7. bab iii alinemen horizontal.docx

33

BAB III

PERHITUNGAN

3.1 ALINYEMEN HORIZONTAL

3.1.1 Menentukan Titik Koordinat

Dari peta kontur yang ada diperoleh data sebagai berikut :

1. Titik A = Xa = 858941,71

= Ya = 520261,05

2. Titik (PI1) = X1 = 859079,41

= Y1 = 520181,31

3. Titik (PI2) = X2 = 859169,89

= Y2 = 520102,45

4. Titik (PI3) = X3 = 859189,94

= Y3 = 519991,33

5. Titik (PI4) = X4 = 8592

6. 71,27

= Y4 = 519900,99

7. Titik (PI5) = X5 = 859162,85

= Y5 = 519788,93

8. Titik B = XB = 859153,79

= YB = 519673,89

22 )05.52026131.520181()71.858941859079,41(

34

3.1.2 Perhitungan Jarak Pandang / Jarak Lurus

d A – PI1 =

=

= 159,1218 m

d PI1 – PI2 =

=√(859169,89 - 859079,41 )2+(520102 , 45−520181 , 31)2

= 120,023 m

d PI2 – PI3 =

=√(859189,94-859169,89)2+(519991 ,33−520102 ,45)2

= 112,914 m

d PI3 – PI4 = √ ( X ₄−X ₃ ) ²+(Y ₄−Y ₃) ²

=√(859271,27-859189,94 )2+(519900 ,19−519991 ,333 )2

= 121,534 m

d PI4 – PI5 =√ ( X ₅−X ₄ ) ²+(Y ₅−Y ₄) ²

=√(859162,85-859271,27 )2+(519788 ,93−519900 ,99 )2

= 155,924 m

d PI5 – B =

=√(859153,79-859162,85 )2+(519673 ,89−519788 ,93 )2

= 115,396m

21

21 )()( AA YYXX

212

212 )()( YYXX

223

223 )()( YYXX

23

23 )()( YYXX BB

35

3.1.3 Perhitungan Sudut Putar Lengkung

Mencari Sudut Putar ( PI1 )

∆ PI1 = arc tan

X 1−XAY 1−YA

± arc tan

X 2−X 1Y 2−Y 1

= arc tan

859079 , 41−858941 ,71520181,31-520261,05 ± arc tan

859169 , 89−859079 , 41520102,45-520181,31

= (-59,925) ± (-48,925)

= 11 °


∆ PI2 = arc tan

X 2−X 1Y 2−Y 1 ± arc tan

X 3−X 2Y 3−Y 2

= arc tan

859169 , 89−859079 , 41520102,45-520181,31 ± arc tan

859189 , 94−859169 , 89519991,33-520102,45

= (-48,925) ± (-10,228)

= 38,69 °


∆ PI3 = arc tan


X 4−X 3Y 4−Y 3

= arc tan

859189 , 94−859169 , 89519991,33-520102,45 ± arc tan

859271 , 27−859189 , 94519900,99-519991,33

= (-10,228) ± (-41,995)

= 31,76 °


∆ PI4 = arc tan


X 5−X 4Y 5−Y 4

= arc tan

859271 , 27−859189 , 94519900,99-519991,33 ± arc tan

859162 , 85−859271, 27519788,93-519900,99

36

= (-41,995) ± (44,054)

= 86,05 °


∆ PI5 = arc tan


XB−X 5YB−Y 5

= arc tan

859162 , 85−859271, 27519788,93-519900,99 ± arc tan

859153 , 79−859162, 85519673,89-519788,93

= (44,054) ± (4,503)

= 39,55 °

3.1.4 Menentukan Bentuk Tikungan

a. Perhitungan tikungan PI1

Direncanakan lengkung Full - Circle ( FC )

Data-data perencanaan

Klasifikasi jalan = 1

Sudut putar β = 110

Jari-jari rencana (Rc) = 500 m

V rencana = 60 km/jam

Lebar jalan = 2 x 3.5 m

Kelandaian Jalan (e) = 10% atau 0,1

Maka :

TC = R tan ( 1/2 ∆ )

= 500 m tan ( ½ 11 )

