PERHITUNGAN GEOMETRI PERHITUNGAN GEOMETRI PERHITUNGAN GEOMETRI

(1)

BAB III

PERHITUNGAN JALAN TRASE I

Data lalu lintas tahun 2007

Satuan mobil penumpang tahun 2008

SMP 2007 = ( 1+ r )^n x SMP 2007

= ( 1+0,05)2_{x 21834}

= 24532.68

Satuan mobil penumpang selama umur rencana jalan 10 th ( th 2017)

SMP 2019 = ( 1+ r )^n x SMP 2008

= ( 1+ 0,065)10_{x 24532.68}

= 48259.49

Dari data data satuan mobil penumpang dapat ditentukan :

a. jenis jalan adalah jalan Arteri kelas 2

b. jumlah lajur = 2 lajur, 2 arah

c. kecepatan rencana adalah 60 km/jam (buku bina marga tabel 2 hal. 10)

d. lebar jalur = 3.25 m (buku bina marga tabel 3 hal. 11)

Jenis kendaraan C LHR EMP SMP

Mobil penumpang 1.1 1.2 4500 1 5400

Bus 1.2 0.9 4200 2 11340

Truk ringan 1.2 L 0.9 1200 3 3240

Truk berat1 .2H 0.9 450 3 1215

Truk 1.2.2 0.9 120 3 324

Trailer 1.2 + 2.2 0.9 45 5 202.5

Trailer 1.2 - 2 0.9 25 5 112.5

Trailer 1.2 – 2.2 0.9 0 5 0

(2)

e. lebar bahu jalan = 1.5 m (bina marga tabel 8 hal 16)

f. lebar median = 0.5 m (bina marga tabel 9 hal. 17)

g. jari-jari minimum = 239 (bina marga tabel 10 hal. 18)

table 2

Kelas jalan 1 2-1 3 4-3 5-4 5

Kecp rencana

km/jam

80 60 50 40 30 20

4.2Perhitugan Azimuth Diketahui :

Koordinat titik A : ( 1000 ; 800 ) , αA=121,258º , d1 = 655,13 m Koordinat titik I : ( 1910 ; 600 ) , α1=90º , d2 = 1190 m Koordinat titik II : ( 2660 ; 600 ) , α2=56,977º , d3 = 715,612 m Koordinat titik III : ( 3400 ; 970 ) , α3=90º , d4 = 500 m Koordinat titik B : ( 4000 ; 900 )

a. Kontrol Alpha ( α )

αA = arc tg ₍ ₎

) (

1 1

A A

X X

Y Y

 

+ 90º = arc tg ₁₉₁₀600_₁₀₀₀800 + 90º = 102.39°

α1 = 90°

α2 = 90° - arc tg ₍ ₎

) (

II III

X X

Y Y

 

= 90° - arc tg ₃₄₀₀970_ ₂₆₆₀600 = 63.45°

α3 = arc tg ₍ ₎

) (

III IV

X X

Y Y

 

+ 90º = arc tg ₄₀₀₀900 ₃₄₀₀970

 

+ 90º = 96.65°

b. Kontrol Jarak ( D )

d I = (XI  XA)2(YI  YA)2

= ₍₆₀₀ ₈₀₀₎2 ₍₁₉₁₀ ₁₀₀₀₎2

 



= 931.72 m → Ok

d II = (XII  XI)2(YII  YI)2 = ₍₆₀₀_ ₆₀₀₎2 _₍₂₆₆₀_ ₁₉₁₀₎2

(3)

d III = (XIII  XII)2 (YIII  YII)2 = ₍₉₇₀_ ₆₀₀₎2 _₍₃₄₀₀_ ₂₆₆₀₎2

= 827.35 m → Ok

d IV = (XIV  XIII)2(YIV  YIII)2

= ₍₉₀₀_ ₉₇₀₎2 _₍₄₀₀₀_ ₃₄₀₀₎2

= 604.07 m → Ok

c. Perhitungan Delta ( ∆ )

