Spiral Circle Spiral (scs) Bina Marga
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SPIRAL CIRCLE SPIRAL (SCS) BINA MARGA
TS
ES XC
YC
SC
CS LC
k
TS
p
LS
LS
R
RC
RC
R=
R R=
s
C
s
ST
𝜃𝑆 =
𝐿𝑆 180 . 2𝑅 𝜋
𝑘 = 𝑋𝐶 − 𝑅 𝑠𝑖𝑛 𝜃𝑆
𝐶 = − 2𝜃𝑆
𝑝 = 𝑌𝐶 − 𝑅 1 − 𝐶𝑜𝑠𝜃𝑆
𝐶 𝐿𝐶 = . 2𝜋𝑅 360
𝑇𝑆 = 𝑅 + 𝑃 tan + 𝑘 2
𝐿𝑆 2 𝑌𝐶 = 6𝑅
𝑅+𝑃 𝐸𝑆 = −𝑅 𝐶𝑜𝑠 2
𝐿𝑆 2 𝑋𝐶 = 𝐿𝑆 − 40𝑅2
𝐿 = 𝐿𝐶 + 2𝐿𝑆
•CONTOH 2 Suatu tikungan direncanakan dengan ketentuan sebagai berikut : (BINA MARGA) d1 = 525 m
PI
0+000 •Kecepatan rencana •e maksimum •Tipe tikungan •R
26o : 80 km/jam : 10 % : Spiral-Circle-Spiral : 409 m
2% 3,6 m
2% 3,6 m
Potongan melintang normal :
SOLUSI dari table dengan R = 409 m didapat : e penuh = 0,076 = 7,6 % dan LS min = 70 m A. MenghitungPanjang Spiral Desain (Ls) 1. Berdasarkandari table dengan R = 409 m didapat : LS min = 70 m dan epenuh = 0,076 = 7,6 % 2. Berdasarkanwaktutempuhmaksimum di lengkungperalihan, 𝑉𝑅 𝐿𝑆 = .𝑇 3,6 T = 3 detik (ditetapkan) 80 𝐿𝑆 = . 3 = 66,667 𝑚 3,6 3. Berdasarkanantisipasigayasentrifugal 𝑉𝑅 3 𝑉𝑅 . 𝑒 𝐿𝑆 = 0,022 − 2,727 𝑅. 𝐶 𝐶 C = 1,5 diambilnilaitengahdari (1-3) R = Rc =409 m (jari-jari di posisi Circle/lingkaran) e = 0,074 = 7,6% 803 80 . 0,076 𝐿𝑆 = 0,022 − 2,727 = 7,307 𝑚 409 . 1,5 1,5
4. Berdasarkantingkatpencapaianperubahankelandaian, 𝑒𝑝 − 𝑒𝑛 𝑉𝑅 𝐿𝑆 = 3,6 . 𝑟𝑒 VR ≤ 70 km/jam re = 0,035 VR ≥ 80 km/jam re = 0,025 0,076 − 0,02 . 80 𝐿𝑆 = = 49,778 𝑚 3,6 . 0,025 Jadi panjang lengkung spiral desain yang digunakan adalah Ls = 70 m
5. Berdasarkanlandai relative 𝐿𝑆 = 𝑚. 𝑒𝑝 + 𝑒𝑛 . 𝑏 Kelandaianrelatif (1/m) untukkecepatan VR = 80 km/jam 𝐿𝑆 = 150. 0,02 + 0,076 . 3,6 = 50,760 𝑚
PerhitunganVariabelTikungan 70 180 𝜃𝑆 = . = 4,903𝑜 2 . 409 𝜋 𝐶 = 26 − 2 . 4,903 = 16,194𝑜 𝐿𝐶 =
16,194 . 2𝜋. 409 = 115,598 𝑚 360
702 𝑌𝐶 = = 1,997 𝑚 6 . 409 702 𝑋𝐶 = 70 − = 69,999 𝑚 40 . 4092 𝑘 = 69,999 − 409 sin 4,903𝑜 = 35,042 𝑚 𝑝 = 1,997 − 409. 1 − 𝐶𝑜𝑠 4,903𝑜 = 0,500 𝑚 26 𝑇𝑆 = 409 + 0,500 tan + 35,042 = 129,583 𝑚 2 409 + 0,500 𝐸𝑆 = − 409 = 11,272 𝑚 26 𝐶𝑜𝑠 2 𝐿 = 115,598 + 2 . 70 = 255,598 𝑚