= 48,145 m

EC = Tc x tan ¼ ∆

= 48,145 x ta ( ¼ x 11 )

= 2,31 m

11.00°

R = 500 m

Lc = 95,993 m

Tc = 48

,145 m

Ec = 2,31 m

37

LC = ( ∆ π180

−R)

= ( 11π180

−500)

= 95,993 m

Kontrol = Lc< 2 Ts

= 95,993 m < 2 x 48,145 m

= 95,993 m < 96,26 m OK

b. Perhitungan tikungan PI2

Direncanakan lengkung Spiral-Circle-Spiral ( S-C-S )



Sudut putar β = 38,690

38





Ls = 60 m

Maka :

θS =

Lsx90π xRc

=

60 x 903. 14 x 150

= 11,4590

θ c = ∆- 2* θc

= 38,69– 2 (11.4590)

= 15,7720

LC= (∆−2∗θs )∗π∗R

180

= (38,69−2∗11,459)∗π∗150

180

= 41,290 m > 20 m OK

Xs = Ls(1− Ls2

40 xRc)

= 60 ( 1- 60²

40∗150² )

= 59,76 m

Ys = Ls ²

6∗R

= 60 ²

6∗150

= 4 m

K = 29,96 mXs = 59,76 m

Ts = 82,976 m

Ls = 60 m

4

1,01

Lc = 41,29

R = 150 m

Lc = 41,29

E = 10,046

38.69°

39

P = Rc.6

Ls2

- Rc . (1 – cos θs )

=

602

6 x150 - 150. (1 – cos 11.459)

= 1,01 m

K = Ls -

Ls3

40 . R 2c - Rc .sin θ s

= 60 -

603

40 .1502− 150 sin 11.459

= 29,96 m

Ts = (Rc + P) Tg ∆/2+ K

= (150 + 1,01) Tg 38,690/2 + 29,96

= 82,976 m

Es = (Rc + P ) . sec ½ ∆ - Rc

= (150 + 1,01 ) sec ½ .38,69 0- 150

= 10,046 m

Lt = Lc + 2 x Ls

= 41,290 m + 2 x 60

= 161,29 m

Kontrol = Lt< 2 Ts

= 161,29 m < 2 x 82,976 m

= 161,29 m < 165,952 m OK

K = 29,96 mXs = 59,76 m

Ts = 82,976 m

Ls = 60 m

4

1,01

Lc = 41,29

R = 150 m

Lc = 41,29

E = 10,046

38.69°

40

c. Perhitungan tikungan PI3




Sudut putar β = 31,76º





Ls = 60 m

Maka :

θS =

Lsx90π xRc

=

60 x 903.14 x 175

= 9,822º

θ c = ∆- 2* θc

= 31,76– 2 (9,8220)

= 12,1160

LC= (∆−2∗θs )∗π∗R

180

41

= (31,76−2∗9,822 )∗π∗175

180

= 37,006 m > 20 m OK

Xs = Ls(1− Ls2

40 xRc)

= 60 ( 1- 60 ²

40∗175 ² )

= 59,824 m

Ys = Ls ²

6∗R

= 60 ²

6∗175

= 3,428 m

P = Rc.6

Ls2

- Rc . (1 – cos θs )

=

602

6 x175 - 175. (1 – cos 9,822)

= 0,864 m

K = Ls -

Ls3

40 . R 2c - Rc .sin θ s

= 60 -

603

40 .1752− 175 sin 9,822

= 29,9671m

Ts = (Rc + P) Tg ∆/2+ K

= (175+ 0,864) Tg 31,760/2 + 29,971

= 80 m

Es = (Rc + P ) . sec ½ ∆ - Rc

= (175 + 0,864 ) sec ½ .31,76 0- 175

31.76°K = 29,97Xs = 59,82

Ls = 60

Ts = 80

3,420,864Lc = 37,006

Ls = 60

R =

175

42

= 7,842 m

Lt = Lc + 2 x Ls

= 37,006 m + 2 x 60

= 157,006 m

Kontrol = Lt< 2 Ts

= 157,006 m < 2 x 80m

= 161,29 m < 160 m OK

d. Perhitungan tikungan PI4

Direncanakan lengkung Spiral-spiral ( S–S )