Δ1 = 90°

Δ2 = 12.39°

Δ3 = 6.65°

d. Kontrol Koordinat

Koordinat titik A ( 1000 ; 1200 )

Titik 1 → X1 = XA + d1 sin αa

= 1000 + ( 931.72 x sin 102.39º )

= 1910 m

Y1 = YA + d1 sin αa

= 1200 + ( 931.72 x cos 102.39º )

= 600 m

Jadi Koordinat Titik 1 adalah ( 1910 ; 600 ) →

ok

Titik 2 → X2 = X1 + d2 sin α1

= 1910 + ( 750 x sin 90º )

= 2660 m

Y2 = Y1 + d2 sin α1

= 600 + ( 750 x cos 90º )

= 600 m

ok

Titik 3 → X3 = X2 + d3 sin α2

(4)

= 3400 m Y3 = Y2 + d3 sin α2

= 600 + ( 827.35 x cos 63.45º )

= 970 m → 1250 m

ok

Titik B → Xb = X3 + d4 sin α3

= 3400 + ( 604.07 x sin 96.65º )

= 4000 m

Yb = Y3 + d4 sin α3

= 970 + (604.07 x cos 96.65º )

= 900 m

Jadi Koordinat Titik B adalah ( 4000 ; 900 ) →

ok

Tabel : 1

TITIK KOORDINAT JARAK (m) AZIMUTH

X Y

A 1000 800

931.72 102.39°

1 1910 600

750 90O

2 2660 600

827.35 63.45°

3 3400 970

604.07 96.65O

(5)

4.3 Perhitungan Potongan

Potongan A - B

Elevasi Kiri Elevasi As Jalan Elevasi Kanan

1-1 158.06 156.91 155.76

2-2 153.71 152.94 152.17

3-3 150.41 149.455 148.5

4-4 147.04 145.855 144.67

5-5 143.7 144.78 145.85

6-6 142.38 143.57 144.75

7-7 141.78 142.87 143.96

8-8 141.81 142.64 143.46

9-9 141.58 142.19 142.8

10-10 141.84 142.22 142.6

11-11 141.1 141.28 141.46

12-12 140.1 140.1 140.1

13-13 137.77 137.59 137.4

14-14 135.5 135.03 134.56

15-15 129.81 129.72 129.62

16-16 127.37 127.35 127.38

17-17 125 125.35 125.45

18-18 126.16 125.72 125.27

19-19 126.46 125.05 125.64

(6)

21-21 126.88 126.495 126.11

22-22 126.15 126.53 126.91

23-23 126.09 126.49 126.88

24-24 126.06 125.54 125.02

25-25 123.4 122.8 122.2

26-26 120.74 120.11 119.48

27-27 118.18 117.69 117.2

28-28 115.34 115.27 115.19

29-29 117.82 118.075 118.33

30-30 121.13 121.36 121.58

31-31 124.47 124.61 124.75

32-32 127.39 127.39 127.39

33-33 130.31 130.59 130.78

34-34 134.22 133.78 133.33

35-35 135.55 135.17 134.78

36-36 136.18 135.71 135.24

37-37 136.71 136.075 135.44

38-38 136.99 136.29 135.59

39-39 137.05 136.315 135.58

40-40 136.84 136.295 135.75

41-41 137.64 136.89 136.14

42-42 138.16 137.37 136.57

43-43 139 138.15 137.3

(7)

45-45 142.05 140.88 139.7

46-46 144.67 143.495 142.32

47-47 145.89 145.28 144.63

48-48 146.52 146.02 145.51

49-49 147.08 146.75 146.41

50-50 146.53 145.55 144.56

51-51 144.4 143 141.6

52-52 141.09 138.55 136

53-53 137.4 135.2 133

54-54 137.39 134.27 131.14

55-55 136.33 133.82 131.3

56-56 136.3 133.57 130.83

57-57 136.38 133.89 131.38

58-58 135.76 133.84 131.92

59-59 135.43 133.695 131.96

60-60 135 133.02 131.04

61-61 132.23 131.34 130.45

62-62 131.67 130.69 129.7

63-63 130.22 129.36 128.5

64-64 128.86 128.08 127.3

65-65 126.02 125.17 124.32

(8)

