TRELIASSo estruturas formadas por barras, ligadas entre si atravs de ns.
Consideramos para efeito de clculo os esforos aplicados nos ns.
Existem alguns tipos de calculo para determinao dos esforos nas barras, como o Mtodo dos Ns, Mtodo Ritter ou Mtodos das sees.
Nesta apostila, sero resolvidos apenas exerccios de trelias pelo Mtodo dos Ns.
Para determinar os esforos internos nas barras das trelias plana, devemos verificar a condio de Isosttica da Trelia, sendo o primeiro passo.
Depois calculamos as reaes de apoio e os esforos normais axiais nos ns. Tais esforos sero denominados de N.
1 Condio de Trelia Isosttica:
2 . n = b + Sendo 2 Calcular as Reaes de Apoio (Vertical e Horizontal):Fx = 0
Fy = 0
M = 0 (Momento fletor)
Por conveno usaremos: no sentido horrio no sentido anti-horrio +
-
3 Mtodos dos Ns
Quando calculamos os esforos, admitimos que as foras saem dos ns e nos prximos ns usamos os resultados das foras do n anterior fazendo a troca de sinais.
Importante lembrar que somente o jogo de sinais devero ser feitos na equao dos ns, pois as foras das reaes horizontais e verticais devem ser inseridos na equao considerando-se exclusivamente os sinais que possuem, ou seja, no fazer jogo de sinais para tais reaes.
Calma, nos exercicios ver que fcil.
Por Conveno os sinais das foras das barras so: + TRAO
- COMPRESSOTrelia Esquemtica
Exerccios
1) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.6 = 9+3
12 = 12 OK2 Passo Reaes de Apoio
Fx = 0
Fy = 0
M = 0 (Momento fletor)
HE = 0
VA+VE = 50+100+50
VA.4-50.4-100.2 = 0 VA+VE = 200 KN VA = 4004 100+VE = 200 KN VA = 100 KN VE = 200-100
VE = 100 KN
3 Passo Mtodo dos Ns
Decomposio das foras
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VA+NAB = 0
NAF = 0
100+NAB = 0
NAB = -100 KN
N B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-50-NBA-NBF.cos45 = 0 NBC+NBF.sen45 = 0
-50-(-100)-NBF.cos45 = 0 NBC+70,7.sen45 = 0
-NBF = -50cos45
NBC = - 50 KN
NBF = 70,7 KNN C
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-100-NCF = 0
-NCB+NCD = 0
NCF = -100 KN
-(-50)+NCD = 0
NCD = - 50 KN
N F
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NFC+NFB.sen45+NFD.sen45 = 0 -NFB.cos45+NFD.cos45-NFA+NFE = 0
-100+70,7.sen45+NFD.sen45 = 0 -70,7.cos45+70,7.cos45-0+NFE = 0
NFD = 50sen45
NFE = 0 KN
NFD = 70,7 KNN E
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NED+100 = 0
0-HE = 0
NED = -100 KN HE = 0 KNN D
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-50-NDF.sen45-NDE = 0 -NDC-NDF.cos45 = 0
-50-70,7.sen45+100 = 0-(-50)-70,7.cos45 = 0
-50-50+100 = 0
50-50 = 0
0 = 0
0 = 0
BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
NAB-100COMPRESSO
NED-100COMPRESSO
NAF0-
NEF0-
NBC -50COMPRESSO
NDC-50COMPRESSO
NBF70,7TRAO
NDF70,7TRAO
NCF-100COMPRESSO
2) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.5 = 7+3
10 = 10 OK2 Passo Reaes de Apoio
Fx = 0
Fy = 0
M = 0 (Momento fletor)HA+HB = 40
VB = 20 KN -HA.2+20.4+40.1 = 0
60+HB = 40
-HA.2+120 = 0HB = 40-60
HA = 1202HB = -20 KN
HA = 60 KN3 Passo Mtodo dos Ns
Decomposio das foras
N B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VB-NBA-NBC.sen26,57 = 0
-HB+NBC.cos26,57 = 0
20-NBA-22,36.sen26,57 = 0 -20+NBC.cos26,57 = 0
NBA = 10 KN
NBC = 20cos26,57
NBC = 22,36 KN
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NAB+NAC.sen26,57 = 0
HA+NAC.cos26,57+NAE = 0
10+NAC.sen26,57 = 0 60+(-22,36).cos26,57+NAE = 0
NAC = -10sen26,57 NAE+60-20 = 0
NAC = -22,36 KN
NAE = -40 KN
N E
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NEC = 0 -NEA+NED = 0
-(-40)+NED = 0
NED = -40 KN
N C
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0NCB.sen26,57-NCA.sen26,57-NCE-NCD.sen26,57=0 -40-NCB.cos26,57-NCA.cos26,57+NCD.cos26,57 = 0 22,36.sen26,57-(-22,36).sen26,57-0-NCD.sen26,57=0 -40-22,36.cos26,57-(-22,36).cos26,57+44,7.cos26,57=0 10+10-NCD.sen26,57=0 -40-20+20+40 = 0
NCD = 20sen26,57
0 = 0
NCD = 44,7 KN
N D
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-20+NDC.sen26,57 = 0
-NDC.cos26,57-NDE = 0
-20+44,7.sen26,57 = 0 -44,7.cos26,57-(-40) = 0
-20+20 = 0 -40+40 = 0
0 = 0
0 = 0
BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
NAB10TRAO
NBC22,36TRAO
NAC-22,36COMPRESSO
NAE-40COMPRESSO
NEC0-
NED-40COMPRESSO
NCD44,7TRAO
3) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.8 = 13+3
16 = 16 OK2 Passo Reaes de ApoioFx = 0
Fy = 0
M = 0 (Momento fletor)HA = 0
VA+VB = 2+2+2
-VB.16+2.12+2.8+2.4=0
VA = 6-3
VB = 4816
VA = 3 t
VB = 3 t3 Passo Mtodo dos Ns
Decomposio das foras
N 1
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N13.sen36,87+VA = 0
HA+N13.cos36,87+N12 = 0
N13.sen36,87+3 = 0 0+(-5).cos36,87+N12 = 0
N13 = -3sen36,87
N12 = 4 t
N13 = -5 t
N 2
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N23 = 0
-N21+N24 = 0
-4+N24 = 0
N24 = 4 t
N 3
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-2-N34.sen36,87-N32-N31.sen36,87+N35.sen36,87 = 0 +N34.cos36,87-N31.cos36,87+N35.cos36,87 = 0-2- N34.sen36,87-0-(-5).sen36,87+N35.sen36,87 = 0 +N34.cos36,87-(-5).cos36,87+N35.cos36,87 = 0 -N34.sen36,87+N35.sen36,87-2+3 = 0 +N34.cos36,87+N35.cos36,87+4 = 0 (-N34+N35).sen36,87 = -1 (+N34+N35).cos36,87 = -4
N34-N35 = 1sen36,87 N34+N35 = -4cos36,87 N34-N35 = 1,67 1 N34+N35 = -5 2
Sistema de Equaes
Substituindo na equao 1 ou 2 1 N34-N35 = 1,67
N34+N35 = -52 N34+N35 = -5
-1,67+N35 = -5
2N34 = -3,33
N35 = -5+1,67 N34 = -3,332
N35 = -3,33 t N34 = -1,67 tN 5
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0 -2-N53.sen36,87-N57.sen36,87-N54 = 0 -N53.cos36,87+N57.cos36,87 = 0 -2- (-3,33).sen36,87-(-3,33).sen36,87-N54 = 0 -(-3,33).cos36,87+N57.cos36,87 = 0 -2+2+2-N54 = 0 2,66+N57.cos36,87 = 0 N54 = 2 t N57 = -2,66cos36,87
N57 = -3,33 tN 4
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0 N43.sen36,87+N45+N47.sen36,87 = 0 -N42+N46-N43.cos36,87+N47.cos36,87 = 0
+(-1,67).sen36,87+2+N47.sen36,87 = 0 -(+4)+N46-(-1,67).cos36,87+(-1,67).cos36,87 = 0
-1+2+N47.sen36,87 = 0 N46-4+1,34-1,34 = 0
N47 = -1sen36,87 N46 = 4 t
N47 = 1,67 t
Por simetria dos carregamentos e das caractersticas das barras (dimenses, ngulos), as barras dos ns 7, 6 e 8 so iguais as dos ns 1, 2 e 3, conforme tabela.
