BAB I PE N DAH U LUAN 1.1. Data Umum ( lihat gambar 1.1 & 1.2 ) • Tipe jembatan • Jumlah jalur • Panjang bentang jembatan • Lebar bentang jembatan • Jumlah gelagar • Panjang bersih gelagar • Jarak antar gelagar • Lebar perkerasan • Dasar pondasi abutmen • Duga air tertinggi maksimum • Duga air rendah minimum
: : : : : : : : : : :
Beton bertulang dengan gelagar balok T 2.00 jalur 18.30 m 6.82 m 4.00 buah 17.40 m 1.40 m 6.00 m 6.50 m dibawah perletakan 1.50 m dibawah perletakan 9.00 m dibawah perletakan
•
Berat volume earth fill
g1
:
1.80
ton/m3
•
Berat volume tanah (keliling)
g2
:
1.60
ton/m3
•
Teg. Tanah ijin saat air tinggi
s'1
:
3.20
kg/cm2
•
Teg. Tanah ijin saat air rendah
s'2
:
3.80
kg/cm2
1.2. Spesifikasi Beban / Muatan a. Beban mati balok : • b.
Beban tetap, Q
1.5 ton/m1
=
Beban hidup terdiri dari beban truck klas II : berdasarkan PPJJR SKBI - 1.3.28. 1987 • Beban roda, T = 70% × a1, a2 = 0.20 m b1 = 0.125 m b2 = 0.50 m
( BM 70% )
10 ton
7 ton
14 T
3.5 T
14 T
7T
1.75T
7T
b2
b1 a1
a2
1.75
7T
1.75T
•
=
3.00
7T
Beban jalur, D Beban garis, P
=
70%
×
12 t/m
=
8.4 ton/m1
Beban merata, q
=
70%
×
2.2 t/m2
=
1.54 ton/m2
( L < 30m ) 1
½P
½q
P
q 5.50
5.50
½P •
½q
Beban kejut, k
=
1.3. Spesifikasi Bahan a. Beton : • Kuat tekan, fc'
b.
20 50 + L
•
Kuat tekan ijin, `fc'
=
18.7 Mpa 14 Mpa
•
Modulus Elastisitas, Ec
=
4700 ´ √ 18.7
•
Berat jenis
=
2.40 t/m
=
240 Mpa
=
2.10E+05 Mpa
=
2.20 t/m3
=
Baja Tulangan : • Kuat leleh, fy •
c.
1+
Modulus Elastisitas, Es
=
1+
20 50 + 18.3
( K-225 ) =
=
1.293
Ingin Merubah MUTU BETON? Ketik disini!
2.03E+04 Mpa
3
( U-24 )
Ingin Merubah MUTU BAJA? Ketik disini!
Asphalt •
Berat jenis
1.4. Peraturan yang digunakan • Pedoman Perencanaan Pembebanan Jembatan Jalan Raya (PPJJR) • Peraturan Muatan Indonesia (PMI) NI. 18 • Peraturan Beton Bertulang Indonesia (PBI) 1971 NI. 2 • Tata Cara Perhitungan Struktur Beton (SNI 03-2847-2002)
18.30 0.15
2.00
0.45
A
2.00
4.35
2.00
2.00
2.00
4.35
2.00
4.35
2.00
2.00
2.00
4.35
C
0.15
0.45
17.40 MAT -1.50 M
ABUTMENT A
ABUTMENT C MAR -9.00 M
2
ABUTMENT A
ABUTMENT C MAR -9.00 M
Gambar 1.1. Long section of bridge 6.62 0.15
6.00
0.15
0.16
0.16
0.50
0.50 0.20 0.20 0.50
1.00
0.85
0.50
0.90
1.10
0.50
0.90
1.40
0.50
0.90
1.40
0.50
1.40
0.85 1.10
6.40
Gambar 1.2. Cross section of bridge
