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BAB III
PERHITUNGAN ALINEMEN HORISONTAL
A. Klasifikasi Medan
A – 1 = = 1,517 %
1 – 2 = = 5,47 %
2 – 3 = = 4,494 %
3 – 4 = = 0,5 %
4 – 5 = = 1,496 %
5 – 6 = = 0,497 %
6 – 7 = = 2 %
7 – 8 = = 1,495 %
8 – 9 = = 5,494%
9 – B = = 0,003 %
Klasifikasi Medan A – B
= 2,0878 %
Tabel 3.1. Klasifikasi MedanRata – rata Kemiringan Melintang ( % ) Jenis Medan
20
0 – 9,99,9 – 24,5
25
DatarBukit
Gunung
Jadi klasifikasi medan A – B adalah datar
Dari Daftar I Standar Perencanaan Geometrik didapat :
1. Kecepatan rencana (Vr) : 100 km/jam
2. Lebar row minimum : 40 m
3. Lebar perkerasan : 2 x 3,50 m
4. Lebar bahu : 3,00 m
5. Lereng melintang perkerasan (en) : 2 %
6. Lereng melintang bahu : 4%
7. Miring tikungan maksimum : 10 %
8. Jari–jari tikungan minimum(Rmin) : 350 m
9. Landai maksimum : 4 %
10. LHR : 6000 – 20.000
11. Lebar median minimum : 1,5 m
12. Landai maksimum : 4 %
13. Klasifikasi medan : Datar
Koordinat Tiap Titik
Koordinat titik A ( 10016,4116 )
Koordinat titik I ( 10016+13,5 ; 4116+14,5 ) → ( 10029,5 ; 4130,5 )
X1 = 13,5 m
Y1 = 14,5 m
Koordinat titik II ( 10029,5+24,8;4130,5+17,3 )
→(10054,3;4147,8)
X2 = 24,8 m
Y2 = 17,8 m
♦ Kordinat titik III ( 10054,3+24,3;4147,8+9,5 ) → ( 10078,6;4157,3 )
21
X3 = 24,3m
Y3 = 9,5 m
Koordinat titik B ( 10078,6+11,3;4157,3+3 ) → ( 10089,9;4160,3 )
Jarak Antar Titik
dA-I = = 19,8116 m
dI-II = = 30,5267 m
dII-III = = 26,09099 m
dIII-B = = 11,6914 m
= dA –I + dI – II + dII – III + dIII – B
= 19,8116 + 30,5267 + 26,0909 + 11,6914
= 88,1206 m
Perhitungan Sudut
Sudut Azimuth A = 46°59’12” = 46,980
I X2
α2 1
Y2 III 4 X4
Y1 3 3
Y3
2
X1 1
II 2 X3
1 = 900 – Azimuth titik A
= 900 – 46,980
= 43,020
22
A
Y4
B
2 = arc tan
= arc tan
= 34,680
3 = arc tan
= arc tan
= 31,620
♦ Tikungan I
1 = 1 + 2
= 43,020 + 34,680
= 77,70
Tikungan II
1 = 2 = 34,680
2 = arc tan
= arc tan
= 31,620
2 = 1 + 2
= 34,680 + 31,620
= 66,30
♦ Tikungan III
3 = 3 = 31,620
4 = arc tan
23
= arc tan
= 14,860
3 = 3 - 4
= 31,620 – 14,860
= 16,760
1. Perencanaan Tikungan I
1 = 77,7°
Rmin = 350 m
Vr = 100 km/jam
Rr = 358
en = 2 %
C = 0,4
Dari tabel Panjang Minimum Spiral dan Kemiringan Melintang
diperoleh nilai :
e = 0,099
Ls = 100 m ...... ( 1 )
Ls min =
= = 86,14 m ...... ( 2 )
Dari tabel Daftar Standar Perencanaan Alinemen didapat :
B = 3 m
=
=
24
Ls =
=
= 85,86 m.............( 3 )
Dari...( 1 ),...( 2 ),...( 3 ) dipilih yang terbesar
Jadi Ls = 100 m
s =
=
= 8,0060
c = 1 – 2 . s
= 77,700 – 2 . 8,0060
= 61,688 0
Lc =
=
= 385,248 m
Diketahui Lc min = 25 m
Lc >Lc min, jadi tikungan yang dipakai tipe S - C – S
Xc =
=
= 99,804 m
25
Yc =
=
= 4,655 m
K = Xc – Rr . Sin s
= 99,804 – 358 .Sin 8,0060
= 49,9429 m
P = Yc - Rr ( 1 – Cos s )
= 4,655 – 358 ( 1 – Cos 8,0060 )
= 1,165 m
Tt = ( Rr + P ) tan ½ 1 + K
= ( 358 + 1,165 ) tan ½ x 77,70 + 49,9429
= 339,2353 m
Et = ( Rr + P ) sec ½ 1 – Rr
= ( 358 + 1,165 ) sec ½ x 77,7 – 358
= 103,182 m
L = 2. Ls + Lc
= 2. 86,14 + 385,248
= 557,528 m
2. Perencanaan Tikungan II
2 = 66,3°
Rmin = 350 m
26
Vr = 100 km / jam
Rr = 358 m
en = 2 %
C = 0,4
Dari tabel Panjang Minimum Spiral dan Kemiringan Melintang
diperoleh nilai :
e = 0,099
Ls = 100 m ...... ( 1 )
Ls min =
= = 86,14 m ...... ( 2 )
Dari tabel Daftar Standar Perencanaan Alinemen didapat :
B = 3 m
=
=
Ls =
=
= 85,86 m.............( 3 )
Dari...( 1 ),...( 2 ),...( 3 ) dipilih yang terbesar
Jadi Ls = 100 m
s =
27
=
= 8,0060
c = 1 – 2 .s
= 66,30 – 2 . 8,0060
= 50,288 0
Lc =
=
= 314,054 m
Diketahui Lc min = 25 m
Lc >Lc min, jadi tikungan yang dipakai tipe S - C – S
Xc =
=
= 99,804 m
Yc =
=
= 4,655 m
K = Xc – Rr . Sin s
= 99,804 – 358 .Sin 8,0060
= 49,9429 m
P = Yc - Rr ( 1 – Cos s )
= 4,655 – 358 ( 1 – Cos 8,0060 )
= 1,165 m
Tt = ( Rr + P ) tan ½ 2 + K
28
= ( 358 + 1,165 ) tan ½ * 66,30 + 49,9429
= 284,526 m
Et = ( Rr + P ) sec ½ 2 – Rr
= ( 358 + 1,165 ) sec ½ . 66,3 – 358
= 70,985 m
L = 2. Ls + Lc
= 2. 85,86 + 314,054
= 485,774 m
3. Perencanaan Tikungan III
3 = 16,76°
Rmin = 350 m
Vr = 100 km / jam
Rr = 358 m
en = 2 %
C = 0,4
Dari tabel Panjang Minimum Spiral dan Kemiringan Melintang
diperoleh nilai :
e = 0,099
Ls = 100 m ...... ( 1 )
Ls min =
= = 86,14 m ...... ( 2 )
Dari tabel Daftar Standar Perencanaan Alinemen didapat :
B = 3 m
29
=
=
Ls =
=
= 85,86 m.............( 3 )
Dari...( 1 ),...( 2 ),...( 3 ) dipilih yang terbesar
Jadi Ls = 100 m
s =
=
= 8,0060
c = 3 – 2 . s
= 16,760 – 2 .8,0060
= 0,748 0
Lc =
=
= 4,671 m
Diketahui Lc min = 25 m
Lc < Lc min, jadi tikungan yang dipakai tipe S – S
Dihitung kembali :
3 = 16,76°
3 = 2 s
30
Maka :
s = ½ 2
= ½ 16,760
= 8,380
Ls =
=
= 104,660
P = – R ( 1 – Cos θs )
= – 358 ( 1 – Cos 8,38º )
= 1,2772 m
K = Ls – – R x Sin s
= 104,66 – – 358 x Sin 8,38
= 52,26228 m
Tt = ( Rr + P ) Tan ½ 3 + K
= ( 358 + 1,277 ) Tan ½ 16,760+ 52,26228
= 105,187 m
Et = ( Rr + P ) Sec ½ 3 – Rr
= ( 358 + 1,277 ) Sec ½ 16,760 - 358
= 5,154 m
31
Tabel 3.2 Data TikunganData Tikungan I Tikungan II Tikungan III
Bentuk
Vr
s
c
Ls
Lc
L
Tt
Et
K
P
Xc
S – C – S
77,70
100 km/jam
8,0060
61,688 0
100 m
385,248m
557,528 m
102,05 m
103,182 m
49,9429 m
1,165 m
99,804 m
S – C – S
66,30
100 km/jam
8,0060
50,288 0
100 m
314,054 m
485,774 m
284,526 m
70,985 m
49,9429 m
1,165 m
99,804 m
S – S
16,760
100 km/jam
8,0060
-
100 m
4,671 m
-
105,187m
5,154 m
52,26228 m
1,2772 m
-
32
Yc
Rr
e
en
4,655 m
358 m
9,9 %
2 %
4,655 m
358 m
9,9 %
2 %
-
358 m
9,9 %
2 %
Sumber : Hasil Perhitungan
I. Diagram Super Elevasi dan Sumbu Putar Jalan
1. Tikungan I Tipe S – C – S
33
P
TSLs SC=CS Ls
ST
R
Tt Et
PYc Yc
K ∆1
θs θs
Xc
θc θc
Gambar 3.1 Tikungan Belok ke kananTipe S – C – S
Kiri
CL ±0,00%
- 2%
Kanan
Ls = 100 m Lc = 385,248 m Ls = 100 m
Gambar 3.2 Diagram Superelevasi Tipe S-C-S
2. Tikungan II Tipe ( S-C-S )
34
9,9 %
- 9,9 %
TS SC CS ST
TS STLS LS
SCCS
Yc Yc
P P
K
Xc
Tt
R
ST
LC Et
θs θc θc θs
Gambar 3.3 Tikungan Belok ke KiriTipe S – C – S
Kanan
CL ±0,00%
-2%
Kiri
Ls = 100 m Lc = 314,054 m Ls = 100 m
Gambar 3.4 Diagram Superelevasi Tipe S-C-S
3. Tikungan III Tipe ( S – S )
35
+9,9 %
- 9,9 %
TS SC CS ST
P
TSLs SC=CS Ls
ST
R
Tt Et
PYc Yc
K ∆1
θs θs
Gambar 3.5 Tikungan Belok ke kanan Tipe S – S
+ 9,9
Kiri
CL ± 0,00%
Kanan
Gambar 3.6 Diagram Superelevasi Tipe S – S
II. Hitungan Stationing Titik – Titik Penting
I
III
dA-I dI-II dIII-B
dII-III
A II
Sta A = 10 + 500
dA-I = 19,8116 m
dI-II = 30,5267 m
dII-III = 26,0909 m
dIII-B = 11,6914 m
36
- 2%
- 9,9
LS = 104,66
TS ST
SC=CS LS = 104,66
B
1. Tikungan I
Sta PP1 = Sta A+ dA-1
= ( 10 + 500 ) + (19,8116)
= 10 + 519,8116
Sta Ts1 = Sta PP1 – Tt1
= (10 + 519,8116) – 339,2353
= 10 +180,6107
Sta Cs1 = Sc1
= Sta Ts1 + Ls1
= (10 +180,6107 ) + 100
= 10 + 280,6107
Sta St1 = ( Sta Sc1 = Cs1 ) + Ls1
= ( 10 + 280,6107) + 100
= 10 + 380,6107
2. Tikungan II
Sta St2 = Sta St1 + (d1-II – Tt1 – Tt2)
= ( 10 + 380,6107 ) + ( 30,5267 – 339,2353 – 284,526 )
= 9 + 212,6239
Sta Sc2 = Sta Ts2 + Ls2
= (9 + 212,6239) + 100
= 9 + 312,6239
Sta St2 = Sta Cs2 + Ls2
= (9 + 312,6239) + 100
= 9 + 412,6239
3. Tikungan III
Sta Ts3 = Sta St1 + St2 + (d1-II – Tt1 – Tt2 – dII-III – Tt3)
= ( 10 + 380,6107 ) + ( 9+412,6239) +( 30,5267 – 339,2353
-284,526-26,0909-105,187 )
37
= 19 + 68,7221
Sta Sc3 = Sta Ts3 + Ls3
= (19 + 68,7221) + 104,66
= 19 + 173,3821
Sta St3 = Sta Cs3 + Ls3
= (19 + 173,3821) + 104,66
