Resolução Exercicios Estradas
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Resolução Exercicios Estradas


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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= 
 
rad
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 
rads 258880,0)08,0(2180
242 =\u22c5\u2212°\u22c5°=\u22c5\u2212\u2206=
\u3c0\u3b8\u3c6 
mestmradRD radc 44,9644,129)258880,0(500 +==\u22c5=\u22c5= \u3c6 
mradsensenRXk scs 99,39)08,0(50095,79 =\u22c5\u2212=\u22c5\u2212= \u3b8 
( ) [ ] 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. 
 
Solução: rad
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 
rads 618626,0)214286,0(2180
602 =\u22c5\u2212°\u22c5°=\u22c5\u2212\u2206=
\u3c0\u3b8\u3c6 
52,161052,216)618626,0(350 +==\u22c5=\u22c5= mradRD radc \u3c6 
mradsensenRXk scs 89,74)214286,0(35031,149 =\u22c5\u2212=\u22c5\u2212= \u3b8 
( ) [ ] 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 
 
rad
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 
deve ser dada no aparelho (colocado sobre o TS) para locar a estaca 200. Dados: Vp=100 
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= 
rad
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 
rads 549283,0)074425,0(2180
402 =\u22c5\u2212°\u22c5°=\u22c5\u2212\u2206=
\u3c0\u3b8\u3c6 
mestmradRD radc 57,91657,329)549283,0(600 +==\u22c5=\u22c5= \u3c6 
mradsensenRXk scs 65,44)074425,0(60026,89 =\u22c5\u2212=\u22c5\u2212= \u3b8 
( ) [ ] 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 
 rad
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 
m, calcular as coordenadas do ponto ST2 em relação ao sistema de coordenadas dado. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 
\u3b8s = 0,083333 rad 
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 
dado. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solução: 
 
Calculando os elementos da transição, temos: 
 
\u3b8s = 0,50 rad 
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: 
 
\u3b8s = 0,625 rad 
Xs = 96,16 m 
Ys = 20,25 m 
k = 49,36 m 
p = 5,13 m 
 
 
 
 
 
TS1 (k ; \u2013Rc\u2013 p) TS2 (k + L ; Rc+ p) 
SC1 (\u2013Xs + k ; Ys \u2013 Rc \u2013 p) SC2 (k + L + Xs ; Rc + p \u2013 Ys) 
CS1 (\u2013Xs + k ; Rc + p \u2013 Ys ) CS2 (k + L + Xs ; Ys \u2013 Rc \u2013 p) 
ST1 (k ; Rc + p) ST2 (k + L ; \u2013Rc \u2013