BAB III - Muhammad Haykal Mussaad's Blog | https ... · Web viewGambar 3.5 Tikungan Belok ke kanan Tipe S – S + 9,9 Kiri CL ± 0,00% Gambar 3.6 Diagram Superelevasi Tipe S – S

BAB III

PERHITUNGAN ALINEMEN HORISONTAL

A. Klasifikasi Medan

A – 1 = = 1,517 %

1 – 2 = = 5,47 %

2 – 3 = = 4,494 %

3 – 4 = = 0,5 %

4 – 5 = = 1,496 %

5 – 6 = = 0,497 %

6 – 7 = = 2 %

7 – 8 = = 1,495 %

8 – 9 = = 5,494%

9 – B = = 0,003 %

Klasifikasi Medan A – B

= 2,0878 %

Tabel 3.1. Klasifikasi MedanRata – rata Kemiringan Melintang ( % ) Jenis Medan

20

0 – 9,99,9 – 24,5

25

DatarBukit

Gunung

Jadi klasifikasi medan A – B adalah datar

Dari Daftar I Standar Perencanaan Geometrik didapat :

1. Kecepatan rencana (Vr) : 100 km/jam

2. Lebar row minimum : 40 m

3. Lebar perkerasan : 2 x 3,50 m

4. Lebar bahu : 3,00 m

5. Lereng melintang perkerasan (en) : 2 %

6. Lereng melintang bahu : 4%

7. Miring tikungan maksimum : 10 %

8. Jari–jari tikungan minimum(Rmin) : 350 m

9. Landai maksimum : 4 %

10. LHR : 6000 – 20.000

11. Lebar median minimum : 1,5 m

12. Landai maksimum : 4 %

13. Klasifikasi medan : Datar

Koordinat Tiap Titik

Koordinat titik A ( 10016,4116 )

Koordinat titik I ( 10016+13,5 ; 4116+14,5 ) → ( 10029,5 ; 4130,5 )

X1 = 13,5 m

Y1 = 14,5 m

Koordinat titik II ( 10029,5+24,8;4130,5+17,3 )

→(10054,3;4147,8)

X2 = 24,8 m

Y2 = 17,8 m

♦ Kordinat titik III ( 10054,3+24,3;4147,8+9,5 ) → ( 10078,6;4157,3 )

21

X3 = 24,3m

Y3 = 9,5 m

Koordinat titik B ( 10078,6+11,3;4157,3+3 ) → ( 10089,9;4160,3 )

Jarak Antar Titik

dA-I = = 19,8116 m

dI-II = = 30,5267 m

dII-III = = 26,09099 m

dIII-B = = 11,6914 m

= dA –I + dI – II + dII – III + dIII – B

= 19,8116 + 30,5267 + 26,0909 + 11,6914

= 88,1206 m

Perhitungan Sudut

Sudut Azimuth A = 46°59’12” = 46,980

I X2

α2 1

Y2 III 4 X4

Y1 3 3

Y3

2

X1 1

II 2 X3

1 = 900 – Azimuth titik A

= 900 – 46,980

= 43,020

22

A

Y4

B

2 = arc tan

= arc tan

= 34,680

3 = arc tan

= arc tan

= 31,620

♦ Tikungan I

1 = 1 + 2

= 43,020 + 34,680

= 77,70

Tikungan II

1 = 2 = 34,680

2 = arc tan

= arc tan

= 31,620

2 = 1 + 2

= 34,680 + 31,620

= 66,30

♦ Tikungan III

3 = 3 = 31,620

4 = arc tan

23

= arc tan

= 14,860

3 = 3 - 4

= 31,620 – 14,860

= 16,760

1. Perencanaan Tikungan I

1 = 77,7°

Rmin = 350 m

Vr = 100 km/jam

Rr = 358

en = 2 %

C = 0,4

Dari tabel Panjang Minimum Spiral dan Kemiringan Melintang

diperoleh nilai :

e = 0,099

Ls = 100 m ...... ( 1 )