BAB II PERENCANAAN BANGUNAN ATAS 2.1. Tiang Sandaran a. Momen lentur, Gaya geser,
M V
h = 160
= =
2 2
× ×
d = 130
100 100
×
1.00
Mn
=
Mu
=
f . b . d2 . k Mn
= =
200 kgm 200 kg
= =
2000 Nm 2000 N
b = 100 Rn
=
Mu
fbd2 rperlu
=
0.85 fc' fy
= 0.8
1
2000
×
10 3
×
100
×
-
√1
-
=
1.4793 Mpa
130 2 2Rn 0.85fc'
3
rmin
=
0.85
´ 18.7 240
=
1.4 fy
=
rpakai
=
As
=
-
2 0.85
0.00583
<
rperlu
100
×
130
( As
=
´ ´
1.4793 18.7
=
84.3 mm2
=
0.00648
240
rperlu rbd
=
1.4
1
-
1
= =
0.00648
0.00648
×
2f 8
Dipakai baja tulangan
100.53
mm2 )
b. Kontrol kapasitas momen balok Dianggap baja tulangan telah mencapai luluh pada saat beton mulai retak (ec = 0.003 ), fs = fy NT
=
a
=
ND As . fy
=
0.85 fc'.b
100.53 0.85
c
=
a b1
=
15.2 0.85
=
fs
=
600
=
600
Mn
=
As
d-c c fy
d -
a 2
= Mn Mu
=
´
´ 240 18.7 ´
=
15.2 mm
100
17.9 mm
130
- 17.9 = 17.9 100.53 = (
=
2953202.71 Nmm
3761.9629 MPa ´
240
)´
> 130
fy
OK -
15.2 2
2953.20271 Nm
1.48
OK
c. Perencanaan tulangan geser : Vu 0.00 N = Vc
=
1 fVc 2
=
1 3 1 2
√ fc' . bd ´
0.65
= ´
1 18.7 3 18726.3
´
100
=
´
130
6086.05 N
=
18726.3 N
Vu > (Teoritis tidak perlu sengkang)
SNI 03-2847-2002, pasal 13.5.4. Batas spasi tulangan geser : smaksimum
=
1
d
=
1
´
130
=
65 mm 4
2
2
atau smaksimum digunakan spasi Avperlu
Avmin
=
=
600 mm
= =
65 mm dengan luas tulangan geser :
75 √ fc' . bws fy 1200
=
bw s
=
´
1
< Avmin
Dipakai baja tulangan s
=
® Æ 6
A v . fy 1 3
=
b
18.7
´
100
1200 1
fy
3 Avperlu
75
´
=
( Av
=
28.3 1 3
´ ´
65
=
7.315 mm2
=
9.028 mm2
240 100
3 Av
´
´
65
240 9.028 mm2
28.27 240 100
=
mm2 ), maka jarak sengkang : 203.575 mm
»
200 mm
2f 8 f6
160
2f 8 100
5
2.2. Perhitungan Plat Kantilever ( lihat gambar 2.1 ) a. Momen lentur (Bending moment) Tabel 2.1. Perhitungan momen lentur No. 1 2
0.10 0.10
3 4
0.10 0.10
5 6
1.00 1.00
7
1.00
P T
g (kg/m3)
Volume (m3) × 0.16 × ( 0.11 × 0.05 × ( 0.15
× ×
0.50 0.70
= ### ) / 2 =
× ×
0.50 0.50
= ### ) / 2 =
× 0.20 × ( 0.10 × 0.65
× ×
0.85 0.85
= ### ) / 2 =
0.06
=
2
×
7000
kg
(Air hujan) (Railling)
×
100 kg/m × = =
(m)
(kgm)
0.0039
16.896 7.731
0.0038
2400 2400
6.00 9.00
0.825 0.750
4.950 6.750
2400 2400
408.00 102.00
0.425 0.283
173.400 28.900
2200
85.80
0.325
27.885
200.00
1.200
240.000
9049.78
0.250
2,262.445
65.00 24.00
0.325 0.880
21.125 21.120
(aspal)
× ×
0.65 2.00
(kg)