= 19 + 278,0421
Sta B = Sta St3 + (dIII – B – Tt3)
= (19 + 278,6421) + (11,6914 – 105,187 )
= 19 + 185,1465
Panjang jalan (A – B)
= Sta B – Sta A
= (19 + 185,1465) – (10 + 500)
= 8 + 314,8535
B. Pelebaran Perkerasan pada Tikungan
L = Jarak gandar 6,09 m
A = Tonjolan depan 1,218 m
c = Kebebasan samping 0,609 m
M = Lebar kendaraan 2,436 m
n = Jumlah jalur 2
Lebar Perkerasan Normal = 2 x 3,5 m
1. Tikungan I ( S – C – S )
Diketahui : R = 358 m
V = 100 km/jam
n = 2
Wn = 7 m
a. Lebar lintasan kendaraan rencana pada tikungan ( U )
38
U = M + R –
= 2,436 + 358 –
= 2,487 m
b. Lebar melintang akibat tonjolan depan ( Td = Fa )
Td = - R
= - 358
= 0,022 m
c. Lebar tambahan akibat kelainan pengemudi ( z )
z =
=
= 0,554 m
d. Lebar perkerasan pada tikungan ( Wc )
Wc= n ( M + c ) + Td ( n – 1 ) + z
= 2 ( 2,436 + 0,609 ) + 0,022 ( 2 – 1 ) + 0,543
= 6,66 m
Wc < Wn
Jadi tidak perlu ada tambahan pelebaran perkerasan.
2. Tikungan II ( S – C – S )
Diketahui : R = 358 m
V = 100 km/jam
n = 2
Wn = 7 m
a. Lebar lintasan kendaraan rencana pada tikungan ( U )
U = M + R –
39
= 2,436 + 358 –
= 2,487 m
b. Lebar melintang akibat tonjolan depan ( Td = Fa )
Td = - R
= - 358
= 0,022 m
c. Lebar tambahan akibat kelainan pengemudi ( z )
z =
=
= 0,554 m
d. Lebar perkerasan pada tikungan ( Wc )
Wc= n ( M + c ) + Td ( n – 1 ) + z
= 2 ( 2,436 + 0,609 ) + 0,022 ( 2 – 1 ) + 0,543
= 6,66 m
Wc < Wn
Jadi tidak perlu ada tambahan pelebaran perkerasan.
3. Tikungan III ( S – S )
Diketahui : R = 358 m
V = 100 km/jam
n = 2
Wn = 7 m
a. Lebar lintasan kendaraan rencana pada tikungan ( U )
U = M + R –
= 2,436 + 358 –
40
= 2,487 m
b. Lebar melintang akibat tonjolan depan ( Td = Fa )
Td = - R
= - 358
= 0,022 m
c. Lebar tambahan akibat kelainan pengemudi ( z )
z =
=
= 0,554 m
d. Lebar perkerasan pada tikungan ( Wc )
Wc= n ( M + c ) + Td ( n – 1 ) + z
= 2 ( 2,436 + 0,609 ) + 0,022 ( 2 – 1 ) + 0,543
= 6,66 m
Wc < Wn
Jadi tidak perlu ada tambahan pelebaran perkerasan.
C. Jarak Pandang Horizontal
1.Tikungan I ( S - C - S )
a. Berdasarkan Jarak Pandang Henti ( JPH )
Diketahui : L = 2 x Ls + Lc
= 2 x 86,14 + 385,248 = 557,528
Vr = 100 km/jam
t = 2,5 detik ( t = 0,5 – 4 detik, dipakai t = 2,5 detik )
f = 0,28 ( dari Tabel Koefisien Gesek )
R = 358
41
S = d1 + d2
=
=
= 210,107 m
S < L
θ =
=
= 16,820
M = Rr x ( 1 – cos θ )
= 358 x (1 – cos 16,820)
= 15,31 m < 3,5 m (lebar Perkerasan)
M > 7
Maka perlu dipasang rambu-rambu lalu lintas.