Sudut putar β = 86,05 0




43


Ls = 45 m

P* = 0,0723817

K* = 0,488892

Maka :

θS = ½ ∆

= ½ 86,05 0

= 43,025 0

Ls = (θS x π x R ) / 90

= (43,025 0 x 3.14 x45)/ 90

= 67,58 m

Dengan θS = 43,025 0 , dari tabel panjang besaran P* dan K* maka diperoleh nilai

besaran – besaran berikut :

P* = 0,0723817

K* = 0,488892

Dengan V = 50 km/jam dan R = 45 m, dari tabel panjang lengkung peraliha

minimum dan superelevasi yang dibutuhkan maka diperoleh nilai besara – besaran

berikut :

Ls = 45 m

e = 0,08

Ls hitung = 67,58 m < Ls tabel = 45 m OK

P = P* x Ls hitung

= 0,0723817 x 67,58 m

= 4,892 m

K = K* x Ls hitung

= 0,488892 x 67,58 m

86.05°K = 33,039Ts = 79,605

P = 4,89

E = 23,46

R = 45

LT = 135,16

44

= 33,039 m

Ts = (Rc + P) tan ( ½ ∆ ) + K

= (45m+ 4,892)tan ( ½ 86,05) + 33,039

= 79,605 m

Es = R+P

cos(12

∆)−R

= 45+4,892

cos(12

86,05)−45

= 23,246 m

Lt = 2 x Ls

= 2 x 67,58 m

= 135,16 m

e. Perhitungan tikungan PI5

45




Sudut putar β = 39,55º





Ls = 60 m

Maka :

θS =

Lsx90π xRc

=

60 x 903. 14 x 130

= 13,22º

θ c = ∆- 2* θc

= 39,55– 2 (13,220)

= 13,110

LC= (∆−2∗θs )∗π∗R

180

= (39,55−2∗913,22 )∗π∗130

180

= 27,746 m > 20 m OK

Xs = Ls(1− Ls2

40 xRc)

= 60 ( 1- 60 ²

40∗130 ² )

= 18,461 m

46

Ys = Ls ²

6∗R

= 60 ²

6∗130

= 4,615m

P = Rc.6

Ls2

- Rc . (1 – cos θs )

=

602

6 x130 - 130. (1 – cos 13,22)

= 1,170 m

K = Ls -

Ls3

40 . R 2c - Rc .sin θ s

= 60 -

603

40 .1302− 130 sin 13,22

= 29,951 m

Ts = (Rc + P) Tg ∆/2+ K

= (130+ 1,170) Tg 39,550/2 + 29,951

= 77,110 m

Es = (Rc + P ) . sec ½ ∆ - Rc

= (130 + 1,170 ) sec ½ .39,55 0- 130

= 9,390 m

Lt = Lc + 2 x Ls

= 29,746 m + 2 x 60

= 149,746 m

Kontrol = Lt< 2 Ts

= 149,746 m < 2 x 77,110m

= 161,29 m < 154,22 m OK

39.55°

R = 13

0

E = 9,

39

K = 29,95

Xs = 18,46

Xs = 18,46

4,61

1,17

47

3.1.5 Pelebaran Perkerasan Tikungan

a. Pelebaran perkerasan PI1

Diketahui data-data sebagai berikut:

Bentuk tikungan = Full Circle


Lebar perkerasan (Bn) = 7 m


48

Jumlah lajur (n) = 2 lajur

Kebebasan samping (c) = 1,0 m

Jari-JariRencana (Rc) = 500 m

Jika,

1000Rc < 6

1000500 < 6

2 < 6 ( Metode Grafis )

Sehingga:

Dari grafik diperoleh :

b’ = 2,44

td = 0,01

z = 0,57

B = n ( b’ + c ) + ( n – 1) td+ Z

= 2 ( 2.44 + 0,8 ) + ( 2-1) 0.01 + 0,57

= 7,06 m

Maka pelebaran perkerasan pada tikungan PI3 adalah:

∆b = B – Bn

= 7,06 – 7

= 0.06 m untuk sebelah kiri dan kanan jalan

Jadi pelebaran yang terjadi adalah 0,06m ( 6cm ) untuk masing – masing lajur.