BARRAFORAS NORMAIS AXIAIS (t)ESFORO
N13 = N87-5COMPRESSO
N12 = N864TRAO
N24 = N644TRAO
N23 = N670-
N34 = N74-1,67COMPRESSO
N35 = N75-3,33COMPRESSO
N542TRAO
4) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.7 = 11+3
14 = 14 OK2 Passo Reaes de ApoioFx = 0
Fy = 0
M = 0 (Momento fletor)HA+HB =0
VB = 0
-HA.3+2.6+2.4+2.2=0
HA = -HB
HA = 243
HB = -8 t
HA = 8 t3 Passo Mtodo dos Ns
Decomposio das foras
N 5
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VB-N51-N56.sen26,57 = 0
-HB+N56.cos26,57 = 0
6-N51-8,94.sen26,57 = 0 -8+N56.cos26,57 = 0
-N51+6-4 = 0 N56 = 8cos26,57
N51 = 2 t
N56 = 8,94 t
N 1
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N15+N16.sen45 = 0
HA+N12+N16.cos45 = 0
2+N16.sen45 = 0 8+N12+(-2,83).cos45 = 0
N16 = -2sen45 N12+8-2 = 0
N16 = -2,83 t
N12 = - 6 tN 6
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-2+N65.sen26,57-N61.sen45-N62-N67.sen26,57=0-N65.cos26,57-N61.cos45+N67.cos26,57=0
-2+8,94.sen26,57-(-2,83).sen45-N62-6,7.sen26,57=0 -8,94.cos26,57-(-2,83).cos45+N67.cos26,57=0
-2+4+2-3-N62 = 0 -8+2+N67.cos26,57 = 0
N62 = 1 t
N67 = 6cos26,57
N67 = 6,7 tN 2
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N26+N27.sen26,57 = 0 -N21+N23+N27.cos26,57 = 0
1+N27.sen26,57 = 0 -(-6)+N23+(-2,23).cos26,57 = 0
N27 = -1sen26,57 N23+6-2 = 0
N27 = -2,23 t
N23 = -4 t
N 3
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N37 = 0 -N32+N34 = 0
-(-4)+N34 = 0
N34 = -4 tN 7
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-2+N76.sen26,57-N72.sen26,57-N73-N74.sen26,57=0 -N76.cos26,57-N72.cos26,57+N74.cos26,57=0
-2+6,7.sen26,57-(-2,23).sen26,57-0-N74.sen26,57=0 -6,7.cos26,57-(-2,23).cos26,57+4,47.cos26,57=0
-2+3+1-N74.sen26,57 = 0 -6+2+4 = 0
N74 = 2sen26,57
0 = 0
N74 = 4,47 t
N 4
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-2+N47.sen26,57 = 0 -N43-N47.cos26,57 = 0
-2+4,47.sen26,57 = 0 -(-4)-4,47.cos26,57 = 0
-2+2 = 0 +4-4 = 0
0 = 0
0 = 0
BARRAFORAS NORMAIS AXIAIS (t)ESFORO
N512TRAO
N568,94TRAO
N16-2,83COMPRESSO
N12-6COMPRESSO
N621TRAO
N676,7TRAO
N27-2,23COMPRESSO
N23-4COMPRESSO
N370-
N34-4COMPRESSO
N744,47TRAO
5) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.5 = 7+3
10 = 10 OK2 Passo Reaes de ApoioFx = 0
Fy = 0 M = 0 (Momento fletor)HA = 0
VA+VB = 10+20
-VB.2,4+20.1,8+10.0,6=0
VA+17,5 = 30
VB = 422,4
VA = 30-17,5
VB = 17,5 KN
VA = 12,5 KN
3 Passo Mtodo dos Ns
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VA+NAC.sen53,13 = 0 HA+NAE+NAC.cos53,13 = 0
12,5+NAC.sen53,13 = 0 0+NAE+(-15,63).cos53,13 = 0
NAC = -12,5sen53,13 NAE = 9,38 KN
NAC = -15,63 KN
N C
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-10-NCA.sen53,13-NCE.sen53,13=0 -NCA.cos53,13+NCD+NCE.cos53,13=0-10-(-15,63).sen53,13-NCE.sen53,13=0 -(-15,63).cos53,13+NCD+3,13.cos53,13=0
-10+12,50-NCE.sen53,13 = 0 9,38+NCD+1,88 = 0
NCE = 2,5sen53,13
NCD = -11,26 KN
NCE = 3,13 KN
N E
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0 NEC.sen53,13+NED.sen53,13 = 0 -NEA+NEB+NED.cos53,13-NEC.cos53,13 = 0
3,13.sen53,13+NED.sen53,13 = 0 -9,38+NEB+(-3,13).cos53,13-(+3,13).cos53,13 = 0
NED = -2,5sen53,13 NEB-9,38-1,88-1,88 = 0
NED = -3,13 KN NEB = 13,14 KN
N D
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-20-NDE.sen53,13-NDB.sen53,13=0 -NDE.cos53,13-NDC+NDB.cos53,13=0-20-(-3,13).sen53,13-NDB.sen53,13=0 -(-3,13).cos53,13-(-11,26)+(-21,88).cos53,13=0
-20+2,50-NDB.sen53,13 = 0 1,88+12,26-13,13 = 0
NDB = -17,50sen53,13
