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```J=0,2 m/s3; b) a curva 2 será uma curva circular sem transições; c) entre
o ST1 e o PC2 existe um trecho em tangente de comprimento 200 m; d) a curva 2 terá o
maior raio possível, respeitadas as condições a, b e c.

Solução:

CÁLCULO DA CURVA 1:
m
JR
VL
LR
VJ
c
s
sc
80
5002,0
6,3
72 3
33
=\u22c5
\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb
=\u22c5=\u21d2\u22c5=

R
L
c
s
s 08,05002
80
2
=\u22c5=\u22c5=\u3b8
452,66 m
\u22062=24º
\u22061=24º
CURVA 2
1000 m
PI2
1000 m
PI1
EST. 0
CURVA 1
F
ESTRADAS DE RODAGEM \u2013 PROJETO GEOMÉTRICO Solução dos Exercícios 40
mLX ssss 95,79216
08,0
10
08,0180
21610
1
4242
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5= \u3b8\u3b8
mLY ssss 13,242
08,0
3
08,080
423
33
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5= \u3b8\u3b8
242 =\u22c5\u2212°\u22c5°=\u22c5\u2212\u2206=
\u3c0\u3b8\u3c6
( ) [ ] mradRYp scs 53,0)08,0cos(150013,2cos1 =\u2212\u22c5\u2212=\u2212\u22c5\u2212= \u3b8
( ) ( ) 38,6738,146
2
24tan53,050099,39
2
tan +==\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb °\u22c5++=\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb \u2206\u22c5++= mpRkTT c

E(TS) = E(PI) \u2013 [TT] = (50 + 0,00) \u2013 (7 + 6,38) = 42 + 13,62
E(SC) = E(TS) + [LS] = (42 + 13,62) + (4 + 0,00) = 46 + 16,32
E(CS) = E(SC) + [D] = (46 + 16,32) + (6 + 9,44) = 53 + 3,06
E(ST) = E(CS) + [LS] = (53 + 3,06) + (4 + 0,00) = 57 + 3,06

CÁLCULO DA CURVA 2:

E(PC2) = E(ST1) + 200 m = (57 + 3,06) + (10 + 0,00) = 67 + 3,06 = 1.343,06 m
T = 452,66 \u2013 TT \u2013 200 = 452,66 \u2013 146,38 \u2013 200 = 106,28 m
mTR 01,500
2
º24tan
28,106
2
tan 2
=
\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb=\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb \u2206=
mD 44,209
180
2401,500 =°
°\u22c5\u22c5= \u3c0

E(PT2) = E(PC2) + D = 1.343,06 + 209,44 = 1.552,50 m = 77 + 12,50
E(F) = E(PT2) + 1000m - T = 1.552,50 + 1.000 \u2013 106,28 = 2.446,22 m =122 + 6,22

Glauco Pontes Filho 41
9. (*) Dada a curva horizontal da figura, calcular os valores de X e Y do ponto P que está na
estaca 100 + 0,00. Dados: Rc = 350 m, E(PI) = 90 + 15,00, Ls = 150 m e \u2206 = 60º.

R
L
c
s
s 214286,03502
150
2
=\u22c5=\u22c5=\u3b8
mLX ssss 31,149216
214286,0
10
214286,01150
21610
1
4242
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5= \u3b8\u3b8
mLY ssss 68,1042
214286,0
3
214286,0150
423
33
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5= \u3b8\u3b8
602 =\u22c5\u2212°\u22c5°=\u22c5\u2212\u2206=
\u3c0\u3b8\u3c6
( ) [ ] mradRYp scs 674,2)214286,0cos(135068,10cos1 =\u2212\u22c5\u2212=\u2212\u22c5\u2212= \u3b8
( ) ( ) 50,181350,278
2
60tan674,235089,74
2
tan +==\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb °\u22c5++=\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb \u2206\u22c5++= mpRkTT c
E(TS) = E(PI) \u2013 [TT] = (90 + 15,00) \u2013 (13 + 18,50) = 76 + 16,50
E(SC) = E(TS) + [LS] = (76 + 16,50) + ( 7 + 10,00) = 84 + 6,50
E(CS) = E(SC) + [D] = (84 + 6,50) + (10 + 16,52) = 95 + 3,02
E(ST) = E(CS) + [LS] = (95 + 3,02) + ( 7 + 10,00) = 102 + 13,02

L = (102 + 13,02) \u2013 (100 + 0,00)
L = 2 est + 13,02 m = 53,02 m

LR
L
sc
026773,0
1503502
02,53
2
22
=\u22c5\u22c5=\u22c5\u22c5=\u3b8
mLX 02,53
216
026773,0
10
026773,0102,53
21610
1
4242
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5= \u3b8\u3b8
mLY 47,0
42
026773,0
3
026773,002,53
423
33
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5= \u3b8\u3b8

ST
CS
L
60º
95+3,02
102+13,02
100+0,00
ESTRADAS DE RODAGEM \u2013 PROJETO GEOMÉTRICO Solução dos Exercícios 42
10. (*) Deseja-se projetar uma curva de transição com J = 0,4 m/s3. Calcular a deflexão que
km/h, \u2206=40º, Rc=600 m, E(PI) = 209 + 3,23.