67-67 109.12 109.11 109.1

68-68 116.5 116.59 116.67

69-69 123.36 123.58 123.22

70-70 127.8 127.54 127.27

71-71 131.36 131.12 130.88

72-72 135 134.35 133.69

73-73 141.66 140.205 138.75

74-74 142.59 141.73 140.96

4.4 Data kondisi medan

Potongan 1 =

40 2 . 157 19 . 160 

x 100 % = 7.5 %

Potongan 2 =

40

88 . 154 43 . 156 

x 100 % = 3.9 %

Potongan 3 =

40 32 . 150 9 . 151 

x 100 % = 3.95 %

Potongan 4 =

40

09 . 147 33 . 148 

x 100 % = 3.1 %)

Potongan 5 =

40 7 . 143 5 . 145 

x 100 % = 4.5 %

Potongan 6 =

40

39 . 142 75 . 144 

x 100 % = 5.9 %

Potongan 7 =

40

79 . 141 96 . 143 

x 100 % = 5.425 %

Potongan 8 =

40

91 . 141 64 . 142 

x 100% = 3.875 %

Potongan 9 =

40 59 . 141 9 . 142 

(9)

(10)

(11)

(12)

(13)

Potongan 73 =

40

75 . 139 66 . 141 

x 100 % = 4.775 %

Potongan 74 =

40

96 . 140 59 . 142 

x 100 % = 4.075 %

4.4 PERHITUNGAN ALINYEMEN HORISONTAL a) Tikungan 1

Dipakai Rmin = 112.04 m (denagn Vr 60 km/jam, teknik perencanaan jalan raya hal 95)

Rc = 358 m (ditentukan sendiri)

E = 0,054

Ls = 50 m ( Tabel Bina Marga )

Vr = 60 km/jam

∆1 = 12.39°

Perhitungan Tikungan 1

Cek rumus modified short form

Qs =

R Ls



648 . 28

=

358 50 648 .

28 

= 4º

Qc = Δ – 2Qs

(14)

Lc =

360 Qc

2πR

=

360 39 . 4

2xπx358

= 27.43 m >20 m, maka digunakan S-C-S

Xc = Ls _

  

 



 ₂

2

40 1

Rc Ls

= 50 _

  

 



 ₂

2

358 40

50 1

= 49.97m

Yc =

Rc Ls



6

2

=

358 6

502 

= 1.16 m

P = Yc Rc



1 cosQs



= 1.16 – 358 ( 1 – cos 4° ) = 0.29 m K = Xc – Rc sin Qs = 49.97 – 358 sin 4° = 25 m

Ts =



RcP



tg  K 2

1 ₌

_

_

₁₂_.₃₉ ₂₅

2 1 29 . 0

358 tg  

= 63.86 m

Es =



Rc P



Rc



_



_

358 2.39m

39 . 12 cos

29 . 0 358

cos12 1 12

 

 

  



L = 2



Ls + Lc

= 2



50 + 27.43 = 127.43 m

b) Tikungan 2

E = 0,064

(15)

Vr = 60 km/jam ∆1 = 26.55°

Qs =

R Ls



648 . 28

=

286 50 648 .

28 

= 5º

Qc = Δ – 2Qs

= 26.55º – (2x5º) = 16.55º

Lc =

360 Qc

2πR

=

360 55 . 16

2xπx286

= 82.61m >25 m, maka digunakan S-C-S

Xc = Ls _

  

 



 ₂

2

40 1

Rc Ls

= 50 _

  

 



 ₂

2

286 40

50 1

= 49.96 m

Yc =

Rc Ls



6

2

=

286 6

502 

= 1.46 m

P = Yc Rc



1 cosQs



= 1.46 – 286 ( 1 – cos 5° ) = 0.37m K = Xc – Rc sin Qs = 49.96 – 286 sin 5° = 25.03 m

Ts =



RcP



tg  K 2

1 ₌

_

_

₂₆_.₅₅ ₂₅_.₀₃

2 1 37 . 0

(16)

= 92.59 m

Es =



Rc P



Rc



_



_

286 8.23.m

55 . 26 cos

37 . 0 286

cos12 1 12

 

 

  



L = 2



Ls + Lc

= 2



50 + 82.61 = 182.61 m

c) Tikungan 3

E = 0,08

Ls = 50 m ( Tabel Bina Marga )

Vr = 60 km/jam

∆1 = 39.85°

Qs =

R Ls



648 . 28

=

205 50 648 .

28 

= 6.99º

Qc = Δ – 2Qs

= 39.85º – (2x6.99º) = 25.87º

Lc =

360 Qc

2πR

=

360 87 . 25

2xπx205

(17)

Xc = Ls _

  

 



 ₂

2

40 1

Rc Ls

= 50 _

  

 



 ₂

2

205 40

50 1

= 49.93 m

Yc =

Rc Ls



6

2

=

205 6

502 

= 2.03 m

P = Yc Rc



1 cosQs



_{= 2.03 – 205 ( 1 – cos 6.99° ) = 0.51m}

K = Xc – Rc sin Qs = 69,933 – 205 sin 6.99° = 24.98 m

Ts =



RcP



tg  K 2

1 ₌

_

_

₃₉_.₈₅ ₂₄_.₉₈

2 1 51 . 0

205 tg  

= 99.48 m

Es =



Rc P



Rc



_



_

205 13.59.m

85 . 39 cos

51 . 0 205

cos12 1 12

 

 

  



L = 2



Ls + Lc

= 2



50 + 92.58 = 192.58 m

d) Perhitungan stationing Diketahui : Sta = 0 + 000

D1 = 931.72 m

D2 = 750 m

D3 = 827.35 m

D4 = 604.07 m

Perhitungan :

- Sta A = 0 + 000 - Sta A-I = Sta A + d1

= ( 0 + 00 ) + 931.72 = 0 + 931.72

(18)

= ( 0+ 931.72 ) - 63.86 = 0+ 867.86

- Sta Sc1 = Sta TS1 + LS

= (0+ 867.86) + 50 = 0 + 917.86

- Sta Cs1 = Sta Sc1 +Lc

= (0+ 917.86) + 27.43 = 0+ 945.29

-Sta St1 = Sta Cs1 +Ls

= (0+ 945.29) + 50 = 0+ 995.29

- Sta II = Sta ST1 + D2 – Ts

= (0+995.29) + 750 – 63.86 = 1 + 681.43

- Sta Ts2 = Sta II+ Ts

=(1 + 681.43) - 92.59 = 2 + 338.84

-Sta SC2 = Sta Ts2 +Ls

= (2 + 338.84) + 50 = 2 + 388.84 - Sta CS2 = Sta SC2 + Lc

= (2 + 388.84) + 82.61 = 2 + 471.45

- Sta ST2 = Sta CS2 + Ls

= (2+ 471.45) + 50 = 2+521.45

- Sta III = Sta ST2 – Ts + D3

= (2+ 459,385) – 92.59 + 827.35 = 3 + 256.21

- Sta III-B = Sta III+ D3

(19)

- Sta TS3 = Sta III-B – Ts3

4.5 Perhitungan jarak pandang henti ( JPH = D)

(20)