0 = 0
NDB = -21,88 KN
N B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VB+NBD.sen53,13 = 0
-NBD.cos53,13-NBE = 0
17,5+(-21,88).sen53,13 = 0 -(-21,88).cos53,13-13,14 = 0
0 = 0
0 = 0
BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
NAC-15,63COMPRESSO
NAE9,38TRAO
NCE3,13TRAO
NCD-11,26COMPRESSO
NED-3,13COMPRESSO
NEB13,14TRAO
NDB-21,88COMPRESSO
6) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.5 = 7+3
10 = 10 OK2 Passo Reaes de ApoioFx = 0
Fy = 0 M = 0 (Momento fletor)HA+HB = 0 VA = 225
-HB.0,9+75.2,4+150.1,2 = 0
HB = -HA
HB = 3600,9HA = - 400 KN
HB = 400 KN
3 Passo Mtodo dos Ns
N B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NBA = 0 HB+NBD = 0
400+NBD = 0
NBD = -400 KN
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VA-NAB-NAD.sen36,87 = 0 -HA+NAC+NAD.cos36,87 = 0
225-0-NAD.sen36,87 = 0 -400+NAC+375.cos36,87 = 0
NAD = 225sen36,87 NAC = 100 KN
NAC = 375 KN
N D
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NDA.sen36,87+NDC = 0 -NDA.cos36,87-NDB+NDE = 0
375.sen36,87+NDC = 0
-375.cos36,87-(-400)+NDE = 0
NDC = -225 KN -300+400+NDE = 0
NDE = -100 KN
N C
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-150-NCD-NCE.sen36,87=0 -NCA+NCE.cos36,87=0-150-(-225)-NCE.sen36,87=0
-100+125.cos36,87 = 0 -150+225-NCE.sen36,87 = 0 -100+100 = 0
NCE = 75sen36,87
0 = 0
NCE = 125 KN
N E
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-75+NEC.sen36,87 = 0 -NEC.cos36,87-NED = 0
-75+125.sen36,87 = 0 -125.cos36,87-(-100) = 0
-75+75 = 0 -100+100 = 0
0 = 0
0 = 0BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
NBA0-
NBD-400COMPRESSO
NAD375TRAO
NAC100TRAO
NDC-225COMPRESSO
NDE-100COMPRESSO
NCE125TRAO
7) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.8 = 13+3
16 = 16 OK2 Passo Reaes de ApoioFx = 0
Fy = 0 M = 0 (Momento fletor)HE = 0 VA+VE = 8
VA.4a-4.3a-4.1a = 0
4+VE = 8
VA.4a-12a-4a = 0
VE = 8-4
VA = 16a4a
VE = 4 KN
VA = 4 KN3 Passo Mtodo dos Ns
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VA+NAB.sen30 = 0 NAB.cos30+NAF = 0
4+NAB.sen30 = 0 +(-8)cos30+NAF = 0
NAB = -4sen30 NAF = 6,9 KN
NAB = -8 KN
N F
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NFB-4 = 0
-NFA+NFG = 0
NFB = 4 KN
-6,9+NFG = 0
NFG = 6,9 KNN B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0NBC.sen30-NBA.sen30-NBF-NBG.sen30 = 0 -NBA.cos30+NBC.cos30+NBG.cos30 = 0 NBC.sen30-(-8).sen30-4-NBG.sen30 = 0 -(-8).cos30+NBC.cos30+NBG.cos30 = 0 NBC.sen30-NBG.sen30+4-4 = 0 6,9+NBC.cos30+NBG.cos30 = 0
NBC.sen30-NBG.sen30 = 0 NBC.cos30+NBG.cos30 = -6,9
(NBC-NBG).sen30 = 0 (NBC+NBG).cos30 = -6,9
NBC-NBG = 0sen30
NBC+NBG = -6,9cos30
NBC-NBG = 0 1 NBC+NBG = -8 2
Sistema de Equaes
Substituindo na equao 1 ou 2 1 NBC-NBG = 0
NBC-NBG = 02 NBC+NBG= -8
-4 - NBG = 0
2NBC = -8
NBG = -4 KN NBC = -82
NBC = -4 KNN C
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0 -NCB.sen30-NCD.sen30-NCG = 0 -NCB.cos30+NCD.cos30 = 0 -(-4).sen30-(-4).sen30-NCG = 0 -(-4).cos30+NCD.cos30 = 0 2 + 2- NCG = 0 3,5+NCD.cos30 = 0
NCG = 4 KN NCD = -3,5cos30
NCD = -4 KN
N G
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NGB.sen30+NGC+NGD.sen30 = 0 -NGB.cos30+NGD.cos30-NGF+NGH = 0
-4.sen30+4+NGD.sen30 = 0 -(-4).cos30+(-4).cos30-6,9+NGH = 0
NGD = -2sen30 +3,5-3,5-6,9+NGH = 0 NGD = -4 KN NGH = 6,9 KN
Por simetria dos carregamentos e das caractersticas das barras (dimenses, ngulos), as barras dos ns H, D e E no precisam ser calculadas.
BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
NAB = NED-8COMPRESSO
NAF = NEH6,9TRAO
NFG = NHG6,9TRAO
NFB = NHD4TRAO
NBC = NDC-4COMPRESSO
NBG = NDG-4COMPRESSO
NCG4TRAO
8) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.10 = 17+3
20 = 20 OK2 Passo Reaes de ApoioFx = 0
Fy = 0 M = 0 (Momento fletor)HJ = 0 VF+VJ = 2000 VF.4a-400.4a-400.3a-400.2a-400.1a = 0
1000+VJ = 2000 VF = 4000a4a
VJ = 2000-1000
VF = 1000 N VJ = 1000 N
3 Passo Mtodo dos Ns
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-400-NAF = 0
NAB = 0
NAF = -400 N
N F
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NFB.sen45+NFA+VF = 0 NFB.cos45 + NFG = 0
NFB.sen45+(-400)+1000 = 0 -848,5.cos45+NFG = 0
NFB = -600sen45 NFG = 600 N
NFB = -848,5 N
N G
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NGB = 0 -NGF+NGH = 0
-600+NGH = 0
NGH = 600 NN B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-400-NBF.sen45-NBG-NBH.sen45=0 -NBA+NBC-NBF.cos45+NBH.cos45 = 0
-400-(-848,5).sen45-0-NBH.sen45=0 -0+NBC-(-848,5).cos45+282,8.cos45=0
-400+600-NBH.sen45=0 NBC+600+200 = 0
NBH = 200sen45
NBC = -800 N
NBH = 282,8 N
N C
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-400-NCH=0
-NCB+NCD = 0
NCH = -400 N
-(-800)+NCD=0
NCD = -800 NPor simetria dos carregamentos e das caractersticas das barras (dimenses, ngulos), as barras dos ns H, D, I, E e J no precisam ser calculadas.
BARRAFORAS NORMAIS AXIAIS (N)ESFORO
NAB = NED0-
NAF = NEJ-400COMPRESSO
NFB = NJD-848,5COMPRESSO
NFG = NJI600TRAO
NGB = NID0-
NGH = NIH600TRAO
NBH = NDH282,8TRAO
NBC = NDC-800COMPRESSO
NCH-400COMPRESSO
9) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.6 = 8+4
12 = 12 OK2 Passo Reaes de ApoioFx = 0
Fy = 0 M = 0 (Momento fletor)HE = 18 KN VE+VF = 0 VE.3,6-9.5,4-9.2,7 = 0
HF = 0
VE = -VF VE.3,6-48,6-24,3 = 0
VF = -20,25 KN
VE = 72,93,6
VE = 20,25 KN
3 Passo Mtodo dos Ns
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NAC = 0
-9+NAB = 0
NAB = 9 KN
N B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0 -NBD-NBC.sen36,87=0
-NBA-NBC = 0
-NBD-(-11,25).sen36,87=0 -NBA-NBC.cos36,87=0
-NBD+6,75=0 -9-NBC.cos36,87 = 0
NBD = 6,75 KN
NBC = -9cos36,87
NBC = -11,25 KN
N C
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NCA-NCE+NCB.sen36,87 = 0 -9+NCB.cos36,87+NCD = 0
0-NCE+(-11,25).sen36,87 = 0 -9+(-11,25).cos36,87+NCD = 0
NCE = -6,75 KN -9-9+NCD = 0 NCD = 18 KN
N D
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0 NDB-NDE.sen36,87-NDF=0 -NDE.cos36,87-NDC = 0