Solução: m
JR
VL
LR
VJ
c
s
sc
31,89
6004,0
6,3
100 3
33
=\u22c5
\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb
=\u22c5=\u21d2\u22c5=
R
L
c
s
s 074425,06002
31,89
2
=\u22c5=\u22c5=\u3b8
mLX ssss 26,89216
074425,0
10
074425,0131,89
21610
1
4242
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5= \u3b8\u3b8
mLY ssss 21,242
074425,0
3
074425,031,89
423
33
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5= \u3b8\u3b8
402 =\u22c5\u2212°\u22c5°=\u22c5\u2212\u2206=
\u3c0\u3b8\u3c6
( ) [ ] mradRYp scs 554,0)074425,0cos(160021,2cos1 =\u2212\u22c5\u2212=\u2212\u22c5\u2212= \u3b8
( ) ( ) 23,31323,263
2
40tan554,060065,44
2
tan +==\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb °\u22c5++=\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb \u2206\u22c5++= mpRkTT c
E(TS) = E(PI) \u2013 [TT] = (209 + 3,23) \u2013 (13 + 3,23) = 196 + 0,00
E(SC) = E(TS) + [LS] = (196 + 0,00) + ( 4 + 9,31) = 200 + 9,31
E(CS) = E(SC) + [D] = (200 + 9,31) + (16 + 9,57) = 216 + 18,87
E(ST) = E(CS) + [LS] = (216 + 18,87) + ( 4 + 9,31) = 221 + 8,18
L = (200 + 0,00) \u2013 (196 + 0,00) = 4 est + 0,00 m = 80 m
LR
L
sc
029618,0
17,1075002
34,56
2
22
=\u22c5\u22c5=\u22c5\u22c5=\u3b8

mLX 335,56
216
029618,0
10
029618,0134,56
21610
1
4242
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb +\u2212\u22c5= \u3b8\u3b8
mLY 56,0
42
029618,0
3
029618,034,56
423
33
=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5=\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212\u22c5= \u3b8\u3b8

ST
CS
L
50º
217+9,17
222+16,34
220+0,00
Glauco Pontes Filho 43
11. (*) A figura mostra trecho de uma via contendo tangentes perpendiculares entre si e duas
curvas circulares com transição, reversas e consecutivas. Dados que Rc = 200 m e Ls = 80

Solução:

Calculando os elementos da transição, temos:

Coordenada X = (TT \u2013 k) + TT = 2 (241,28) \u2013 39,95 = 442,61 m
Coordenada Y = -(k + TT) = - (39,95 + 241,28) = -281,23 m
SC1
CS1
ST1= TS2
SC2
CS2
ST2
C2
C1
TS1
E
N
Xs = 99,93 m
Ys = 2,78 m
k = 49,99 m
p = 0,69 m
TT = 252,35 m
SC1
CS1
ST1= TS2
SC2
CS2 ST2
C2
C1
TS1
E
k
TT
TT-k
TT
TT
k
ESTRADAS DE RODAGEM \u2013 PROJETO GEOMÉTRICO Solução dos Exercícios 44
12. (*) A figura mostra trecho do eixo da planta de um autódromo formado por 3 tangentes
paralelas concordadas entre si por curvas circulares com transição. Sabendo que Rc = 50 m
e Ls = 50 m, calcular as coordenadas do ponto ST2 em relação ao sistema de coordenadas

Solução:

Calculando os elementos da transição, temos:

Xs = 48,76 m
Ys = 8,18 m
k = 24,79 m
p = 2,06 m

Coordenada E = k = 24,76 m

Coordenada N = 4 (Rc+p) = 4 (50 + 2,06) = 208,24 m

100 m
TS1 E
TS2
SC1
CS1
ST1 SC2
CS2 ST2
C2
C1
N
Rc+p
Rc+p
Rc+p
Rc+p
k
100 m
TS1 E
TS2
SC1
CS1
ST1 SC2
CS2
ST2
C2
C1
N
Glauco Pontes Filho 45
13. (*) A figura mostra uma pista de teste composta por duas curvas horizontais de raio Rc = 80
m, concordadas com duas tangentes de comprimento 150 m através de curvas de transição
de comprimento Ls = 100 m. Calcular as coordenadas dos pontos TS, SC, CS e ST em
relação ao sistema de eixos da figura, que tem como origem o centro de uma das curvas.

Solução:

Calculando os elementos da transição, temos: