bab iii tugas jalan raya

BAB III

PERHITUNGAN GEOMETRIK JALAN RAYA

A. PenentuanJenis Medan

Penentuan jenis medan berdasarkan perbedaan elevasi

tanah asli pada trase jalan yang direncanakan, yang dapat

dilihat pada potongan memanjang trase jalan.

Tabel 3.1 Klafisikasi Medan

Klasifikasi Medan

Golongan Medan LerengMelintang

Datar (D) 0 % - 9.9 %

Perbukitan (B) 10 % - 24.9 %

Pegunungan (G) 25 % keatas

Tabel 3.2 HasilHitunganKemiringan Lereng

- Potongan A-A

0.5 Kemiringan = 59.5– 60 x

100%

30

30 m = 1.67 %

- Potongan I-I

Kemiringan = 60.5 – 61.5 x

100%

1 30

30 m = 3.33 %

- Potongan II-II

Kemiringan = 63.5 – 64 x

100%24

0.5 30

30 m

= 1.67 %

- Potongan III-III

kemiringan = 66.5-67 x 100

%

0.5 30

30 m = 1.67 %

- Potongan IV-IV

kemiringan = 65- 64 x 100 %

1 30

30 m = 3.33 %

- Potongan V-V

Kemiringan = 62.5–62.5

x 100%

30

30 m = 0 %

- Potongan VI-VI

Kemiringan = 64 – 62.5 x

100%

1.5 30

30 m = 5 %

- Potongan VII-VII

Kemiringan = 66 - 65 x

100%

1 30

25

30 m = 3.33 %

- Potongan VIII-VIII

Kemiringan = 65– 64.5 x

100%

0.5 30

30 m = 1.67 %

- Potongan IX-IX

kemiringan = 64 - 64 x 100

%

30

30 m

= 0 %

- Potongan X-X

Kemiringan = 65.5 –65.5 x

100%

30

30 m = 0 %

- Potongan XI-XI

Kemiringan = 67 – 67 x

100%

30

30 m = 0 %

- Potongan XII-XII

Kemiringan = 66.5 –66.5 x

100%

26

30

30 m = 0 %

- Potongan XIII-XIII

Kemiringan = 64.5 – 63 x

100%

0.5 30

30 m = 5 %

- Potongan B-B

Kemiringan = 64 – 63 x

100%

1 30

30 m = 3.33 %

Kemiringan rata-rata = 28.33

15

= 1.8887 %

Dari

hasilperhitunganmakadiperolehjenisjalanmendatar

dantermasuk dalamkatagoridatarberdasarkan table 3.1

B. Perhitungan Geometri Jalan

95◦

y1 z1 C D z3 y3

A x1 x2

x3 B

27

STA JARAK

X Y ZSTA I A-C 200 17.623 200.775 STA II 0 0 297.661C-D STA III 197.075 81.217 213.154

Perhitungankoordinat

Titik STAJarakKoordinat Koordinat Koordinat

X Y X Y ( X , Y ) A 0 0 2200 1500 (2200 ; 1500)

C 20017.62

3 24001517.6

2 ( 2400 ; 1517.62 )

D 197.07581.21

72597.07

51591.3

3 (2597.075 ;1591.33)

B 0 02597.07

51591.3

3(2597.075 ; 1591.33)

C. Alinyemen Horisontal

1.Menentukan Sudut (Pertemuan antara Dua

Tangen)

Diketahui :

Koordinat A (2200 ;1500)

A terletakpadatangen azimuth 95

a. SudutTikungan I ( β1 )Y

95 º

17.623 C 90º X28

A 200 95 º β

β1 = arc tan (17.623 / 200)

= 5° 2’ 8.2’’

b. SudutTikungan II ( β2 ) Y

81.217

X 90º Dβ B

197.075

β2 = arc tan (81.217 / 197.075)

= 24° 20’ 15.31’’

2.Perhitungan Tikungan I

a. Tikungan I

Diketahui :

Kecepatanrencana (Vr) = 80 km/jam

Lebarperkerasan (B) = (2 x 3.5) m

Lerengmelintangperkerasan (en) = 2 %

Lerengmelintangbahu = 6 %

Landaimaksimumtikungan (e maks) = 10 %

Jari-jarilengkung minimum (R min)= 210 m

Landaimaksimum = 5 %

29

1 = 5° 2’ 8.2’’ = 5.035º

R renc 1 = 318 m padatabel 4.7,

makadidapatnilai :

Ls1 = 70 m

e = 0.088

D = 4.5º

θs1 = 90 Ls1

π R renc 1 =

90 × 70π × 318

= 6.306 °

θc1 = β1 - 2θs1 = 5.04° (2 × 7.217 ° ) = 7.572°

Lc1=θc2

360°×2π× R renc 2 =

7.572 °360°

×2π×318 =42.14 m

Lc1>20 m (ok bisadipakailengkung S-C-S)

L1=2Ls1+Lc1 = (2× 70 ) +42.14 =182.14 m

P1 =Ls1

2

6 R renc 1

- R renc 1 (1- cosθs1 )

P1 =7 02

6×318-( 318 × (1- cos 6.306º ) )

P1 =0.644 m

k= Ls1-Ls1

3

40 R renc 12 -( R renc 1 sin θs1 )

k= 70-7 03

40×31 82 -( 318 ×sin 6.306 °)

k= 34.986 m

30

Es1= ( R renc 1+P1 ) sec12

β1-Rc

Es1= (318+ 0.644 ) sec(12

× 5.035 °)-318

Es1=2.015 m

Ts1 = Sin 121(Rc + Es ) + k

Ts1 = Sin 12

5.035 ° (318+ 2.015 ) + 34.986

Ts1 = 49.796m

Data lengkunguntuklengkung spiral-circle-spiral

padatikunganpertamadalah :

V = 80 km/jam 1 = 5 °2'8.2 ''

θs1 = 6.306° R renc 1 =318 m

Es1 =2.015m TS1 = 49.796m

L1 = 182.14m e = 0.088

Ls1 = 70 m Lc1 =42.14m

p = 0.644m k = 34.086m

en = 2 % B = 3.5 m

θc1 = 7.572°

Landairelatif

1m

=(e+en )Ls1

×B=(0.0 88 +0.02 )7 0

×3.5=0.0054 m

b. Pelebaran Tikungan I

Diketahui : B = (2 x 3.5) m, b = 2.5 m,

p = 6.5 m, A = 1.5 m

31

R i =Rrenc -12

B= 318 -(12 ×3.5)=316.25 m

R c = √(R i + 12

b)2

+ ( p + A )2

R c = √(316.25 + 12

2.5)2

+ ( 6.5 + 1.5 )2= 317.5 m

B = [√ {√Rc2 - 64 + 1.25}

2

+ 64 - √Rc2 - 64 + 1.25]

B = [√{√317.52 - 64 + 1.25 }2+ 64 – √317.52 - 64 + 1.25 ]=2.6 m

Z=0.105V

√R i

=0.105×8 0

√316.25=0.472 m

C=1 m

Bt =n ( B+C) +Z= [2× (2.6+1 ) ]+0.472 =7.67 m

∆b= Bt - Bn.........Bn (Lebarperkerasan normal)=2×3.5=7 .0 m

∆b=7. 67 -7 .00 =0.67 m

c. Jarak Pandang Lengkung HorizontalTikungan I

R = Rrenc +12

B= 318 +(12 ×3.5)=319.75 m

S= (0.278 V renc t ) +Vrenc

2

254 f m

fm=Koefisiengesek ban denganmukajalan=0.3

S= (0.278× 80×2.5 )+8 02

254×0.3=139.59 m

32

θ=90 Sπ R

=90×139.59π× 319.75

=12.507 °

L=20 4.026 m

Karena S < L, maka

M=R (1-cosθ ) =319.75 × (1-cos12.507 °) =7 .588 m.

d.StasioningTikungan I

Titik A terletak pada STA 20 + 500

d1 = 200.775m,

d2 = 297.661 m,

d3 = 213.154m

LS1 = 70 m,

LC1 = 42.14m,

TS1 = 49.796m

STA A = 20 + 500

STA TS1 = STA A + (d1 - TS1)

= 20 + 500 + (200.775–49.796) = 20 +

588.571

STA SC1 = STA TS1 + LS1

= 20 + 650.979+ 70 = 20 + 658.571

3. Perhitungan Tikungan II

a. Tikungan II

Diketahui :

33

Kecepatanrencana (Vr) = 80 km/jam

Lebarperkerasan (B) = (2 x 3.5) m

Lerengmelintangperkerasan (en) = 2 %

Lerengmelintangbahu = 6 %

Landaimaksimumtikungan (e maks) = 10 %

Jari-jarilengkung minimum (R min)= 210 m

Landaimaksimum = 5 %

2 = 24° 20’ 15.31’’

R renc 1 = 286 m padatabel 4.7,

makadidapatnilai :

Ls2 = 70 m

e = 0.093

D = 5.00º

θs2 =90 Ls2

π R renc 2

=90×70π× 286

=7.012 °

θc2 =β2-2θs2 =24 ° 20’ 15.31 ’’- (2× 7.012 °) =10.314 °

Lc2=θc2

360°×2π× R renc 2 =

1 0.314 °360°

×2π×286 =51.484 m

Lc2>20 m (ok bisadipakailengkung S-C-S)

L2 =2Ls2+Lc2 = (2× 70 )+51.484 =191.484 m

P2 =Ls2

2

6 R renc 2

- R renc 2 (1- cosθs2 )

P2 =7 02

6×286-(286 × (1-cos7.012 ° ))

P2 = 0 .716 m

34

k= Ls2-Ls2

3

40 R renc 22 -( R renc 2 sin θs2 )

k=50-703

40×2862 -( 286 ×sin 7.012 °)

k= 34.98 m

Es2= ( R renc 1+ P1 ) sec12

β1-Rc

Es2= (286 +0.716 )sec(12 ×24 °20 ' 15.31 '')- 286

Es2=7.306 m

Ts2 = Sin 121(Rc + Es ) + k

Ts2 = Sin 12

24 °20 ' 15 .31 ''(286 + 7.3 06 ) + 34.98

Ts2= 96.807m

Data lengkunguntuklengkung spiral-circle-spiral

padatikungankeduaadalah :

V = 80 km/jam 2 = 24° 20’ 15.13’’

θs2 = 7.012 ° R renc 2 = 286 m

Es2 = 7.306m TS2 = 96.807m

L2 = 191.484m e = 0.093

Ls2 = 70 m Lc2 = 51.484m

p = 0.716m k = 34.98m

en = 2 % B = 3.5 m

θc2 = 1 0.314°

Landairelatif35

1m

=(e+en )Ls1

×B=(0.0 93 +0.02 )7 0

×3.5=0.00565 m

b. Pelebaran Tikungan II

Diketahui :

b = 2.5 m, p = 6.5 m, A = 1.5 m

R i = Rrenc -12

B= 286 -(12

×3. 5)= 284.25 m

R c = √(R i + 12

b)2

+ ( p + A )2

R c = √(284.25 + 12

2.5)2

+ ( 6.5 + 1.5 )2= 285.5 m

B = [√ {√Rc2 - 64 + 1.25}

2

+ 64 - √Rc2 - 64 + 1.25]