Ls min =

= = 86,14 m ...... ( 2 )

Dari tabel Daftar Standar Perencanaan Alinemen didapat :

B = 3 m

=

=

24

Ls =

=

= 85,86 m.............( 3 )

Dari...( 1 ),...( 2 ),...( 3 ) dipilih yang terbesar

Jadi Ls = 100 m

s =

=

= 8,0060

c = 1 – 2 . s

= 77,700 – 2 . 8,0060

= 61,688 0

Lc =

=

= 385,248 m

Diketahui Lc min = 25 m

Lc >Lc min, jadi tikungan yang dipakai tipe S - C – S

Xc =

=

= 99,804 m

25

Yc =

=

= 4,655 m

K = Xc – Rr . Sin s

= 99,804 – 358 .Sin 8,0060

= 49,9429 m

P = Yc - Rr ( 1 – Cos s )

= 4,655 – 358 ( 1 – Cos 8,0060 )

= 1,165 m

Tt = ( Rr + P ) tan ½ 1 + K

= ( 358 + 1,165 ) tan ½ x 77,70 + 49,9429

= 339,2353 m

Et = ( Rr + P ) sec ½ 1 – Rr

= ( 358 + 1,165 ) sec ½ x 77,7 – 358

= 103,182 m

L = 2. Ls + Lc

= 2. 86,14 + 385,248

= 557,528 m

2. Perencanaan Tikungan II

2 = 66,3°

Rmin = 350 m

26

Vr = 100 km / jam

Rr = 358 m

en = 2 %

C = 0,4


diperoleh nilai :

e = 0,099

Ls = 100 m ...... ( 1 )

Ls min =

= = 86,14 m ...... ( 2 )


B = 3 m

=

=

Ls =

=

= 85,86 m.............( 3 )


Jadi Ls = 100 m

s =

27

=

= 8,0060

c = 1 – 2 .s

= 66,30 – 2 . 8,0060

= 50,288 0

Lc =

=

= 314,054 m


Lc >Lc min, jadi tikungan yang dipakai tipe S - C – S

Xc =

=

= 99,804 m

Yc =

=

= 4,655 m

K = Xc – Rr . Sin s

= 99,804 – 358 .Sin 8,0060

= 49,9429 m

P = Yc - Rr ( 1 – Cos s )

= 4,655 – 358 ( 1 – Cos 8,0060 )

= 1,165 m

Tt = ( Rr + P ) tan ½ 2 + K

28

= ( 358 + 1,165 ) tan ½ * 66,30 + 49,9429

= 284,526 m

Et = ( Rr + P ) sec ½ 2 – Rr

= ( 358 + 1,165 ) sec ½ . 66,3 – 358

= 70,985 m

L = 2. Ls + Lc

= 2. 85,86 + 314,054

= 485,774 m

3. Perencanaan Tikungan III

3 = 16,76°

Rmin = 350 m

Vr = 100 km / jam

Rr = 358 m

en = 2 %

C = 0,4


diperoleh nilai :

e = 0,099

Ls = 100 m ...... ( 1 )

Ls min =

= = 86,14 m ...... ( 2 )


B = 3 m

29

=

=

Ls =

=

= 85,86 m.............( 3 )


Jadi Ls = 100 m

s =

=

= 8,0060

c = 3 – 2 . s

= 16,760 – 2 .8,0060

= 0,748 0

Lc =

=

= 4,671 m


Lc < Lc min, jadi tikungan yang dipakai tipe S – S

Dihitung kembali :

3 = 16,76°

3 = 2 s

30

Maka :

s = ½ 2

= ½ 16,760

= 8,380

Ls =

=

= 104,660

P = – R ( 1 – Cos θs )

= – 358 ( 1 – Cos 8,38º )