b. Berdasarkan Jarak Pandang Menyiap (JPM)
a = 2,052 + 0,0036 x V
= 2,052 + 0,0036 x 100
= 2,412 m/dt2
t1 = 2,12 + 0,026 x V
= 2,12 + 0,026 x 100
= 4,72 m/dt
t2 = 6,56 + 0,048 x V
= 6,56 + 0,048 x 100
= 11,36 m/dt
d1 =
42
=
= 104,064 m
d2 = 0,278 x V x t2
= 0,278 x 100 x 11,36
= 315,808 m
d3 = 90 m (30 – 100 dipakai 100 m )
d4 =
=
= 210,538 m
S = d1 + d2 + d3 + d4
= 104,064 + 315,8808 + 90 + 210,538
= 720,410 m
S > L
c. Kebebasan samping
θ =
=
= 57,6770
M = R x ( 1 – cos θ )
= 358 x (1 – cos 57,6770)
= 166,585 m
M > 40 (row minimum) maka pada tikungan perlu dipasang rambu –
rambu lalu lintas, dilarang menyiap.
2. Tikungan II ( S - C - S )
43
a. Berdasarkan Jarak Pandang Henti ( JPH )
Diketahui : L = 2 x Ls + Lc
= 2 x 86,14 + 385,248 = 557,528
Vr = 100 km/jam
t = 2,5 detik ( t = 0,5 – 4 detik, dipakai t = 2,5 detik )
f = 0,28 ( dari Tabel Koefisien Gesek )
R = 358
S = d1 + d2
=
=
= 210,107 m
S < L
θ =
=
= 16,820
M = Rr x ( 1 – cos θ )
= 358 x (1 – cos 16,820)
= 15,31 m < 3,5 m (lebar Perkerasan)
M > 7
Maka perlu dipasang rambu-rambu lalu lintas.
b. Berdasarkan Jarak Pandang Menyiap (JPM)
a = 2,052 + 0,0036 x V
44
= 2,052 + 0,0036 x 100
= 2,412 m/dt2
t1 = 2,12 + 0,026 x V
= 2,12 + 0,026 x 100
= 4,72 m/dt
t2 = 6,56 + 0,048 x V
= 6,56 + 0,048 x 100
= 11,36 m/dt
d1 =
=
= 104,064 m
d2 = 0,278 x V x t2
= 0,278 x 100 x 11,36
= 315,808 m
d3 = 90 m (30 – 100 dipakai 100 m )
d4 =
=
= 210,538 m
S = d1 + d2 + d3 + d4
= 104,064 + 315,8808 + 90 + 210,538
= 720,410 m
S > L
c. Kebebasan samping
θ =
45
=
= 57,6770
M = R x ( 1 – cos θ )
= 358 x (1 – cos 57,6770)
= 166,585 m
M > 40 (row minimum) maka pada tikungan perlu dipasang rambu –
rambu lalu lintas, dilarang menyiap.
3. Tikungan III ( S – S )
a. Berdasarkan Jarak Pandang Henti ( JPH )
Diketahui : V = 100 km/jam
t = 2,5 detik ( t = 0,5 – 4 detik, dipakai t = 2,5 detik )
f = 0,28 ( dari Tabel Koefisien Gesek )
L = 2 x Ls
= 2 x 89,617
= 179,234 m
S = d1 + d2
=
=
= 80,52 m
S < L
θ =
=
= 6,44
46
M = Rr x ( 1 – cos θ )
= 358 x (1 – cos 6,440)
= 2,26 m
M < 7
Maka tidak perlu dipasang rambu-rambu lalu lintas.
b. Berdasarkan Jarak Pandang Menyiap (JPM)
a = 2,052 + 0,0036 x V
= 2,052 + 0,0036 x 100
= 2,412 m/dt2
t1 = 2,12 + 0,026 x V
= 2,12 + 0,026 x 100
= 4,72 m/dt
t2 = 6,56 + 0,048 x V
= 6,56 + 0,048 x 100
= 11,36 m/dt
d1 =
=
= 104,064 m
d2 = 0,278 x V x t2
= 0,278 x 100 x 11,36
= 315,808 m
d3 = 90 m (30 – 100 dipakai 100 m )
d4 =
=
= 210,538 m
47
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