6,75-(-22,5).sen36,87-NDF=0 -NDE.cos36,87-18 =0
6,75+13,5-NDF=0 NDE = -18cos36,87
NDF = 20,25 KN
NDE = -22,5 KN
N E
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VE+NEC+NED.sen36,87 = 0 HE+NED.cos36,87 = 0
20,25+(-6,75)+(-22,5).sen36,87 = 0 HE+(-22,5).cos36,87 = 0
20,25-6,75-13,5 = 0 HE = 18 KN Ok0 = 0
N F
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NFD-VF = 0 -HF = 0
20,25-20,25 = 0 HF = 0 Ok
0 = 0 BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
NAB9TRAO
NAC0-
NCE-6,75COMPRESSO
NCD18TRAO
NBD6,75TRAO
NBC-11,25COMPRESSO
NDF20,25TRAO
NDE-22,5COMPRESSO
10) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.6 = 9+3
12 = 12 OK2 Passo Reaes de ApoioFx = 0
Fy = 0 M = 0 (Momento fletor)HC+HF = 0 VC = 100 KN HF.4,5-100.7,2 = 0
HF = -HC
HF = 7204,5HC = -160 KN
HF = 160 KN
3 Passo Mtodo dos Ns
N D
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NDA-100 = 0 NDE = 0
NDA = 100 KN
N A
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0NAB.sen22,62-NAD-NAE.sen22,62 = 0 NAB.cos22,62+NAE.cos22,62 = 0NAB.sen22,62-100-NAE.sen22,62 = 0 (NAB+NAE).cos22,62 = 0 (NAB-NAE).sen22,62 = 100 NAB+NAE = 0cos22,62 NAB-NAE = 100sen22,62
NAB-NAE = 260 1 NAB+NAE = 0 2
Sistema de Equaes
Substituindo na equao 1 ou 2 1 NAB-NAE = 260
NAB+NAE = 02 NAB+NAE= 0
130+NAE = 0
2NAB = 260
NAE = -130 KN NAB = 2602
NAB = 130 KN
N E
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NEB+NEA.sen22,62 = 0 NEF-NED-NEA.cos22,62 = 0
NEB+(-130).sen22,62 = 0 NEF-0-(-130).cos22,62 = 0
NEB = 50 KN NEF = -120 KN
N F
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
NFC+NFB.sen39,81 = 0 -NFE-HF-NFB.cos39,81 = 0
NFC+(-52).sen39,81 = 0 -(-120)-160-NFB.cos39,81 = 0
NFC = 33,3 KN NFB = -40cos39,81
NFB = -52 KN
N B
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0+NBC.sen22,62-NBA.sen22,62-NBE-NBF.sen39,81=0 -NBA.cos22,62+NBC.cos22,62+NBF.cos39,81 = 0
173.sen22,62-130.sen22,62-50-(-52).sen39,81=0 -130.cos22,62+NBC.cos22,62+(-52).cos39,81=0
66,5-50-50+33,5=0 NBC.cos22,62-120-40 = 0
0=0
NBC = 160cos22,62
NBC = 173 KNN C
Foras Verticais (V)
Foras Verticais (H)
FV= 0
FH = 0VC-NCF-NCB.sen22,62=0 -NCB.cos22,62+HC = 0
100-(+33,3)-(+173).sen22,62=0 -(+173).c os22,62+160 = 0
100-33,3-66,7=0
-160+160 = 0
0=0
0 = 0BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
NDA100TRAO
NDE0-
NAE-130COMPRESSO
NAB130TRAO
NEB50TRAO
NEF-120COMPRESSO
NFC33,3TRAO
NFB-52COMPRESSO
NBC173TRAO
11) Calcule as reaes de apoio e as foras normais nas barras atravs do Mtodo dos Ns.