= 1,2772 m

K = Ls – – R x Sin s

= 104,66 – – 358 x Sin 8,38

= 52,26228 m

Tt = ( Rr + P ) Tan ½ 3 + K

= ( 358 + 1,277 ) Tan ½ 16,760+ 52,26228

= 105,187 m

Et = ( Rr + P ) Sec ½ 3 – Rr

= ( 358 + 1,277 ) Sec ½ 16,760 - 358

= 5,154 m

31

Tabel 3.2 Data TikunganData Tikungan I Tikungan II Tikungan III

Bentuk

Vr

s

c

Ls

Lc

L

Tt

Et

K

P

Xc

S – C – S

77,70

100 km/jam

8,0060

61,688 0

100 m

385,248m

557,528 m

102,05 m

103,182 m

49,9429 m

1,165 m

99,804 m

S – C – S

66,30

100 km/jam

8,0060

50,288 0

100 m

314,054 m

485,774 m

284,526 m

70,985 m

49,9429 m

1,165 m

99,804 m

S – S

16,760

100 km/jam

8,0060

-

100 m

4,671 m

-

105,187m

5,154 m

52,26228 m

1,2772 m

-

32

Yc

Rr

e

en

4,655 m

358 m

9,9 %

2 %

4,655 m

358 m

9,9 %

2 %

-

358 m

9,9 %

2 %

Sumber : Hasil Perhitungan

I. Diagram Super Elevasi dan Sumbu Putar Jalan

1. Tikungan I Tipe S – C – S

33

P

TSLs SC=CS Ls

ST

R

Tt Et

PYc Yc

K ∆1

θs θs

Xc

θc θc

Gambar 3.1 Tikungan Belok ke kananTipe S – C – S

Kiri

CL ±0,00%

- 2%

Kanan

Ls = 100 m Lc = 385,248 m Ls = 100 m

Gambar 3.2 Diagram Superelevasi Tipe S-C-S

2. Tikungan II Tipe ( S-C-S )

34

9,9 %

- 9,9 %

TS SC CS ST

TS STLS LS

SCCS

Yc Yc

P P

K

Xc

Tt

R

ST

LC Et

θs θc θc θs

Gambar 3.3 Tikungan Belok ke KiriTipe S – C – S

Kanan

CL ±0,00%

-2%

Kiri

Ls = 100 m Lc = 314,054 m Ls = 100 m

Gambar 3.4 Diagram Superelevasi Tipe S-C-S

3. Tikungan III Tipe ( S – S )

35

+9,9 %

- 9,9 %

TS SC CS ST

P

TSLs SC=CS Ls

ST

R

Tt Et

PYc Yc

K ∆1

θs θs

Gambar 3.5 Tikungan Belok ke kanan Tipe S – S

+ 9,9

Kiri

CL ± 0,00%

Kanan

Gambar 3.6 Diagram Superelevasi Tipe S – S

II. Hitungan Stationing Titik – Titik Penting

I

III

dA-I dI-II dIII-B

dII-III

A II

Sta A = 10 + 500

dA-I = 19,8116 m

dI-II = 30,5267 m

dII-III = 26,0909 m

dIII-B = 11,6914 m

36

- 2%

- 9,9

LS = 104,66

TS ST

SC=CS LS = 104,66

B

1. Tikungan I

Sta PP1 = Sta A+ dA-1

= ( 10 + 500 ) + (19,8116)

= 10 + 519,8116

Sta Ts1 = Sta PP1 – Tt1

= (10 + 519,8116) – 339,2353

= 10 +180,6107

Sta Cs1 = Sc1

= Sta Ts1 + Ls1

= (10 +180,6107 ) + 100

= 10 + 280,6107

Sta St1 = ( Sta Sc1 = Cs1 ) + Ls1

= ( 10 + 280,6107) + 100

= 10 + 380,6107

2. Tikungan II

Sta St2 = Sta St1 + (d1-II – Tt1 – Tt2)

= ( 10 + 380,6107 ) + ( 30,5267 – 339,2353 – 284,526 )

= 9 + 212,6239

Sta Sc2 = Sta Ts2 + Ls2

= (9 + 212,6239) + 100

= 9 + 312,6239

Sta St2 = Sta Cs2 + Ls2

= (9 + 312,6239) + 100

= 9 + 412,6239

3. Tikungan III

Sta Ts3 = Sta St1 + St2 + (d1-II – Tt1 – Tt2 – dII-III – Tt3)

= ( 10 + 380,6107 ) + ( 9+412,6239) +( 30,5267 – 339,2353

-284,526-26,0909-105,187 )

37

= 19 + 68,7221

Sta Sc3 = Sta Ts3 + Ls3

= (19 + 68,7221) + 104,66

= 19 + 173,3821

Sta St3 = Sta Cs3 + Ls3

= (19 + 173,3821) + 104,66

= 19 + 278,0421

Sta B = Sta St3 + (dIII – B – Tt3)

= (19 + 278,6421) + (11,6914 – 105,187 )

= 19 + 185,1465

Panjang jalan (A – B)

= Sta B – Sta A

= (19 + 185,1465) – (10 + 500)

= 8 + 314,8535

B. Pelebaran Perkerasan pada Tikungan

L = Jarak gandar 6,09 m

A = Tonjolan depan 1,218 m

c = Kebebasan samping 0,609 m

M = Lebar kendaraan 2,436 m

n = Jumlah jalur 2

Lebar Perkerasan Normal = 2 x 3,5 m

1. Tikungan I ( S – C – S )

Diketahui : R = 358 m

V = 100 km/jam

n = 2

Wn = 7 m

a. Lebar lintasan kendaraan rencana pada tikungan ( U )

38

U = M + R –

= 2,436 + 358 –

= 2,487 m

b. Lebar melintang akibat tonjolan depan ( Td = Fa )

Td = - R

= - 358

= 0,022 m

c. Lebar tambahan akibat kelainan pengemudi ( z )

z =

=

= 0,554 m

d. Lebar perkerasan pada tikungan ( Wc )

Wc= n ( M + c ) + Td ( n – 1 ) + z

= 2 ( 2,436 + 0,609 ) + 0,022 ( 2 – 1 ) + 0,543

= 6,66 m

Wc < Wn

Jadi tidak perlu ada tambahan pelebaran perkerasan.

2. Tikungan II ( S – C – S )


V = 100 km/jam

n = 2

Wn = 7 m


U = M + R –

39

= 2,436 + 358 –

= 2,487 m


Td = - R

= - 358

= 0,022 m


z =

=

= 0,554 m


Wc= n ( M + c ) + Td ( n – 1 ) + z

= 2 ( 2,436 + 0,609 ) + 0,022 ( 2 – 1 ) + 0,543

= 6,66 m

Wc < Wn


3. Tikungan III ( S – S )


V = 100 km/jam

n = 2

Wn = 7 m


U = M + R –

= 2,436 + 358 –

40

= 2,487 m


Td = - R

= - 358

= 0,022 m


z =

=

= 0,554 m


Wc= n ( M + c ) + Td ( n – 1 ) + z

= 2 ( 2,436 + 0,609 ) + 0,022 ( 2 – 1 ) + 0,543

= 6,66 m

Wc < Wn


C. Jarak Pandang Horizontal

1.Tikungan I ( S - C - S )

a. Berdasarkan Jarak Pandang Henti ( JPH )

Diketahui : L = 2 x Ls + Lc

= 2 x 86,14 + 385,248 = 557,528

Vr = 100 km/jam

t = 2,5 detik ( t = 0,5 – 4 detik, dipakai t = 2,5 detik )

f = 0,28 ( dari Tabel Koefisien Gesek )

R = 358

41

S = d1 + d2

=

=

= 210,107 m

S < L

θ =

=

= 16,820

M = Rr x ( 1 – cos θ )

= 358 x (1 – cos 16,820)

= 15,31 m < 3,5 m (lebar Perkerasan)

M > 7

Maka perlu dipasang rambu-rambu lalu lintas.

b. Berdasarkan Jarak Pandang Menyiap (JPM)

a = 2,052 + 0,0036 x V

= 2,052 + 0,0036 x 100

= 2,412 m/dt2

t1 = 2,12 + 0,026 x V

= 2,12 + 0,026 x 100

= 4,72 m/dt

t2 = 6,56 + 0,048 x V

= 6,56 + 0,048 x 100

= 11,36 m/dt

d1 =

42

=

= 104,064 m

d2 = 0,278 x V x t2

= 0,278 x 100 x 11,36

= 315,808 m

d3 = 90 m (30 – 100 dipakai 100 m )

d4 =

=

= 210,538 m

S = d1 + d2 + d3 + d4

= 104,064 + 315,8808 + 90 + 210,538

= 720,410 m

S > L

c. Kebebasan samping

θ =

=

= 57,6770

M = R x ( 1 – cos θ )

= 358 x (1 – cos 57,6770)

= 166,585 m

M > 40 (row minimum) maka pada tikungan perlu dipasang rambu –

rambu lalu lintas, dilarang menyiap.

2. Tikungan II ( S - C - S )

43


Diketahui : L = 2 x Ls + Lc

= 2 x 86,14 + 385,248 = 557,528

Vr = 100 km/jam



R = 358

S = d1 + d2

=

=

= 210,107 m

S < L

θ =

=

= 16,820

M = Rr x ( 1 – cos θ )

= 358 x (1 – cos 16,820)

= 15,31 m < 3,5 m (lebar Perkerasan)

M > 7

Maka perlu dipasang rambu-rambu lalu lintas.


a = 2,052 + 0,0036 x V

44

= 2,052 + 0,0036 x 100

= 2,412 m/dt2

t1 = 2,12 + 0,026 x V

= 2,12 + 0,026 x 100

= 4,72 m/dt

t2 = 6,56 + 0,048 x V

= 6,56 + 0,048 x 100

= 11,36 m/dt

d1 =

=

= 104,064 m

d2 = 0,278 x V x t2

= 0,278 x 100 x 11,36

= 315,808 m

d3 = 90 m (30 – 100 dipakai 100 m )

d4 =

=

= 210,538 m

S = d1 + d2 + d3 + d4

= 104,064 + 315,8808 + 90 + 210,538

= 720,410 m

S > L


θ =

45

=

= 57,6770

M = R x ( 1 – cos θ )

= 358 x (1 – cos 57,6770)

= 166,585 m



3. Tikungan III ( S – S )


Diketahui : V = 100 km/jam



L = 2 x Ls

= 2 x 89,617

= 179,234 m

S = d1 + d2

=

=

= 80,52 m

S < L

θ =

=

= 6,44

46

M = Rr x ( 1 – cos θ )

= 358 x (1 – cos 6,440)

= 2,26 m

M < 7

Maka tidak perlu dipasang rambu-rambu lalu lintas.


a = 2,052 + 0,0036 x V

= 2,052 + 0,0036 x 100

= 2,412 m/dt2

t1 = 2,12 + 0,026 x V

= 2,12 + 0,026 x 100

= 4,72 m/dt

t2 = 6,56 + 0,048 x V

= 6,56 + 0,048 x 100

= 11,36 m/dt

d1 =

=

= 104,064 m

d2 = 0,278 x V x t2

= 0,278 x 100 x 11,36

= 315,808 m

d3 = 90 m (30 – 100 dipakai 100 m )

d4 =

=

= 210,538 m

47

S = d1 + d2 + d3 + d4

= 104,064 + 315,8808 + 90 + 210,538

= 720,410 m

S > L


θ =

=

= 57,6770

M = R x ( 1 – cos θ )

= 358 x (1 – cos 57,6770)

= 166,585 m



48

Documents

BAB III - Muhammad Haykal Mussaad's Blog | https ... · Web viewGambar 3.5 Tikungan Belok ke kanan Tipe S – S + 9,9 Kiri CL ± 0,00% Gambar 3.6 Diagram Superelevasi Tipe S – S