b. Pelebaran perkerasan PI2


Bentuk tikungan = Spiral-Circle-Spiral


Lebar perkerasan (Bn) =7 m



49



Jika,

1000Rc > 6

1000150 > 6

6,67 > 6 (MetodeAnalitis)

Sehingga:

Z = 0,105 V

√R

= 0,105 70 km / jam

√150

= 0,6 m

Td = √ R ²+ A (2 p+A )- R

= √150²+1,2 (26,1+ A 1,2 )- 150

= 0.054 m

b‘ = 2,4 +( R - √ R ²−P ² )

= 2,4 + ( 150 - √150²−6 ,1²

= 2.254 m

B = n ( b’ + c ) + ( n – 1) td+ Z

= 2 ( 2.254 + 0,8 ) + ( 2-1) 0,054 + 0,6

= 7,302 m


b = B – Bn

= 7,302 – 7

50


Jadi pelebaran yang terjadi adalah 0,302 m ( 30,2 cm ) untuk masing – masing

lajur.

c. Pelebaran perkerasan PI3


Bentuk tikungan = Spiral – Circle - Spiral







Jika,

1000Rc < 6

1000175 < 6

5,71 < 6 ( Metode Grafis )

Sehingga:

Dari grafik diperoleh :

b’ = 2,51

td = 0,04

z = 0,47

B = n ( b’ + c ) + ( n – 1) td+ Z

= 2 ( 2.51 + 0,8 ) + ( 2-1) 0.04 + 0,47

= 7,13 m


∆b = B – Bn

= 7,13 – 7

51


Jadi pelebaran yang terjadi adalah 0,06m ( 13 cm ) untuk masing – masing lajur.

d. Pelebaran perkerasan PI4


Bentuk tikungan = Spiral-Spiral

Klasifikasi jalan = Kolektor






Jika,

1000Rc > 6

100045 > 6


Sehingga:

Z = 0,105 V

√R

= 0,105 50 km / jam

√45

= 0,783 m

Td = √ R ²+ A (2 p+A )- R

= √45²+1,2 (2∗6,1+1,2 )- 45

= 0.178 m

b‘ = 2,4 +( R - √ R ²−P ² )

= 2,4 + ( 45 - √45²−6 , 1²

52

= 2.82 m

B = n ( b’ + c ) + ( n – 1) td+ Z

= 2 ( 2.82 + 0,8 ) + ( 2-1) 0,178 + 0,783

= 8,201 m


b = B – Bn

= 8,201 – 7

= 1,201 m untuk sebelah kiri dan kanan jalan

Jadi pelebaran yang terjadi adalah 1,201 m ( 1201 cm ) untuk masing – masing

lajur.

e. Pelebaran perkerasan PI5


Bentuk tikungan = Spiral- Circle -Spiral

Klasifikasi jalan = Kolektor






Jika,

1000Rc > 6

1000130 > 6


Sehingga:

Z = 0,105 V

√R

53

= 0,105 60 km / jam

√130

= 0,552 m

Td = √ R ²+ A (2 p+A )- R

= √130²+1,2 (2∗6,1+1,2 )- 130

= 0.062 m

b‘ = 2,4 +( R - √ R ²−P ² )

= 2,4 + ( 130 - √130²−6 ,1²

= 2,54 m

B = n ( b’ + c ) + ( n – 1) td+ Z

= 2 ( 2.54 + 0,8 ) + ( 2-1) 0,062 + 0,552

= 7,294 m


b = B – Bn

= 7,294 – 7

= 0,294 m untuk sebelah kiri dan kanan jalan

Jadi pelebaran yang terjadi adalah 0,294 m ( 29,4 cm ) untuk masing – masing

lajur.

3.1.6 Perhitungan Jarak Pandangan Henti (JPH)

a. Lengkung PI1

Data yang diketahui :

V rencana : 60 km/jam

Waktu reaksi (t) : 2.5 det

Koefisien gesekan melintang dihitung dengan menggunakan persamaan berikut:

Fm = V * (-0,000625) + 0,19

54

= 60 km/jam * (-0,000625) + 0,19

= 0,153 m

D1 = 0.278* V*T

= 0.278 x 60 x 2.5

= 41.700 m

D2 =

V 2

254 x fm

=

602

254 x 0 . 153

= 92,635 m

Maka jarak pandangan henti pada tikungan PI1 adalah

DJPH = D1 + D2

= 41.700 + 92,635

= 134,335 m

Kontrol : dJPH > Dmin

133.436 > 75 OK

b. Lengkung PI2





Fm = V * (-0,000625) + 0,19

= 70 km/jam * (-0,000625) + 0,19

= 0,146 m

D1 = 0.278 V.T

= 0.278 x 70 x 2.5

55

= 48,65 m

D2 =

V 2

254 x fm

=

702

254 x 0 . 146

= 132,132 m


DJPH = D1 + D2

= 48,65+ 132,132

= 180,782 m

c. Lengkung PI3





Fm = V * (-0,000625) + 0,19

= 70 km/jam * (-0,000625) + 0,19

= 0,146 m

D1 = 0.278 V.T

= 0.278 x 70 x 2.5

= 48,65 m

D2 =

V 2

254 x fm

=

702

254 x 0 . 146

= 132,132 m


DJPH = D1 + D2

56

= 48,65+ 132,132

= 180,782 m

d. Lengkung PI4





Fm = V * (-0,000625) + 0,19

= 50 km/jam * (-0,000625) + 0,19

= 0,159 m

D1 = 0.278* V*T

= 0.278 x 50 x 2.5

= 34,75 m

D2 =

V 2

254 x fm

=

502

254 x 0 . 159

= 61,90 m


DJPH = D1 + D2

= 34,75 + 61,90

= 96,65 m

e. Lengkung PI5


57




Fm = V * (-0,000625) + 0,19

= 60 km/jam * (-0,000625) + 0,19

= 0,153 m

D1 = 0.278* V*T

= 0.278 x 60 x 2.5

= 41.700 m

D2 =

V 2

254 x fm

=

602

254 x 0 . 153

= 92,635 m


DJPH = D1 + D2

= 41.700 + 92,635

= 134,335 m

Kontrol : dJPH > Dmin

133.437 > 75 OK

3.1.7 Perhitungan Jarak Pandang Menyiap (JPM)

a. Lengkung PI1


KcptnMendahului(m) : 15 km/jam

t1 = 2.12 + 0.026* V= 2.12 + 0.026 * 60= 3.680 dtk

58

t2 = 6.56 + 0.048* V= 6.56 + 0.048* 60= 9.440 dtk

a = 2.052 + 0.0036* V= 2.052 + 0.0036* 60= 2.268 km/jam

d1 = 0.278* t1( V – m + ½ a t1 )= 0.278 * 3.42 ( 60 – 15 + ½ 2.268 3.68 )= 50.30 m

d2 = 0.278* V* t2= 0.278 *60* 9.44= 157.459 m

d3 = 30 m

d4 = 2/3 d2

= 2/3(157.459)= 105 m

DJPM = d1+ d 2 + d3 + d4

= 50.306 + 157.459 + 30 + 104.973= 342,731 m

b. Lengkung PI2



t1 = 2.12 + 0.026* V= 2.12 + 0.026 *70= 3,94 dtk

t2 = 6.56 + 0.048* V

59

= 6.56 + 0.048*70= 9,94 dtk

a = 2.052 + 0.0036* V= 2.052 + 0.0036 *70= 2.304 km/jam

d1 = 0.278* t1( V – m + ½ a t1 )= 0.278 x 3.94 ( 70 – 15 + ½ 2.304 x3.94 )= 65,21 m

d2 = 0.278* V* t2= 0.278 x 70 x 8.960= 193,043 m

d3 = 55 m

d4 = 2/3 d2

= 2/3 193,043= 128,695 m

dJPM = d1+ d 2 + d3 + d4

= 65,21 + 193,04 + 55 +128,695= 441,945 m

c. Lengkung PI3



t1 = 2.12 + 0.026* V= 2.12 + 0.026 *70= 3,94 dtk

t2 = 6.56 + 0.048* V= 6.56 + 0.048*70= 9,94 dtk

a = 2.052 + 0.0036* V= 2.052 + 0.0036 *70= 2.304 km/jam

d1 = 0.278* t1( V – m + ½ a t1 )= 0.278 x 3.94 ( 70 – 15 + ½ 2.304 x3.94 )= 65,21 m