1 Passo Condio de Isosttica
2.n = b+
2.10 = 17+3
20 = 20 OK2 Passo Reaes de ApoioFx = 0
Fy = 0 M = 0 (Momento fletor)HA = 0 VA+VB = 30 KN -VB.10+15.6+15.3 = 0
VA+13,5 = 30
VB = 13510
VA = 30-13,5
VB = 13,5 KN
VA = 16,5 KN3 Passo Mtodo dos Ns
N 1
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
VA+N12.sen45 = 0 HA+N13+N12.cos45 = 0
16,5+N12.sen45 = 0 0+N13+(-23).cos45 = 0
N12 = -16,5sen45 N13 = 16,3 kN
N12 = -23 KN
N 3
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N32 = 0 -N31+N32.cos71,57+N34 = 0
-16,3+0.cos.71,57+N34 = 0
N34 = 16,3 kN N 4
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N42 = 0 -N43+N42.cos71,57+N45 = 0
-16,3+0.cos.71,57+N45 = 0
N45 = 16,3 kNN 2
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-15-N21.sen45-N23.sen71,57-N24.sen71,57-N25.sen45=0 -N21.cos45-N23.cos71,57+N26+N25.cos45+N24.cos71,57 = 0
-15-(-23).sen45-0-0-N25.sen45=0 -(-23).cos45-0+N26+2,29.cos45+0=0
N25 = 1,62sen45 N26+16,26+1,62 = 0
N25 = 2,29 KN
N26 = -17,88 KN
N 6
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
-15-N65 = 0 -N62+N67 = 0
N65 = -15 KN -(-17,88)+N67 = 0
N67 = -17,88 KNN 5
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N52.sen45+N56+N57.sen45 = 0 -N52.cos45-N54+N57.cos45+N58 = 0
2,29.sen45+(-15)+N57.sen45 = 0 -2,29.cos45-16,3+18,9.cos45+N58 = 0
N57.sen45+1,62-15 = 0 N58-1,62-16,3+13,36 = 0
N57 = 13,38sen45 N58 = 4,5 KN
N57 = 18,9 KNN 8
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N87 = 0 -N85+N87.cos71,57+N89 = 0
-4,5+0.cos.71,57+N89 = 0
N89 = 4,5 KN
N 9
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N97.sen71,57+VB = 0 -N97.cos71,57-N98+N910 = 0
N97.sen71,57+13,5 = 0 -(-14,23).cos71,57-4,5+N910 = 0
N97 = -13,5sen71,57
N910 = 0
N97 = -14,23 KN N 10
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0
N107 = 0 -N107-N109 = 0
-0-N109 = 0
N109 = 0N 7
Foras Verticais (V)
Foras Verticais (H)
FV = 0
FH = 0-N75.sen45-N78.sen71,57-N79.sen71,57-N710.sen45 = 0 -N76-N75.cos45-N78.cos71,57+N79.cos71,57+N710.cos45 = 0
-18,9.sen45-0-(-14,23).sen71,57-0=0 -(-17,88)-(+18,9).cos45-0+(-14,23).cos71,57+0=0
-13,5+13,5 = 0 17,88-13,36-4,5 = 0
0 = 0
0 = 0
BARRAFORAS NORMAIS AXIAIS (KN)ESFORO
N12-23COMPRESSO
N1316,3TRAO
N3416,3TRAO
N420-
N320-
N4516,3TRAO
N522,29TRAO
N26-17,88COMPRESSO
N67-17,88COMPRESSO
N65-15COMPRESSO
N584,5TRAO
N5718,9TRAO
N870-
N894,5TRAO
N97-14,23COMPRESSO
N9100-
N1070-
Bibliografia
INTRODUO ISOSTTICA, autor Eloy Ferraz Machado Junior, 2007
Apostila MECNICA DOS MATERIAIS, autor Ricardo Gaspar, 2005
n = n de ns
b = quantidade de barras
= n de reaes (Verticais e Horizontais)