Dari hasilperhitungan di dapat data : V = 70 km/jam = 26o e = 7,6 % R = 409 m Es = 11,272 m
𝜃𝑆 c Lc Yc L
TS = 129,583 m TS
SC
= 4,903oXc = 16,194ok = 115,598 m = 1,997 m = 255,598 m
T
PI
S
0+000
Ls=70 m
R
LC = 1
15,59 8
=1
= 69,999 m = 35,042 m p = 0,500 m Ts= 129,583 m
29, 5
=26o CS
83
m
m
L = s 70 m
RC RC
R
ST
•Diagram Superelevasi TS
daerah spiral
SC
ST CS daerah spiral
daerah full circle Lc
Ls
Ls
Sisi luar
ep = 7,6%
e penuh = 7,6%
en = 2% e = 0% en = 2%
as jalan
e penuh = 7,6% Sisi dalam
ep = 7,6% A
B C
D
en = 2% e n = 2% 3,6 m
Pot. A
en = 2% 3,6 m
e = 0% 3,6 m
en = 2% 3,6 m
Pot. B
3,6 m
ep = 7,6% en = 2% 3,6 m
Pot. C
ep = 7,6% 3,6 m
3,6 m
Pot. D
•Kelandaian Relatif
TS
SC daerah spiral Ls ep = 7,6%
(en+ep).b e = 0% atau h en = 2%
ep = 7,6% en = 2% e = 0% en = 2% ep = 7,6%
ep
ep = 7 ,6 %
en+ep
en
LS
e n = 2%
b 3,6 m
Kelandaianrelatif 1 ℎ = 𝑑𝑖𝑚𝑎𝑛𝑎, ℎ = 𝑒𝑛 + 𝑒𝑝 . 𝑏 𝑚 𝐿𝑠 𝑒𝑛 + 𝑒𝑝 . 𝑏 1 = 𝑚 𝐿𝑠 1 0,02 + 0,076 . 3,6 1 = = 0,0049 = 𝑚 70 204
en = 2%
b 3,6 m
=7
0m
ep = 7 ,6 %
Potongan Melintang •Potongan A di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en
en+ep
en = 2%
e n = 2% 3,6 m
3,6 m
Pot. A
ep = 7,6% A
Sta. A = Sta. PI – Ts = 0+525,000 – 129,583 = 0 + 395,417
•Potongan B di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep
en x A-B
en = 2%
e = 0% 3,6 m
3,6 m
ep = 7,6%
Pot. B A
𝑥𝐴−𝐵 𝑒𝑛 = 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
B
𝑥𝐴−𝐵 =
𝑒𝑛 . 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
( 0,02) . 70 = = 14,583 𝑚 (0,02 + 0,076)
Sta. B = Sta. A + xA-B = 0 + 395,417 + 14,583 = 0+410,000
• Potongan C di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep en = 2%
en x A-C
3,6 m
en = 2% 3,6 m
ep = 7,6%
Pot. C A
𝑥𝐴−𝐶 𝑒𝑛 + 𝑒𝑛 = 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
C
𝑥𝐴−𝐶 =
2𝑒𝑛 . 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
=
(2. 0,02) . 70 = 29,167 𝑚 (0,02 + 0,076)
Sta. C = Sta. A + xA-C = 0+395,417 + 29,167 = 0+424,584
•Potongan D di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep
en
ep = 7,6% ep = 7,6%
ep = 7,6%
3,6 m
A
D
Sta. D = Sta. A + LS = = 0+395,417 + 70 = 0+465,417
3,6 m
Pot. D
•Potongan dan Sta. berapa? di 20 m setelah Sta. A TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep
en x A-A'
ep = 7,6%
3,6 m
A
𝑥𝐴−𝐴′ 𝑦 = 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
e = 0,7%
𝑦=
A'
en = 2% 3,6 m
Pot. A’
𝑒𝑛 + 𝑒𝑝 . 𝑥𝐴−𝐴′ (0,02 + 0,076) . 20 = = 0,027 = 2,7 % 𝐿𝑆 70
e = y – en = 2,7% - 2% = 0,7 % Sta. A’ = Sta. A + xA-A’ = 0+395,417 + 20 = 0+415,417
Data Stationing TS = 129,583 m TS 0+000
SC
0+395,417
Ls=70 m R
T
PI
S
=1
29, 5
=26o 0+465, 417 CS 0+581, 015 LC = 1 15,59 8m
RC
RC
83
m
ST
Ls = 70 m R
0+6 51,0
15
SPIRAL SPIRAL (SS) BINA MARGA
TC
EC SCS
k p
TS
LS
LS
RC
R
R=
ST
R
R= s
s
𝜃𝑆 =
2
𝐿𝑆 2 𝑌𝐶 = 6𝑅 𝐿𝑆 3 𝑋𝐶 = 𝐿𝑆 − 40𝑅 2 𝑘 = 𝑋𝐶 − 𝑅 𝐶𝑜𝑠 𝜃𝑆 𝑝 = 𝑌𝐶 − 𝑅 1 − 𝐶𝑜𝑠𝜃𝑆 𝑇𝑆 = 𝑅 + 𝑝 tan + 𝑘 2 𝐸𝑆 =
𝑅+𝑝 −𝑅 𝐶𝑜𝑠 2
𝐿 = 2. 𝐿𝑆
TS
ES XC
YC
SC
CS LC
k
TS
p
LS
LS
R
RC
RC
R=
R R=
s
C
s
ST
𝜃𝑆 =
𝐿𝑆 180 . 2𝑅 𝜋
𝑘 = 𝑋𝐶 − 𝑅 𝑠𝑖𝑛 𝜃𝑆
𝐶 = − 2𝜃𝑆
𝑝 = 𝑌𝐶 − 𝑅 1 − 𝐶𝑜𝑠𝜃𝑆
𝐶 𝐿𝐶 = . 2𝜋𝑅 360
𝑇𝑆 = 𝑅 + 𝑃 tan + 𝑘 2
𝐿𝑆 2 𝑌𝐶 = 6𝑅
𝑅+𝑃 𝐸𝑆 = −𝑅 𝐶𝑜𝑠 2
𝐿𝑆 2 𝑋𝐶 = 𝐿𝑆 − 40𝑅2
𝐿 = 𝐿𝐶 + 2𝐿𝑆
•CONTOH 2 Suatu tikungan direncanakan dengan ketentuan sebagai berikut : (BINA MARGA) d1 = 525 m
PI
0+000 •Kecepatan rencana •e maksimum •Tipe tikungan •R
26o : 80 km/jam : 10 % : Spiral-Circle-Spiral : 409 m
2% 3,6 m
2% 3,6 m
Potongan melintang normal :
SOLUSI dari table dengan R = 409 m didapat : e penuh = 0,076 = 7,6 % dan LS min = 70 m A. MenghitungPanjang Spiral Desain (Ls) 1. Berdasarkandari table dengan R = 409 m didapat : LS min = 70 m dan epenuh = 0,076 = 7,6 % 2. Berdasarkanwaktutempuhmaksimum di lengkungperalihan, 𝑉𝑅 𝐿𝑆 = .𝑇 3,6 T = 3 detik (ditetapkan) 80 𝐿𝑆 = . 3 = 66,667 𝑚 3,6 3. Berdasarkanantisipasigayasentrifugal 𝑉𝑅 3 𝑉𝑅 . 𝑒 𝐿𝑆 = 0,022 − 2,727 𝑅. 𝐶 𝐶 C = 1,5 diambilnilaitengahdari (1-3) R = Rc =409 m (jari-jari di posisi Circle/lingkaran) e = 0,074 = 7,6% 803 80 . 0,076 𝐿𝑆 = 0,022 − 2,727 = 7,307 𝑚 409 . 1,5 1,5