0.880 0.837
1
× ×
Momen
19.20 9.24
### 2 2
Lengan
2400 2400
###
###
W
0.05 6.00 kg/m
1000
Total momen lentur, M =
2,811.202 28,112.02 Nm
16 15
50
H = 100 kg/m1 T = 7000 kg
1
40
25 50
2 50
3 4 5
20
6
10
11
20
85
50
90
96
50
Gambar 2.1. Cantifuer Slab load b. Gaya geser (Shear Force) Berat tiang sandaran = 1+2+3+4+railling Slab kantilever dan perkerasan = 5+6+7
= =
67.44 kg 595.80 kg 6
= = = =
Beban roda Beban genangan air hujan Total gaya lintang, V
9,049.78 65.00 9,778.02 97,780.20
kg kg kg N
c. Perhitungan baja tulangan Mu = 28,112.02 Nm Vu = 97,780.20 N hf Rn
=
300 mm
=
Mu
=
fbd2 rperlu
=
rmin
® d =
300
28,112.02
×
0.8
0.85 fc' fy
1
=
0.85
´ 18.7 240
=
1.4 fy
=
rpakai
=
As
=
1.4
-
1000
×
-
√1
-
260 mm
=
0.5198 Mpa
260 2 2Rn 0.85fc' ´ ´
1
-
2 0.85
0.00583
>
rperlu
1000
×
260
( As
=
201.06
1000
=
132.568 mm
-
=
=
10 3
×
1
40
0.5198 18.7
=
0.00220
240
rmin rbd
= =
0.00583
0.00583
f 16
Dipakai baja tulangan Sperlu
=
×
201.06 1,516.67
×
f 16
Jadi dipakai tulangan
−
=
1,516.67 mm2 mm2 ), dengan jarak tulangan :
120 mm
d. Kontrol terhadap geser beton
tc
V
= 7 8
bd
97,780.20
= 7 8
×
1000
= ×
0.43 Mpa
<
0.45 fc'
OK
260
2.3. Perhitungan Plat bagian dalam (inner slab), ( lihat gambar 2.2 )
CL
7
6.62 0.15 0.16
6.00
0.15 0.16
CL 6.62 0.15
6.00
0.15
0.16
0.16
0.50 1.75 0.50 0.20 0.10
0.20 0.50
0.90
0.85
0.50
0.90
1.10
0.50
1.40
0.90
0.50
1.40
0.90
0.50
0.85
1.40
1.10
6.40
Gambar 2.2. Wheel position
50
beban roda 20
6 20
82
a
52
sb sa
Beban roda, T = 7000 kg Bidang kontak = 0.82 × 0.52 m Penyebaran beban roda : T = 7000 × 1.29 = 21,223.69 kg/m2 0.82 × 0.52 = 0.212 Mpa
ly b
1/2 lx
1/2 lx
lx
lx a/ lx
=
1.40 m
,
ly
=
### 1.40
=
0.37
b/ lx
=
### 1.40
=
0.59
=
µ
8
ly b
1/2 lx
1/2 lx
lx
a. Momen lentur akibat beban hidup Dari PBI 1971 diperoleh : MO = 1 × 21,223.69 11 Mlx = MO = 1,404.62 sa 0.92 Mly
Mlx
= 1
+
× =
= 4a 3 lx
1
b. Momen lentur akibat beban mati Berat slab = 0.20
0.52
>
ly sa
>
3 × ½ 2.1
= = =
¾x a + ¾x r x ¾ x 0.52 + ¾ x 0.92 m
=
1,404.62 kgm
1,535.11 kgm
=
15,351.1 Nm
1,535.11 4 x 0.52 + 3 x 1.40
=
1,026.66 kgm
×
1.40
ly
×
2400
=
480 kg/m2
Berat perkerasan
=
0.06
×
2200
=
132 kg/m2
Berat air hujan
=
0.05
×
1000
=
50 kg/m2
=
662 kg/m2
=
1
×
662
=
10 1 3
×
92.7
Total qDL Mxm Mym
=
1
=
10 1 3
×
qDL
×
Mxm
c. Momen total Mx = 15,351.1 My = 10,266.6