60

d2 = 0.278* V* t2= 0.278 x 70 x 8.960= 193,043 m

d3 = 55 m

d4 = 2/3 d2

= 2/3 193,043= 128,695 m

dJPM = d1+ d 2 + d3 + d4

= 65,21 + 193,04 + 55 +128,695= 441,945 m

d. Lengkung PI4



t1 = 2.12 + 0.026 *V= 2.12 + 0.026* 50= 3,42 dtk

t2 = 6.56 + 0.048 *V= 6.56 + 0.048* 50= 8,96 dtk

a = 2.052 + 0.0036* V= 2.052 + 0.0036 * 30= 2.232 km/jam

d1 = 0.278* t1( V – m + ½ a t1 )= 0.278 x 3,42 ( 50 – 15 + ½ x 2.232 x 3,42 )= 33,878 m

d2 = 0.278* V* t2= 0.278 x 50 x8,96= 124,544 m

d3 = 30 m

d4 = 2/3 d2

= 2/3x 124,544

61

= 82,029 m

DJPM = d1+ d 2 + d3 + d4

= 33,878 + 124,544 + 30 + 82,029= 205.818 m

e. Lengkung PI5



t1 = 2.12 + 0.026* V= 2.12 + 0.026 * 60= 3.680 dtk

t2 = 6.56 + 0.048* V= 6.56 + 0.048* 60= 9.440 dtk

a = 2.052 + 0.0036* V= 2.052 + 0.0036* 60= 2.268 km/jam

d1 = 0.278* t1( V – m + ½ a t1 )= 0.278 * 3.42 ( 60 – 15 + ½ 2.268 3.68 )= 50.30 m

d2 = 0.278* V* t2= 0.278 *60* 9.44= 157.459 m

d3 = 30 m

d4 = 2/3 d2

= 2/3(157.459)= 105 m

DJPM = d1+ d 2 + d3 + d4

= 50.306 + 157.459 + 30 + 104.973= 342,731 m

3.1.8 Perhitungan Kebebasan Samping Pada Tikungan

a. Tikungan PI1

Diketahui :

62

Kecepatan Rencana ( V ) = 60 km/jam

R rencana (Rc) = 500 m

Jarak Pandang Henti (S) = 134,335 m

Jarak Pandang Menyiap (S,) = 342,731 m

PajangLengkung Horizontal (Lc) = 95,993 m

B = 7 m

a. Berdasarkan Jarak Pandangan Henti Didapatkan

S > L

134,335 < 95,993

Dari persamaandiperoleh :

R’ = Rc – ¼ (lebar jalan)

= 500 – ¼ (7)

= 498,25 m

m = R’ (1−cos90 x L

πR ) + ½ (S-L) sin( 90 x LπR )

= 498,25 (1−cos90 x95,9933,14 x500 )+ ½(134,335-95,993)sin( 90 x 95,993

3,14 x 500 ) = 4,131 m

b. Berdasarkan jarak pandangan menyiap didapat

S, > Lc

342,731 > 95,993

Dari persamaan diperoleh :

R’ = R – ¼ (lebar jalan)

= 500 – ¼ (6)

= 498,25 m

63

m = R’ (1−cos90 x LπR ’ ) + ½ (s-L) sin( 90 x L

πR ’ )=498,25 (1−cos

90 x95,9933,14 x500 )+ ½(342,731-95,993)sin( 90 x 95,993

3,14 x 500 )= 14,118 m

b. Tikungan PI2

Diketahui :





PajangLengkung Horizontal (Lt) = 161,29 m

B = 7 m


S > L

180,782 < 161,29



= 150 – ¼ (7)

= 148,25 m

m = R’ (1−cos90 x L

πR ) + ½ (S-L) sin( 90 x LπR )

= 148,25 (1−cos90 x161,293,14 x150 )+ ½(180,782-161,29)sin( 90 x 161,29

3,14 x 150 ) = 25,905 m


S, > Lc

441,945 > 161,29

64



= 150– ¼ (6)