4. Berdasarkantingkatpencapaianperubahankelandaian, 𝑒𝑝 − 𝑒𝑛 𝑉𝑅 𝐿𝑆 = 3,6 . 𝑟𝑒 VR ≤ 70 km/jam re = 0,035 VR ≥ 80 km/jam re = 0,025 0,076 − 0,02 . 80 𝐿𝑆 = = 49,778 𝑚 3,6 . 0,025 Jadi panjang lengkung spiral desain yang digunakan adalah Ls = 70 m
5. Berdasarkanlandai relative 𝐿𝑆 = 𝑚. 𝑒𝑝 + 𝑒𝑛 . 𝑏 Kelandaianrelatif (1/m) untukkecepatan VR = 80 km/jam 𝐿𝑆 = 150. 0,02 + 0,076 . 3,6 = 50,760 𝑚
PerhitunganVariabelTikungan 70 180 𝜃𝑆 = . = 4,903𝑜 2 . 409 𝜋 𝐶 = 26 − 2 . 4,903 = 16,194𝑜 𝐿𝐶 =
16,194 . 2𝜋. 409 = 115,598 𝑚 360
702 𝑌𝐶 = = 1,997 𝑚 6 . 409 702 𝑋𝐶 = 70 − = 69,999 𝑚 40 . 4092 𝑘 = 69,999 − 409 sin 4,903𝑜 = 35,042 𝑚 𝑝 = 1,997 − 409. 1 − 𝐶𝑜𝑠 4,903𝑜 = 0,500 𝑚 26 𝑇𝑆 = 409 + 0,500 tan + 35,042 = 129,583 𝑚 2 409 + 0,500 𝐸𝑆 = − 409 = 11,272 𝑚 26 𝐶𝑜𝑠 2 𝐿 = 115,598 + 2 . 70 = 255,598 𝑚
Dari hasilperhitungan di dapat data : V = 70 km/jam = 26o e = 7,6 % R = 409 m Es = 11,272 m
𝜃𝑆 c Lc Yc L
TS = 129,583 m TS
SC
= 4,903oXc = 16,194ok = 115,598 m = 1,997 m = 255,598 m
T
PI
S
0+000
Ls=70 m
R
LC = 1
15,59 8
=1
= 69,999 m = 35,042 m p = 0,500 m Ts= 129,583 m
29, 5
=26o CS
83
m
m
L = s 70 m
RC RC
R
ST
•Diagram Superelevasi TS
daerah spiral
SC
ST CS daerah spiral
daerah full circle Lc
Ls
Ls
Sisi luar
ep = 7,6%
e penuh = 7,6%
en = 2% e = 0% en = 2%
as jalan
e penuh = 7,6% Sisi dalam
ep = 7,6% A
B C
D
en = 2% e n = 2% 3,6 m
Pot. A
en = 2% 3,6 m
e = 0% 3,6 m
en = 2% 3,6 m
Pot. B
3,6 m
ep = 7,6% en = 2% 3,6 m
Pot. C
ep = 7,6% 3,6 m
3,6 m
Pot. D
•Kelandaian Relatif
TS
SC daerah spiral Ls ep = 7,6%
(en+ep).b e = 0% atau h en = 2%
ep = 7,6% en = 2% e = 0% en = 2% ep = 7,6%
ep
ep = 7 ,6 %
en+ep
en
LS
e n = 2%
b 3,6 m
Kelandaianrelatif 1 ℎ = 𝑑𝑖𝑚𝑎𝑛𝑎, ℎ = 𝑒𝑛 + 𝑒𝑝 . 𝑏 𝑚 𝐿𝑠 𝑒𝑛 + 𝑒𝑝 . 𝑏 1 = 𝑚 𝐿𝑠 1 0,02 + 0,076 . 3,6 1 = = 0,0049 = 𝑚 70 204
en = 2%