lx 2
×
+
926.80
=
16,277.86 Nm
+
308.93
=
10,575.57 Nm
×
1.40
2
×
=
1.40 lx ½
×
1.40
10,266.6 Nm
=
92.7 kgm
=
927 Nm
=
30.9 kgm
=
309 Nm
d. Perhitungan baja tulangan Arah melintang ( lx ) Mux hf Rn
= =
16,277.86 Nm 200 mm ® d =
=
Mu
fbd2 rperlu
=
0.85 fc' fy
=
16,277.86 0.8
1
200 ×
-
40
10 3
×
1000
×
-
√1
-
= =
160 mm 0.7948 Mpa
160 2 2Rn 0.85fc' 9
rmin
=
0.85
´ 18.7 240
=
1.4 fy
=
rpakai
=
As
=
1.4
1 =
´ ´
1
-
2 0.85
0.00583
>
rperlu
1000
×
160
( As
=
113.10
1000
=
121.176 mm
-
0.7948 18.7
=
0.00340
240
rmin rbd
= =
0.00583
0.00583
f 12
Dipakai baja tulangan Sperlu
=
×
113.10 933.33
×
f 12
Jadi dipakai tulangan
−
120 mm
-
40
=
933.33 mm2 mm2 ), dengan jarak tulangan :
Arah memanjang ( ly ) Muy hf
= =
200 mm
Rn
=
Mu
10,575.57 Nm
=
fbd2 rperlu
rmin
=
® d =
200
10,575.57
×
0.8
0.85 fc' fy
1
=
0.85
´ 18.7 240
=
1.4 fy
=
rpakai
=
As
=
1000
×
-
√1
-
=
0.5164 Mpa
160 2 2Rn 0.85fc' ´ ´
-
2 0.85
0.5164 18.7
0.00583
>
rperlu
1000
×
160
( As
=
113.10
1000
=
121.176 mm
=
0.00219
=
0.00583
0.00583
Dipakai baja tulangan Sperlu
=
1
-
=
160 mm
240
rmin rbd
1.4
10 3
×
1
=
=
113.10 933.33
Jadi dipakai tulangan
× f 12 ×
f 12
−
=
933.33 mm2 mm2 ), dengan jarak tulangan :
120 mm 10
2.4. Perhitungan Gelagar ( lihat gambar 2.3 dan gambar 2.4 )
CL 6.62 0.15
6.00
0.15
0.16
0.16
0.50
0.50 0.20 0.10
0.20 0.50
0.90
0.11
0.85
0.50
0.90
1.10
0.50
0.90
1.40
0.50
0.90
1.40
0.50
1.40
0.85 1.10
6.40 0.31
0.90
1.00
1.29 1.21
0.50
A
A
=
1.21
× 2
5.10
3.09 m2
=
Gambar 2.3. Momen Influence line of girder a. Beban mati ( dead load ) Tiang sandaran = 0.1 Pipa sandaran Perkerasan Air hujan Pelat lantai Gelagar Total qDL
= = = = =
2 0.06 0.05 0.20 0.50
´
0.16
´ ´ ´ ´ ´
2.00 2200 1000 2400 1.00
Balok melintang ( diafragma ), Tb =
´ 2 ´ ´ ´ ´ ´
1.00
´
2400
6.00 3.09 3.09 3.09 2400
´
1.29
´
4.00
0.20
´
0.50
´ 2
´
1.29
= = = = = = =
2400
´
1.35
=
24.768 kg/m 30.960 407.286 154.275 1,481.040 4,800.000 6,898.329
kg/m kg/m kg/m kg/m kg/m kg/m
162.000 kg
C L 18.30 0.15
2.00
2.00
1 A
2.00
2
2.00
3
2.00
2.00
2.00
2.00
2.00
4 C
0.15
11
C L 18.30 0.15
2.00
2.00
2.00
1
2.00
2
2.00
3
2.00
2.00
2.00
2.00
0.15
4
A
C 1.09
0.45
1.09
1.09
4.35
1.09 4.35
4.35
4.35
0.45
17.40
Gambar 2.4. Section of girder for bending moment b. Momen lentur akibat beban mati MqDL ®
Mx
=
