tjr ku

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titik A B C D x 0 100 600 1400 y 0 300 500 400 PENYELESAIAN JARAK D A-B 316.2278 D B-C 538.5165 D C-D 806.2258 AZIMUT A 71.565 B 21.801 C 7.125 MENGHITUNG SUDUT MENENTUKAN JARI-JARI TIKUNGAN / R MIN V REN 60 E MAK 0.1 F MAK -0.039 + 0.192 0.153 R MIN V^ 3600 (E MAK + F MAK)*127 32.131 112.0413 R C = 205 > R MIN = 112.0413 V R = 60 R C = 205 E M = 0.08 TABEL LS = 50 CEK RUMUS SHORTT LS = 0.022 V^3 - 2.727 V*E 57.95122 - RC*C C C DIAMBIL DR 0,3-1 C = 0.4 BERDASARKAN WAKTU TEMPUH MAKS 3 DTK LS = V 3 50 3.6 DIKETAHUI = B= 3.75 VR = 60 M MAKS 125 TABEL

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TEKNIK JALAN RAYA

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Page 1: TJR KU

titik A B C Dx 0 100 600 1400y 0 300 500 400

PENYELESAIANJARAKD A-B 316.2278D B-C 538.5165D C-D 806.2258

AZIMUT A 71.565B 21.801C 7.125

MENGHITUNG SUDUT

MENENTUKAN JARI-JARI TIKUNGAN / R MINV REN 60E MAK 0.1F MAK -0.039 + 0.192 0.153

R MIN V^ 3600(E MAK + F MAK)*127 32.131 112.0413

R C = 205 > R MIN = 112.0413V R = 60R C = 205 E M = 0.08 TABEL

LS = 50

CEK RUMUS SHORTTLS = 0.022 V^3 - 2.727 V*E 57.95122 -

RC*C CC DIAMBIL DR 0,3-1C = 0.4

BERDASARKAN WAKTU TEMPUH MAKS 3 DTKLS = V 3 50

3.6

DIKETAHUI =B= 3.75VR = 60M MAKS 125 TABEL

Page 2: TJR KU

EN 0.02

TIKUNGAN 1 RC= 205LS = (E M + E N)*B*M MAKS = 46.875

MERENCANAKN KOMPONEN TIKUNGAN 1DIKETAHUIV R = 60LEBAR JALAN 2 X 3,75E N = 0.02

*A 49.758D A-B 316.2278D B-C 538.5165

DARI TABEL 1 DILIHAT LS 50 E 0.08XS = LS (1- LS^/40 RC^) 49.92564YS = LS^2/ 6RC 2.03252

SUDUT APIT SUDUT SPIRAL (TETA S)TETA S = 90/PHI LS/RC 28.66242 0.243902 6.990834

MENENTUKAN SUDUT APIT BUSUR CIRCLE (TETA C)TETA C = TETA 1- 2 TETA S 35.77633

PERGESERAN TERHADAP TG ALINYEMEN ASLI (P)P = LS^2 RC(1-COS TETA S) 2.03252 1.524045 0.508475

6 RC

K = LS-(LS^3/40RC^2)-RC SIN TETA S 49.92564 24.95067 24.97497

NILAI TSTS = (RC+P)TAN (0,5 DELTA)+K 120.2773

ES (RP +P)SEC(0,5 TETA1)-RC 226.5309 -205 21.53094

LC TETE C/ 18PHI RC 127.9401

Page 3: TJR KU

STATIONING JARAK PANDANG HENTITIKUNGAN 1 V=STA TS= STA A + D A-B - TS = T =

195.9505 FM =STA SC= STA TS + LS

245.9505 JH=STA CS = STA SC +LC

373.8906STA ST = STA CS + LS

423.8906

TIKUNGAN 2STA TS = STA TIK 1 + D B-C - TS 1 - TS 1 JARAK PANDANG MENYIAP

721.8525 V = STA SC = STA TS + LS M=

771.8525STA CS = STA SC + LC A =

899.7926STA ST = STA CS + LS T1 =

949.7926T2 =

D1 =

D2 =

D3 =D4 =

32.724 25.22722

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DAERAH BEBAS SAMPINGJARAK PANDANG HENTI LT 1 =

603 E =

0.33

VR*T .+ V^2T 2,54 FM102.9492 PERLEBARAN JALUR LALU LINTAS

L =A =C =MIU =

JARAK PANDANG MENYIAP JUMLAH LAJUR =60 VR =15 WN =

R RENCANA =2.052 + 0.0036 X V2.268 1) JARAK LINTASAN TERLUAR-DALAM KENDARAAN

2.12 + 0.026 X V U = MIU + RC- AKAR(RC^+C^)3.68 2.4945126.56 + 0.048 X V 2) LEBAR TAMBAHAN AKIBAT TONJOLAN DEPAN9.44 FA = AKAR (RC^ + A(2L+A) - RC

0.278 T1 X (V-M+A/2X T1) 0.04755550.30607 3) LEBAR JALUR TAMBAHAN AKIBAT KELAINAN

0.278 X V X T2 Z = 157.4592

(ANTARA 30-100 M)2/3 D2 4) LEBAR TIKUNGAN104.9728 WC = N (MIU+C)+(N-1)FA+Z

8.4875675) TAMBAHAN LEBAR PERKERASANW = WC - 2 X 3,750.987567

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DAERAH BEBAS SAMPINGLC + 2LS227.9401

R *(1-COS 90 X JPHPHI X R

6.435126

PERLEBARAN JALUR LALU LINTAS6.51.51.52.5

JUMLAH LAJUR = 260

3.75R RENCANA = 205

1) JARAK LINTASAN TERLUAR-DALAM KENDARAANU = MIU + RC- AKAR(RC^+C^)

2) LEBAR TAMBAHAN AKIBAT TONJOLAN DEPANFA = AKAR (RC^ + A(2L+A) - RC

3) LEBAR JALUR TAMBAHAN AKIBAT KELAINAN0.105 X V

AKAR(RC)0.440011

4) LEBAR TIKUNGANWC = N (MIU+C)+(N-1)FA+Z

5) TAMBAHAN LEBAR PERKERASANW = WC - 2 X 3,75

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