B = [√{√285.52 - 64 + 1.25 }2+ 64 – √285.52- 64 + 1.25 ]=2.61 m

Z=0.105V

√R i

=0.105×8 0

√284 .25=0.498 m

C=1 m

Bt =n ( B+C) +Z= [2× (2.6 1+1 ) ]+0.498 =7.7215 m

∆b= Bt - Bn.........Bn (Lebarperkerasan normal)=2×3.5=7 m

∆b=7. 7215 –7.00 =0.7215 m

c. Jarak Pandang Lengkung HorizontalTikungan II

R i =Rrenc -12

B= 286 -(12

×3.5)=284.25 m

36

S= (0.278 V renc t ) +Vrenc

2

254 f m

f m=Koefisiengesek ban denganmukajalan=0.3

S= (0.278× 80×2.5 )+8 02

254×0.3=139.59 m

θ=90 Sπ R

=90× 139.59π× 284,25

=14.076 °

L= 191.484 m

Karena S < L, maka

M=R (1-cosθ ) =284 × (1-cos 1 4.076 °) = 8.53 m

STA C = STA A + d1

= 20 + 500 + 200.775= 20 + 700.775

STA CS1 = STA SC1 + LC1

= 20 + 720.979+ 42.14= 20 + 742.915

STA ST1 = STA CS1 + LS1

= 20 +763.119+ 70 = 20 + 812.915

d. Stasioning Tikungan II

Titik C padaSTA 20 + 700.775

d1 = 200.775m,

d2 = 297.661 m,

d3 = 213.154m

LS2 = 70 m,

LC2 = 51.484m,

TS2 = 96.807m

STA C = 20 + 700.775

37

STA TS2 = STA C + (d2 - TS2)

= 20 + 700.775+ (297.661-96.807) = 20 +

876.952

STA SC2 = STA TS2 + LS2

= 20 +971.629+ 70 = 20 + 946.952

STA D = STA C + d2

=20 + 700.775+ 297.661= 20 + 998.436

STA CS2 = STA SC2 + LC2

= 20+ 998.436+ 51.484= 21 + 49.920

STA ST2 = STA CS2 + LS2

=21 + 119.92+ 70 = 21 + 119.920

STA B = STA D + d3

= 21 + 68.436+ 213.154= 21 + 211.59

D. Alinyemen Vertikal

1.Perhitungan Kelandaian Jalan

Titik A = STA 20 + 500

Elevasi = Elevasi tanah asli - galian 0.75 m

= 59.75- 0.75 = 59 m

PPV 1 = STA 20 + 700

Elevasi = 65 m

PPV 2 = STA 21 +0.50

Elevasi = 65.5 m

- Titik B = STA 21 + 211.59

Elevasi = 63.5 m

38

Menurut PPJR 1990 dalam daftar 1 untuk jalan raya kelas II

B dengan jenis medan datar mempunyai syarat dengan

landai relatif maksimal = 5 %

g1 = elevasi PPV1 - elevasi Ajarak

× 100% = 65 – 59200

× 100% =3 %

g2 = elevasi PPV2 – elevasi PPV1jarak

× 100% = 65 – 653 5 0

× 100% = 0 %

g3 = elevasi B - elevasi PPV 2jarak

× 100% =63.5 – 65144.686

× 100% = -1.04 %

4. Perhitungan Lengkung yang dipakai

a. PPV 1 (LengkungCembung)

Perbedaanlandai = Δ1 = g1 – g2 = 3 % – 0 % = 3%

Untuk Δ1 = 3 % danVrenc = 80 km/jam

denganmelihatgrafikpanjanglengkungverticalcembungdi

dapat LV1 = 90 m

EV1 = ∆1 × LV1

800 =

3 × 90800

= 0.33 m

b. PPV 2 (LengkungCembung)

39

Perbedaanlandai = Δ2 = g2 – g3 = 0 % – (- 1.04%) =

1.04%

Untuk Δ2 = 1.04 % danVrenc = 80 km/jam

denganmelihatgrafikpanjanglengkungverticalcembungdi

dapat LV2 = 46 m

EV2 = ∆2 × LV2

800 =

1.04 × 45800

= 0.06 m

5.Stasioning

Titik A terletak pada STA 20 + 500 dengan elevasi sumbu

jalan 59 m

PPV 1 = STA 20 + 700

Elevasi = 65 m

PLV 1 = STA PPV 1 – (0.5LV1)

= STA 20 +700– (0.5 × 90)

= STA 20 + 655

Elevasi = Elevasi PPV 1 – (g1× 0.5LV1)

= 65 – (3 % × 0.5 × 90)

= 63.65 m

PTV 1 = STA PPV 1 + (0.5LV1)

= STA 20 + 700+ (0.5 × 90)

= STA 20 + 745

Elevasi = Elevasi PPV 1 + (g2 × 0.5LV1)

= 65 + (0% × 0.5 × 90)

= 65 m40

PPV 2 = STA 21 + 0.50

Elevasi = 65 m

PLV 2 = STA PPV 2 – (0.5LV2)

