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58
3.2.3 Tikungan PI
3
Data tikungan : ∆
3
= 17
o
43’48” Rren = 300 m
1. Mencari superelevasi
775 ,
4 300
39 ,
1432 39
, 1432
= =
= Rren
D
1 ,
9 091
, 822
, 6
775 ,
4 1
, 2
822 ,
6 775
, 4
1 ,
. 2
2 2
max max
2 max
2 max
= =
× ×
+ ×
− =
× ×
+ ×
− =
D D
e D
D e
e
tjd
... 1
, 091
,
max
ok e
e
tjd
=
2. Penghitungan lengkung peralihan Ls
1. Berdasarkan waktu tempuh maximum 3 detik untuk melintasi lengkung
peralihan, maka panjang lengkung :
m t
Vr Ls
667 ,
66 3
6 ,
3 80
6 ,
3
= ×
= ×
=
2. Berdasarkan rumus modifikasi Short :
m c
e Vr
c R
Vr Ls
tjd ren
238 ,
44 4
, 091
, 80
727 ,
2 4
, 300
80 022
, 727
, 2
022 ,
3 3
= ×
− ×
× =
× −
× ×
=
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59 3.
Berdasarkan tingkat pencapaian perubahan kelandaian :
m Vr
e e
Ls
normal
111 ,
71 80
025 ,
6 ,
3 02
, 1
, Re
6 ,
3
max
= ×
× −
= ×
× −
=
4. Berdasarkan rumus bina marga
m m
e en
w Ls
tjd
696 ,
77 200
091 ,
02 ,
2 5
, 3
2 2
= ×
+ ×
× =
× +
× =
Dipakai nilai Ls yang terbesar yaitu 77,696m
3. Perhitungan besaran tikungan
12 25
7 300
2 14
, 3
2 360
696 ,
77 2
2 360
= ×
× ×
× =
× ×
= Rr
Ls Qs
π
24 53
2 12
25 7
2 48
1743 17
2
3
= ×
− =
× −
∆ =
∆ Qs
c
m Rr
c Lc
143 ,
15 180
300 14
, 3
24 53
2 180
= ×
× =
× ×
∆ =
π
Syarat tikungan S-S ...
20 143
, 15
ok Lc
=
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60
12 52
8 48
43 17
2 1
3 2
1
= =
∆ =
Qs
m Ls
839 ,
92 90
300 14
, 3
12 52
8 =
× ×
=
696 ,
77 839
, 92
min
Ls Ls
m Rr
Ls Ls
Xs
636 ,
20 300
40 839
, 92
839 ,
92 40
2 3
2 3
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
× ×
=
m Rr
Ls Ys
788 ,
4 300
6 839
, 92
6
2 2
= ×
= ×
=
m Qs
Rr Rr
Ls P
201 ,
1 12
52 8
cos 1
300 300
6 839
, 92
cos 1
6
2 2
= −
− ×
= −
− ×
=
m Qs
Rr Rr
Ls Ls
K
359 ,
46 12
52 8
sin 300
300 40
839 ,
92 839
, 92
sin 40
2 3
2 3
= ×
− ×
− =
× −
× −
=
m K
P Rr
Ts 364
, 93
359 ,
46 48
43 17
tan 201
, 1
300 tan
2 1
3 2
1
= +
× +
= +
∆ ×
+ =
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61
m Rr
P Rr
Es
846 ,
4 300
48 43
17 cos
201 ,
1 300
cos
2 1
3 2
1
= −
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
+ =
− ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ∆
+ =
839 ,
92 364
, 93
Ls Ts
Tikungan S-S bisa digunakan
4. Perhitungan pelebaran perkerasan
m P
Rr Rr
b b
696 ,
2 6
, 7
300 300
6 ,
2
2 2
2 2
= −
− +
= −
− +
=
m Rr
A P
A Rr
Td 0605
, 300
1 ,
2 6
, 7
2 1
, 2
300 2
2 2
= −
+ ×
+ =
− +
+ =
m Rr
Vr z
485 ,
300 80
105 ,
105 ,
= ×
= ×
=
m z
Td n
c b
n B
538 ,
7 485
, 0605
, 1
2 8
, 696
, 2
2 1
= +
− +
+ =
+ −
+ +
=
m W
B E
tambahan lebar
E
538 ,
5 ,
3 2
538 ,
7 =
× −
= −
= =
BW maka pada tikungan PI
2
diperlukan pelebaran perkerasan 0,538 m
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5. Perhitungan kebebasan samping