b 3,6 m
=7
0m
ep = 7 ,6 %
Potongan Melintang •Potongan A di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en
en+ep
en = 2%
e n = 2% 3,6 m
3,6 m
Pot. A
ep = 7,6% A
Sta. A = Sta. PI – Ts = 0+525,000 – 129,583 = 0 + 395,417
•Potongan B di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep
en x A-B
en = 2%
e = 0% 3,6 m
3,6 m
ep = 7,6%
Pot. B A
𝑥𝐴−𝐵 𝑒𝑛 = 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
B
𝑥𝐴−𝐵 =
𝑒𝑛 . 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
( 0,02) . 70 = = 14,583 𝑚 (0,02 + 0,076)
Sta. B = Sta. A + xA-B = 0 + 395,417 + 14,583 = 0+410,000
• Potongan C di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep en = 2%
en x A-C
3,6 m
en = 2% 3,6 m
ep = 7,6%
Pot. C A
𝑥𝐴−𝐶 𝑒𝑛 + 𝑒𝑛 = 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
C
𝑥𝐴−𝐶 =
2𝑒𝑛 . 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
=
(2. 0,02) . 70 = 29,167 𝑚 (0,02 + 0,076)
Sta. C = Sta. A + xA-C = 0+395,417 + 29,167 = 0+424,584
•Potongan D di sta. berapa? TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep
en
ep = 7,6% ep = 7,6%
ep = 7,6%
3,6 m
A
D
Sta. D = Sta. A + LS = = 0+395,417 + 70 = 0+465,417
3,6 m
Pot. D
•Potongan dan Sta. berapa? di 20 m setelah Sta. A TS
SC daerah spiral Ls
ep = 7,6% ep
en = 2% e = 0% en = 2%
en+ep
en x A-A'
ep = 7,6%
3,6 m
A
𝑥𝐴−𝐴′ 𝑦 = 𝐿𝑆 𝑒𝑛 + 𝑒𝑝
e = 0,7%
𝑦=
A'
en = 2% 3,6 m
Pot. A’
𝑒𝑛 + 𝑒𝑝 . 𝑥𝐴−𝐴′ (0,02 + 0,076) . 20 = = 0,027 = 2,7 % 𝐿𝑆 70
e = y – en = 2,7% - 2% = 0,7 % Sta. A’ = Sta. A + xA-A’ = 0+395,417 + 20 = 0+415,417
Data Stationing TS = 129,583 m TS 0+000
SC
0+395,417
Ls=70 m R
T
PI
S
=1
29, 5
=26o 0+465, 417 CS 0+581, 015 LC = 1 15,59 8m
RC
RC
83
m
ST
Ls = 70 m R
0+6 51,0
15
SPIRAL SPIRAL (SS) BINA MARGA
TC
EC SCS
k p
TS
LS
LS
RC
R
R=
ST
R
R= s
s
𝜃𝑆 =
2
𝐿𝑆 2 𝑌𝐶 = 6𝑅 𝐿𝑆 3 𝑋𝐶 = 𝐿𝑆 − 40𝑅 2 𝑘 = 𝑋𝐶 − 𝑅 𝐶𝑜𝑠 𝜃𝑆 𝑝 = 𝑌𝐶 − 𝑅 1 − 𝐶𝑜𝑠𝜃𝑆 𝑇𝑆 = 𝑅 + 𝑝 tan + 𝑘 2 𝐸𝑆 =
𝑅+𝑝 −𝑅 𝐶𝑜𝑠 2
𝐿 = 2. 𝐿𝑆
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