= STA21 + 0.50– (0.5 × 46)

= STA 21 + 027

Elevasi = Elevasi PPV 2 – (g2 × 0.5LV2)

= 65 – (0% × 0.5 × 46)

= 65 m

PTV 2 = STA PPV 2 + (0.5LV2)

= STA 21 + 0.50+ (0.5 × 46)= STA 21 + 073

Elevasi = Elevasi PPV 2 + (g3 × 0.5LV2)

= 65 + (-1.04 % × 0.5 × 46)

= 64.8 m

E. PerhitunganElevasiSumbuPusatJalan

1. Potongan A-A (STA 20 + 500)

Elevasi tanah asli=59.75m

Elevasi sumbu jalan =Elevasi tanah asli – galian 0.75 m

= 59.75- 0.75

= 59 m

2. Potongan I-I (STA 20 + 550)

Elevasi tanah asli = 61 m

Elevasi sumbu jalan =Elv PPV 1 – [g1 (STA PPV 1 – STA I-

I)]

= 65 – {3 % [(20 + 700) – (20 + 550)]}

= 60.5 m

3. Potongan TS1 – TS1 (STA 20 + 588.571)

Elevasi tanah asli= 63 m

Elevasi sumbu jalan =PPV 1 – [g1 (STA PPV 1 – STA I-I)]

= 65 – {3 % [(20 + 700) – (20 +

588.571)]

41

= 61.65 m

4. Potongan II-II (STA 20 + 600)

Elevasi tanah asli= 63.5 m

Elevasi sumbu jalan =Elv PPV 1 – [g1 (STA PPV 1 – STA

II-II)]

=65 – {3 % [(20 + 700) – (20 + 600)]}

=62m

5. Potongan III-III (STA 20 + 650)



II-II)]

=65– {3 % [(20 + 700) – (20 + 650)]}

=63.5 m

6. Potongan SC1 - SC1 (STA 20 + 658.571)



SC1-SC1)]

=65– {3 % [(20 + 700) – (20 +

658.571)]}

=63.76 m

7. Potongan IV-IV (STA 20 + 700)



II-II)]

= 65 – {3 % [(20 + 700) – (20 + 700)]}

= 65 m

8. Potongan C-C (STA 20 + 700.775)


Elevasi sumbu jalan =Elv PPV 1 = Elv PPV 2

= 65 m

42

9. Potongan CS1-CS1 (STA 20 + 742.915)



=65m

10. Potongan V-V (STA 20 + 750)


Elevasi sumbu jalan =Elv PPV 1 = Elv PPV 2 =66.5 m

11. Potongan VI-VI(STA 20 + 800)



=65 m

12. Potongan ST1-ST1 (STA 20 + 812.915)



=65m

13. Potongan VII-VII(STA 20 + 850)



=65 m

14. Potongan TS2-TS2(STA 20 + 876.952)



=65 m

15. Potongan VIII-VIII(STA 20 + 900)



=65 m

16. Potongan SC2-SC2(STA 20 + 946.952)


43


=65 m

17. Potongan IX-IX(STA 20 + 950)



=65 m

18. Potongan D-D(STA 20 + 998.436)



=65 m

19. Potongan X-X(STA 21 + 000)



=65 m

20. Potongan CS2-CS2(STA 21 + 049.92)



=65 m

21. Potongan XI-XI(STA 21 + 050)



=65 m

22. Potongan XII-XII(STA 21 + 100)



XII-XII)]

= 65 – {-1.04 % [(21+ 050) – (21 +

100)]}

=64.5 m

23. Potongan ST2-ST2(STA 21 + 119.92)

44



ST2-ST2)]

= 65 – {-1.04 % [(21+ 050) – (21 +

119.92)]}

=64.2 m

24. Potongan XIII-XIII(STA 21 + 150)



XIII-XIII)]

= 65– { -1.04% [(21+ 050) – (21 +

150)]}

=63.96 m

25. Potongan B-B(STA 21 + 211.59)



B-B)]

= 65 – {-1.04 % [(21+050) – (21 +

211.59)]}

=63.6 m

45

bab iii tugas jalan raya

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