Tugas Geometirk Jalan Januari 2010

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    Nama : Rizal Zubad FirdausiSemester : V (lima)NIM : 07.10203.020Mata Kuliah : Geometrik Jalan Raya

    Data lapangan: D1 1000 m D2 700 m D3= 500 m D4= 1000 m

    PI 1Diketahui: V = 70 Km/jam e= 0.1 Lebar Jalan standart = 2 x3.75 m

    9 f= 0.15

    Jenis = Full Circle 0.02

    Ditanya : Geometrik Jalan tersebut?

    Jawab :R min =

    127(e+f)

    = 156.21 = 157

    D max = 181913.53 (e+f)

    = 9.17

    Lc min = t . v

    = 6 x 70 x 1000

    3600= 116.67 m

    Rc = Lc min . 360 Dari tabel 4.7 bina marga diperoleh:

    Ls = 60 m= 743.1 e = 0.04

    Tc == 58.48

    Ec =

    = 2.3

    Lc =180

    = 116.67

    Landai Relatif = Lebar jalan . (e normal + e)

    Ls= 0

    Data lengkung Circle dari jalan tersebut adalah:V = 70

    = 9 Rc = 743.1 mTc = 58.48 mLc = 116.67 m

    =

    e normal =

    V

    V

    . 2 .

    R . tg

    T . Tg

    . . Rc

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    Ec = 2.3 mLs = 116.67 me = 3.7 %

    Tugas Geometrik Jalan Raya Smt V

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    PI 2

    Diketahui: V = 70 Km/jam e= 0.1 Lebar Jalan standart = 2 x3.75 m

    14 f= 0.15Jenis = Sp-Cr-Sp 0.02

    Ditanya : Geometrik Jalan tersebut?

    Jawab :R min = 156.21 = 157 mD max = 9.17Lc min = 116.67 m

    R = Lc min x 360 Dari tabel 4.7 bina marga diperoleh:Ls = 60 m

    = 477.71 m e = 0.055

    = 90 . Ls

    . R

    = 3.6

    P = - R (1 - cos s)6 . R

    = 0.31

    K = - R . Sin s

    = 29.98 m

    == 6.8

    E = (R + P) sec - R= 3.9 m

    T = (R + P) tg + K)= 88.67 m

    Lc = L = Lc + 2 Ls180 = 176.67

    = 56.67 m

    Data-data lengkung Jalan dg jenis Spiral-Circle-Spiral:V = 70 Km/jamR = 477.71 mLs = 60 me = 5.5 %

    = 3.6 P = 0.31 mK = 29.98 m

    = 6.8

    =e normal =

    x 2 x

    s

    Ls

    Ls - Ls

    40 . R

    c - 2 s

    c . . R

    s

    c

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    E = 3.9 mT = 88.67 mLc = 56.67 mL = 176.67 m

    Tugas Geometrik Jalan Raya Smt V

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    PI 3

    Diketahui: V = 70 Km/jam e= 0.1 Lebar Jalan standart = 2 x3.75 m

    24 f= 0.15Jenis = Spiral 0.02

    Ditanya : Geometrik Jalan tersebut?

    Jawab :R min = 156.21 = 157 mD max = 9.17Lc min = 116.67 m

    Rs = Lc min x 360 Dari tabel 4.7 bina marga diperolehe = 0.078

    = 278.66 = 279 m

    = 1/2 .

    1/2 . 24

    = 12.0

    Ls =90

    = 116.7 m

    P = - Rs (1 - cos s)

    6 . Rs= 2.05

    K = - Rs . Sin s

    = 58.22 m

    Es = (Rs + P) sec - Rs= 8.32 m

    Ts = (Rs + P) tg + K)= 117.89 m

    L = 2 . Ls= 233.33 m

    Data-data lengkung Jalan dg jenis Spiral-Circle-Spiral:R = 278.66 m

    = 24 Ls = 116.7 m

    = 12.0 P = 2.05 mK = 58.22 mEs = 8.32 m

    =e normal =

    x 2 x

    s

    s . . Rs

    Ls

    Ls - Ls40 . Rs

    s

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    Ts = 117.89 mL = 233.33 me = 0.078

    Tugas Geometrik Jalan Raya Smt V