm My= 0.00(1a)= 0.00 tm Mtotal= 23.13 tm ln= 270 graus Xln= 58.67 cm 1/r= 0.03888 x 10-3 Def con 2.2813 x 10-3 Def aco 0.7126 x 10-3 Õ2X D V€­€°€‚‚€‚†"€‚€‚‚‚ƒ‚ƒ‚ƒƒ‚‚‚‚ÿ Csf= 1.4 Csc= 1.4 Csa= 1.15 Fluencia= 2 Verificacao de Pilares segundo a Norma Brasileira - NB1/78 Forca Normal Nk(t) *FIM* 147.6 Momento inicial Mxk(tm) *0* 0 Momento inicial Myk(tm) *0* 0 Compr. flambagem Lxe(m) 10 Compr. flambagem Lye(m) 10 Lambda x = 43 Lambda y = 87 Situacao - Compressao Centrada caso 2/2 Mx= 0.00(a)= 0.00 tm 1) okÞ?v @ N€¿€°€ƒ‚‚‚‚‚‚‚‚‚‚‚€‚‚†"€‚‚ÿ My= 2.95(a)+ 1.58(f)= 4.53 tm 2) ok Mtotal= 4.53 tm incl 90 graus Esforcos resistentes do Pilar Mx= 0.00(1a)- 0.00(2a)= 0.00 tm My= 16.26(1a)- 10.89(2a)= 5.37 tm Mtotal= 5.37 tm ln= 0 graus Xln= 30.55 cm 1/r= 0.07284 x 10-3 Def con 2.2250 x 10-3 Def aco 0.4701 x 10-3 Csf= 1.4 Csc= 1.4 Csa= 1.15 Fluencia= 2 Inicialmente vamos entender o que foi verificado pelo sistema.#ÜX ™G \€¹€°€‚‚‚‚ƒ‚ƒƒ‚ƒƒ‚ƒ‚ƒƒ‚ƒƒ‚‚ƒ‚ƒ‚ƒ€€‚ƒ‚ÿDuas verificações foram feitas:1) esforços acidentais1.1) excentricidade no sentido X (paralelo aos 80 cm)h/30 ( 2,67 cm ) > 2 cm Exa = 2,67Mxa = 2,67 cm x 147,6 t = 394 tcm ou 3,94 tm1.2) excentricidade no sentido Y (paralelo aos 40 cm)h/30 ( 1,33 cm ) < 2 cm Eya = 2,00Mya = 2,00 cm x 147,6 t = 295 tcm ou 2,95 tm2) fluência1.1) Mxf = 0 (x < 80)1.2) Eyf = E1g { exp [ Fg ] - 1 } Fe - Fgd>v ý& €|€°€ƒ‚‚ÿ (ref: FUSCO solicitações normais - pg 330) sendo:X'™U1#2€NP¤€€°€‚ÿÿÿFg = 1,25 (0.95 x 147,6 t) = 175 td3ý¹1#2€fP¤€€°€‚ÿÿÿE1g = ( Mkyi / Nk + Eya ) = ( 0 + 2 ) = 2 cmT#U@1#2€FP¤€€°€‚ÿÿÿFe = carga de flambag¹@ em de EulerG¹`@4#8€&P¤€€°€€‚ÿÿÿFe = 2 Ec IcI@©@1#2€0P¤€€°€‚ÿÿÿ Lye 2c2`@ A1#2€dP¤€€°€‚ÿÿÿEc = 0,9 21000 Fck(k/cm2) + 35 = 263.924 k/cm2Z)©@fA1#2€RP¤€€°€‚ÿÿÿIc = 80 x 40 3 / 12 = 426.667 cm 4_( AÅA7#>€PP¤$€€°€€€‚ÿÿÿFe = 2 263,92 426.667 = 1.112 tEfA B2#4€&P¤€€°€ƒ‚ÿÿÿ 1000 2t=ÅA~B7#>€zP¤$€€°€€€‚ÿÿÿEyf = 2,0 { exp [ 2 175 ] -1 } = 1,07 cmO BÍB2#4€:P¤€€°€ƒ‚ÿÿÿ 1.112 - 175R!~BC1#2€BP¤€€°€‚ÿÿÿMky f = 147,6 x 1,07 = 1,58 tm%óÍBDD2 2€ç€°€‚‚ƒ‚ƒ‚‚ƒ‚ƒ‚‚‚‚ÿ3) esforços atuantes totais a serem verificados1.1) Mkx = 0 + 3.94 + 0 = 3.94 tm Mky = 0 + 0 + 0 = 0 tm1.2) Mkx = 0 + 0 + 0 = 0 tm Mky = 0 + 2.95 + 1.58 = 4.53 tm4) resultados apresentados pelo sistema SISTRUT|2CÀDJ#d€dPE€€°ÿ0€€°†‘€!�±€ƒƒƒ‚ÿÿÿ4.1 )Inclinação da Linha Neutra = 270 grausWDDE<#H€6PE.€€°†‘€!�±€ƒƒ‚ÿÿÿ Xln = 58.67 cm]!ÀDtE<#H€BPE.€€°†‘€!�±€ƒƒ‚ÿÿÿ 1/r = 0,03888 x 10 -3c(E×E;#F€PPE,€€°†‘€!�±€ƒ‚ÿÿÿ Def . Concreto 2.2813 x 10 -3[!tE2F:#D€BPE*€€°†‘€!