NAB
NAB
NAF
NAF
VA
VA
50
50
NBC
NBF
NBC
NBF
NBA
NBF
NBA
100
100
NCB
NCD
NCB
NCD
NCF
NCF
NFC
NFD
NFB
NFB
NFC
NFD
NFB
NFD
NFA
NFE
NFA
NFE
NED
NED
NEF
HE
NEF
HE
VE
VE
50
50
NDC
NDC
NDF
NDF
NDE
NDE
NDF
VB
VB
HB
HB
NBC
NBC
NBC
NBA
NBA
NAC
NAB
NAB
NAC
NAC
HA
HA
NAE
NAE
NEC
NEC
NED
NEA
NEA
NED
NCB
NCB
NCB
40
NCD
NCA
NCD
NCA
40
NCD
NCA
NCE
NCE
NDC
20
NDC
20
NDC
NDE
NDE
N13
N13
N13
HA
HA
N12
N12
VA
VA
N23
N23
N21
N24
N21
N24
2
N35
2
N35
N35
N31
N31
N34
N34
N32
N31
N34
N32
+
2
2
N57
N53
N57
N53
N54
N53
N57
N54
N43
N47
N45
N43
N45
N47
N46
N43
N47
N42
N42
N46
VB
VB
N56
HB
HB
N56
N56
N51
N51
N16
N15
N15
N16
N16
HA
HA
N12
N12
N65
2
2
N65
N65
N67
N61
N67
N61
N62
N61
N67
N62
N27
N26
N26
N21
N27
N27
N21
N23
N23
N37
N37
N34
N32
N32
N34
N76
2
2
N76
N76
N74
N72
N74
N72
N73
N72
N74
N73
N47
N47
2
2
N47
N43
N43
NAC
HA
NAC
NAC
HA
NAE
NAE
VA
VA
10
10
NCD
NCA
NCD
NCE
NCE
NCA
NCA
NCE
NEC
NED
NEC
NED
NEB
NEC
NED
NEA
NEA
NEB
20
20
NDC
NDB
NDC
NDE
NDB
NDE
NDE
NDB
NBD
NBD
NBD
NBE
NBE
VB
VB
NBA
NBA
NBD
HB
HB
NBD
VA
VA
HA
NAC
HA
NAD
NAD
NAC
NAB
NAD
NAB
NDC
NDA
NDC
NDA
NDE
NDA
NDB
NDE
NDB
150
150
NCA
NCA
NCEW
NCE
NCD
NCD
NCE
75
NEC
NEC
75
NEC
NED
NED
NAB
NAB
NAB
NAF
NAF
VA
VA
NFB
NFB
NFG
NFA
NFA
NFG
4
4
NBC
NBC
NBC
NBA
NBA
NBG
NBG
NBF
NBA
NBG
NBF
+
NCD
NCB
NCB
NCD
NCG
NCB
NCD
NCG
NGB
NGD
NGC
NGB
NGC
NGD
NGB
NGD
NGF
NGH
NGF
NGH
400
400
NAB
NAB
NAF
NAF
NFB
NFA
NFA
NFB
NFB
NFG
NFG
VF
VF
NGB
NGB
NGF
NGH
NGF
NGH
400
400
NBC
NBA
NBA
NBH
NBF
NBH
NBF
NBC
NBH
NBF
NBG
NBG
400
400
NCB
NCB
NCD
NCD
NBG
NCH
Por se tratarem de foras de reao horizontal e estarem na mesma linha de ao, bem como as foras externas de 9 KN serem aplicadas no segmento AE, a reao horizontal HE sofre sozinha a ao dos 18 KN enquanto a HF no solicitada.
Os clculos mostraro essa teoria.
9
NAB
9
NAB
NAC
NAC
NBA
NBA
NBC
NBC
NBD
NBC
NBD
NCB
NCA
NCA
9
9
NCB
NCB
NCD
NCD
NCE
NCE
NDB
NDB
NDC
NDC
NDE
NDE
NDE
NDF
NDF
NEC
NED
NEC
HE
NED
NED
HE
VE
VE
HE confirmada
NFD
NFD
HF
HF
VF
VF
HF confirmada
NDA
NDA
NDE
NDE
100
100
NAB
NAB
NAB
NAE
NAE
NAE
NAD
NAD
+
NEA
NEB
NEA
NEB
NEF
NEA
NED
NED
NEF
NFB
NFC
NFB
NFC
HF
NFE
HF
NFB
NFE
NBC
NBA
NBC
NBC
NBF
NBF
NBA
NBF
NBA
NBE
NBE
VC
VC
NCB
HC
HC
NCB
NCB
NCF
NCF
N12
HA
HA
N12
N12
N13
N13
VA
VA
N32
N31
N31
N32
N32
N34
N34
N42
N42
N31
N32
N43
N34
N45
15
15
N21
N25
N26
N26
N23
N24
N21
N25
N25
N24
N23
N21
N23
N24
15
15
N62
N67
N62
N67
N65
N65
N52
N57
N56
N52
N56
N57
N52
N57
N54
N58
N54
N58
N87
N85
N85
N87
N87
N89
N89
N97
N97
N910
N97
N98
N910
N98
VB
VB
N107
N107
N107
N109
N109
N76
N710
N76
N75
N710
N75
N78
N79
N78
N79
N710
N75
N78
N79
Top Related