JPH =

4.6 Perhitungan jarak pandang menyiap

(21)

t2 = 6.56 + 0.048 x VR

= 6.56 + 0.048 x 60 = 9.44 s

d2 = 0.278 x VRx t2

= 0.278 x 60 x 9.44 = 157.456 m

d3 = 30 – 100 m ( direncanakan 30 m )

d4 = 157.456 104.971

3 2 3

2

2  

D

Jd = d1+d2+d3+d4

= 54.398 + 157.456 + 30 + 104.971 = 346.825

4.7 Pelebaran pada tikungan

Tikungan I b = 2.5 m

Ri = Rrencana – b

= 358 – 2.5 = 355.5 m Rc = Ri + ½ b

= 355.5 + ½ 2.5 = 356.75 m

B =



Rc P A xb



P A Rc p A xb

2 1 ) ( )

( 2

1 )

( 2 2 2 2 2 2

2

   

 



B =





2.5

2 1 ) 5 . 2 5 . 6 ( 5 . 355 )

5 . 2 5 . 6 ( 5 . 2 2 1 ) 5 . 2 5 . 6 ( 5 .

355 2 2 _x 2 2 2 2 2 _x

 

 

 



B = 358.62 – 355.41 B = 3.21 m

ε

= B – b

= 3.21 – 2.5 = 0.71 m

(22)

b = 2.5 m Tikungan III

(23)

4.8 Alinyemen vertikal

a. Stasioning

Perhitungan Lengkung yang dipakai :

Titik A Sta 0 + 0 Elevasi 160 m

PPV 1 Sta 0 + 500 Elevasi 160 m PPV 2 Sta 1 + 000 Elevasi 142 m

PPV 3 Sta 1 + 500 Elevasi 120 m

b. Perhitungan Landai

g1 = 1000% 3.6%

500 142 160

 



( naik )

g2 = 100% 4.4%

500 120 142

 



( naik )

g3 = 100% 2%

500 130 120

  



( turun )

g4 = 100% 3.2%

500 146 130

  



( turun )

g5 = 100% 3.2%

500 130 146

 



( naik )

g6 = 100% 0%

430 130 130

 



( datar )

g7 = 100% 7.06%

170 142 130

  



( turun )

c. Perhitungan Jarak Pandang Henti ( JPH ) Rumus yang digunakan :

JPH = 0,278



V



t + V

_

_f _G

_

 

(24)

₀_,₃₃ ₀_.₀₇₀₆

_

60

 

= 42.61 m

d. Perhitungan Panjang Lengkung ( Lv ) dan Tinggi Lengkung ( Ev )

Rumus yang digunakan :

Lv cembung =

399 JPHn

 

Lv cekung =

xJhn JPHn

5 . 3 120

(25)

Ev =

800 Lvn

 

Dengan : ∆ = perbedaan landai

JPH n = panjang jarak pandang henti Lv n = panjang lengkung

1. Lengkung 1

∆1 = g2 – g1 = 4.4 – 3.6 = 0.8 % ( Cekung )

2. Lengkung 2

∆2 = g3 – g2 = -2 - 4.4 = -6.4% ( Cembung )

3. Lengkung 3

∆3 = g4 – g3 = -3.2 – (-2) = -1.2 % ( Cembung )

4 Lengkung 4

∆4 = g5 – g4 = 3.2 – (-3.2) = 6.4 % ( cekung )

(26)

∆5 = g6 – g5= 0 – 3.2 = -32 % ( cembung )

Lv5 =

399 35 . 42 2 .

3 _ 2

= 14.38 m m

Ev5 =

800 38 . 14 2 . 3 

= 0.05752 m

6. Lengkung 6

∆6 = g7 – g6 = -7.06 – 0 = -7.06 % ( Cembung )

Lv6 =

399 42 . 42 06 .

7 _ 2

= 31.84 m

Ev6 =

800 84 . 31 06 .

7 

= 0.28 m

7. Lengkung 7

∆7 = g8 – g7 = 0 – (-7.06) = -7.06 % ( Cembung )

Lv6 =

399 42 . 42 06 .

7 _ 2

= 31.84 m

Ev6 =

800 84 . 31 06 .

7 