�±€‚ÿÿÿ Def. Aço 0,7126 x 10 -3A×EsF<#H€ PE.€€°†‘€!�±€ƒƒ‚ÿÿÿ&2F™F# €€€ ‚ÿ4sFÍF/ .€ €€†"€€ ‚ÿ&™FóF# €€€ ‚ÿSðÍFFIc ”€á€°€‚‚ƒ€€‚ƒƒ‚ƒ‚‚‚ƒ‚ƒ‚ƒ‚‚ƒƒƒ‚ƒƒƒƒ‚ƒƒƒƒ‚ƒƒƒƒ‚ƒƒƒƒ‚ƒƒƒƒ‚‚ƒ‚‚ƒ‚‚‚‚‚ÿ- concreto1.4 x Nk c = 0.85 0.16 (40 x ( 0.80 x 58.67 ) ) 1.4Nk c = 130,27 t- açoyd = 2% + Fyd / Es = 3,03%0.7 Fyd / Es = 0,7246%Barra 20 As = 3,15 cm2 / barradeformaçãotensãoforça2,28132,0496,45 t1,56201,8965,97 t0,84261,6375,15 t000 t0,71261,494,08 t1.4 x Nk s = 2 x (+6,45 +5,97 +5,15 +0 -4,08)Nks = 19,27 t carga normal = 130,27 + 19,27 = 149,54 t (147,6 devido a arredondamento e erro permitido 1,3%)ÛóFYK8 >€·€°€‚‚‚‚‚‚‚‚‚ƒƒ‚ƒƒƒƒ‚‚ƒ‚ÿ momentos = {130,27 x (40-0,8x58.67/2) }+{2x (+6,45x37 +5,97x18,5 +5,15x0 -0x18.5 +4,08x37)/1.4} momentos = 21,53 + 7,14 = 28,67 tm (28,94 devido à arredondamento)1/r = (c + s) / d = (2.2813% + 0.7126%) / 77 = 0,03888%ex2 = Lxe2 / 2 x 1/r = 10002 / 2 x 0,03888% = 3,94 cmMx 2a ordem = 147,6 x 3,94 = 5,81 tmConclusão:Momento atuante: 3,94 tmMomento resistente:28,67 - 5,81 = 22,86 tmOk! 22,86 > 3,944.2) Inclinação da Linha Neutra = 0 grausª~FIL, (€ü€°€ƒ‚ƒ‚ƒ‚ƒ‚‚ÿ Xln = 30.55 cm 1/r = 0,07284 x 10 -3 Def . Concreto 2.2250 x 10 -3 Def. Aço 0,4701 x 10 -31YK4L- *€ €°€†"€‚ÿ+×L_NT v€¯€°€‚‚ƒ€€‚ƒƒƒ‚ƒ‚‚‚ƒ‚ƒ‚ƒ‚‚ƒƒƒ‚ƒƒƒƒ‚ƒƒƒƒ‚‚ƒ‚‚ƒ‚‚‚‚ÿ- concreto1.4 x Nk c = 0.85 0.16 (80 x ( 0.80 x 30.55 ) ) 1.4Nk c = 135,67 t- açoyd = 2% + Fyd / Es = 3,03%0.7 Fyd / Es = 0,7246%Barra 20 As = 3,15 cm2 / barradeformaçãotensãoforça2,2252,0396,43 t0,47010,9873,11 t1.4 x Nk s = 5 x (+6,43 -3,11)Nks = 11,85 t carga normal = 135,67 + 11,85 = 147,52 t (147,6 devido à arredondamento) momentos = {135,67 x (20-0,8x30.55/2) }+{5x (+6,43x17 +3,11x17)/1.4}�[4L €2 2€·€°€‚‚‚‚‚‚ƒƒ‚ƒƒƒƒ‚ÿ momentos = 10,55 + 5,79 = 16,34 tm (16,26 devido à arredondamento)1/r = (c + s) / d = (2.225% + 0.4701%) / 37 = 0,07284%ey2 = Lye2 / 2 x 1/r = 10002 / 2 x 0,07284% = 7,38 cmMy 2a ordem = 147,6 x 7,38 = 10,89 tmConclusão:Momento atuante: 2.95 + 1.58 = 4.53 tmMomento resistente:16,34 - 10,89 = 5,45 tmOk! 5,45 > 4,53_N € (_N4€$ €€€‚‚ÿ1 €ÿÿÿÿ1ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÏÆ XHelv'š‘š¶?'HelvTimes New RomanTimeSymbolomanz’ÏSymbArial“Ï‹ð‰Vþ�Ariaÿÿÿÿÿÿ /&;)i24ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÍÿÿÿÿpilaresverificação pela normaÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿððÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿðÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ/&;)LzúSšŽ•/Ä^ÿÿÁÿÿÿÿCD - 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