= 148,25 m

m = R’ (1−cos90 x L

πR ) + ½ (s-L) sin( 90 x LπR )

=148,25 (1−cos90 x161,293,14 x150 )+ ½(441,945-161,29)sin( 90 x 161,29

3,14 x 150 )= 14,118 m

c. Tikungan PI3

Diketahui :






B = 7 m


S > L

180,782 > 157,006



= 175 – ¼ (7)

= 173,25 m

m = R’ (1−cos90 x L

πR ) + ½ (S-L) sin( 90 x LπR )

= 173,25 (1−cos90 x157,006

3,14 x175 )+ ½(180,782-157,006)sin( 90 x 157,0063,14 x 175 )

65

= 22,297 m


S, > Lc

441,945 > 157,006



= 175– ¼ (6)

= 173,25 m

m = R’ (1−cos90 x L

πR ) + ½ (s-L) sin( 90 x LπR )

=173,25 (1−cos90 x157,006

3,14 x175 )+ ½(441,945-157,006)sin( 90 x 157,0063,14 x 175 )

= 78,929 m

d. Tikungan PI4

Diketahui :

Kecepatan Rencana (V) = 50km/jam

R rencana (R) = 45 m




B = 7 m

a. BerdasarkanJarak Pandang HentiDidapatkan

S < L

96,65 < 135,16



= 45 – ¼ (7)

66

=43,25 m

m = R’ ¿

= 43,25 (1−cos90 x135,163,14 x 45 )

= 40,267 m

b. Berdasarkan jarak pandang menyiap didapat

S,> L

270,45 > 135,16



= 45 – ¼ (7)

= 43,25 m

m = R’ (1−cos90 x LπR ’ ) + ½ (S-L) sin( 90 x L

πR ’ )= 43,25(1−cos

90 x135,163,14 x 45 )+½(270,45-135,16)sin( 90 x 135,16

3,14 x 45 )= 107,751 m

e. Kebebasan samping pada tikungan PI5

Diketahui :

Kecepatan Rencana (V) = 60 km/jam

R rencana (R) = 130 m



PanjangLengkung Horizontal (Lc) = 149,746 m

67

B = 7 m

a. BerdasarkanJarakPandanganHentiDidapatkan

S < L

134,335 m < 149,746 m



=130 – ¼ (7)

= 128,25 m

m = R’ ¿

=128,25 (1−cos90 x149,746

3,14 x130 )= 20,689 m

b. Berdasarkanjarakpandangmenyiapdidapat

S, > L

342,731 m > 149,746 m



= 130– ¼ (7)

= 128,25 m

m = R’ (1−cos90 x LπR ’ ) + ½ (s-L) sin( 90 x L

πR ’ )= 128,25 (1−cos

90 x149,7463,14 x130 )+½(342,731-149,746)sin( 90 x 149,746

3,14 x 130 )= 73,24 m

68

3.1.9 Perhitungan Diagram Superelevasi

a. Tikungan PI1

Dikeahui data-data sebagai berikut:

Bentuk tikungan = Full- Circle




Elevasi maksimum (emax) = 8%

Elevasi Normal (enormal) = 2%

Landai maksimum (mmax) = 1/120


Elevasi as jalan = 856

Ls’ (fiktif ) = ( emaks) B x mmax

= ( 8%) 7 x 1/120

= 0,0047 m

Perhitungan Superelevasi pada Badan Jalan

Ekb Dalam = Elevasi As jalan - 12

B x emax

= 856 - 12

7 x 8%

= 855,72

Ekb Tengah = Elevasi As jalan - enormal

= 856 – 2%

= 855,98

Ekb Luar = Elevasi As jalan + 12

B x emax

= 856 + 12

7 x 8%

2%

0%

2%

10%

2%

2%

3,5 m 3,5 m

E = 0%

IV

LS LS

EN = 2%

IV III II IIIIIII

LC

69

= 856,28

Gambar diagram Superelevasi PI 1

b. Tikungan PI2


Bentuk tikungan = Spiral-Circle-Spiral










= ( 10% ) 7 x 1/120

E = 0%

IV

SCTS STCS

EN = 2%

IV III II IIIIIII

2%

0%

2%

10%

2%

2%

70

= 0,0058



B x emax

= 857 - 12

7x 10%

= 856,65


= 857 - 2%

= 856,98


B x emax

= 857 + 12

7 x 10%

= 857,35

71


c. Tikungan PI3












= (10%) 7 x 1/120

= 0,0058 m



B x emax

= 819 - 12

7 x 10%

= 818,65


= 856 – 2%

= 818,98


B x emax

E = 0%

IV

SCTS STCS

EN = 2%

IV III II IIIIIII

2%

0%

2%

10%

2%

2%

3,5 m 3,5 m

72

= 819 + 12

7 x 10%

= 819,35


d. Diagram SuperelevasiTikungan PI4

Dikeahui data-data sebagaiberikut:

Bentuk tikungan = Spiral - Spiral




Elevasimaksimum (emax)= 8%

Landaimaksimum (mmax) = 1/120

Jari-jarirencana (Rc) = 45 m


Ls’ (fiktif ) = ( emaks ) B x mmax

2%

0%

2%

8%

2%

2%

3,5 m 3,5 m

E = 0%

TS ST

EN = 2%

III II IIIIIII

73

= ( 8% ) 7 x 1/120

= 0,0047



B x emax

= 814 - 12

7x 8%

= 813,72


= 814 – 2%

= 813,98


B x emax

= 814 + 12

6 x 8%

= 814,28

`

74


e. Tikungan PI5










Elevasi as jalan = 798,5


= (10%) 7 x 1/120

= 0,0058 m



B x emax

= 798,5 - 12

7 x 10%

= 798,15


= 798,5 – 2%

= 798,48

E = 0%

IV

SCTS STCS

EN = 2%

IV III II IIIIIII

2%

0%

2%

10%

2%

2%

3,5 m 3,5 m

75


B x emax

= 798,5 + 12

7 x 10%

= 798,85


3.1.10 Perhitungan Stationing

Sta A = Sta 36 + 000

Sta TC = Sta A + (dA-P1) - Tc

= (Sta 36+000) + 159,122 – 48,145

= (Sta 36+110,98)

Sta CT = Sta TC + Lc

76

= Sta 36+110,98+ 95,99

= Sta (36 + 206,970)

Sta TS = Sta CT + 19,21(jarak dari gambar)

= Sta (36+ 206,970) – 19,21

= Sta (36 +226,18)

Sta SC = Sta TS + k

= Sta (36 + 226,18) + 29,96

= Sta (36 + 256,14)

Sta CS = Sta SC + Lc

= Sta (36 + 256,14) + 41,29

= Sta (36 + 297,43)

Sta ST = Sta CS + k

= Sta (36 + 297,43) + 29,96

= Sta (36 + 327,39)

Sta TS = Sta ST +10,52 (jarak dari gambar)

= Sta (36 +327,39) + 10,52

= Sta (36 + 337,90)

Sta SC = Sta TS + k

= Sta (36 + 337,90) + 29,97

= Sta (36 + 367,87)


= Sta (36 + 367,87) + 37,00

= Sta (36 + 404,87)

Sta ST = Sta CS + k

77

= Sta (36 + 404,87) + 29,97

= Sta (36 + 434,840)

Sta TS = Sta ST+ 29,83 (jarak dari gambar)

= Sta (36 +434,840) + 29,83

= Sta (36 + 464,67)

Sta ST = Sta TS + Ls

= sta (36 + 464,67) + 60

= sta (36 + 524,67)

Sta TS = Sta ST + 67,25 (jarak dari gambar)

= Sta (36 +524,67) + 67,25

= Sta (36 + 591,92)

Sta SC = Sta TS + k

= Sta (36 + 591,92) + 29,95

= Sta (36 + 621,87)


= Sta (36 + 621,87) + 29,74

= Sta (36 + 651,60)

Sta ST = Sta CS + k

= Sta (36 + 651,60)+ 29,95

= Sta (36 + 681,56)

Sta B = Sta ST + 68,44 (jarak dari gambar)

= Sta (36 + 685,507) + 68,44

= Sta (36 + 750)

7. bab iii alinemen horizontal.docx

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