Momen pada potongan 1, x = MqDL
=
1
´
1 2
qDL
x L
. L2
1.09 6898.329
m ´
1
=
1 2
17.4 2
1.09 17.4
´
162.000
´
x L
( M1.DL )
2 MTb
-
1
-
1.09
2.18
1.09
m
=
1 2
´
6898.329
´
17.4 2
MTb
=
1 2
´
162.000
´
2.18
2.18 17.4
1
-
2.18 17.4
M2.DL
MqDL
=
1 2
´
3.27 6898.329
m ´
=
88.290 kgm
= =
61407.192 kgm 614071.92 Nm
=
114441.90 kgm
=
176.580 kgm
= =
114618.48 kgm 1146184.8 Nm
=
159368.99 kgm
( M2.DL )
MqDL
Momen pada potongan 3, x =
61318.902 kgm
17.4
M1.DL
Momen pada potongan 2, x =
=
( M3.DL )
17.4 2
3.27 17.4
1
-
3.27 17.4
12
MTb
=
1 2
´
162.000
´
3.27
M3.DL
Momen pada potongan 4, x = MqDL
=
1
´
4.36 6898.329
17.4 2
4.36
2 MTb
=
1 2
17.4 ´
162.000
´
1
-
4.36
´
1540 ´ 3 d. Momen lentur akibat beban hidup Beban merata, q
=
Mx(P)
=
P.L
Mx(q)
=
1 2
=
´
4.36
1
-
x L
q
. L2
x L
1
11169.24
1.09 ´
3.09
=
=
1
´
1583.89
1.09
´
1
17.4 2
-
17.4
1.09
Mx(P)
=
11169.24
´
17.4
353.160 kgm
= =
196453.34 kgm 1964533.4 Nm
=
11411.817 kgm
=
14079.12 kgm
= =
25490.94 kgm 254909.4 Nm
=
21298.326 kgm
17.4 1
M1.LL
2.18
=
1583.89 kg/m
1.09
2
Momen pada potongan 2, x =
196100.18 kgm
( M1.LL )
17.4 Mx(q)
=
x L
m
17.4
159633.86 kgm 1596338.6 Nm
11169.24 kg
=
x L
Momen pada potongan 1, x = Mx(P)
8400 3 3.09
= =
17.4
M4.DL
c. Beban hidup ( live load ) Koefisien kejut = 1.29 Beban garis. P = 1.29
264.870 kgm
( M4.DL )
m ´
=
m 2.18
-
1.09 17.4
( M2.LL ) 1
-
2.18
13
1
17.4 Mx(q)
=
1
´
1583.89
´
17.4 2
-
2.18
2
17.4
17.4 1
-
2.18
Mx(P)
=
11169.24
Mx(q)
=
1 2
3.27 ´
´
1583.89
3.27 17.4 ´
1
17.4 2
-
3.27 17.4
3.27 17.4
1
-
3.27 17.4
M3.LL
Momen pada potongan 4, x = Mx(P)
=
11169.24
4.36 ´
4.36 17.4
Mx(q)
=
1
´
1583.89
´
47574.74 kgm 475747.4 Nm
=
29659.529 kgm
=
36591.90 kgm
= =
66251.43 kgm 662514.3 Nm
=
36495.424 kgm
=
45025.56 kgm
= =
81520.98 kgm 815209.8 Nm
( M4.LL )
m
17.4
= = ( M3.LL )
m
17.4
26276.42 kgm
17.4
M2.LL Momen pada potongan 3, x =
=
1
17.4 2
-
4.36
2
17.4
4.36 17.4 1
-
4.36 17.4
M4.LL Tabel. 2.2. Momen lentur total (Nm) Pembebanan Beban mati, DL Beban hidup, LL Total e. Momen pada tumpuan Mtump = 1 Mmax
M.1 614071.92 254909.4 868981.27
=
3
1 3
´
M.2 1146184.8 475747.4 1621932.23
2779743.20
M.3 1596338.6 662514.3 2258852.87
=
M.4 1964533.4 815209.8 2779743.20
926581.07
Nm
f. Gaya geser ( shearing force ) Beban mati terbagi merata
=
Balok melintang
=
1 2 0.90
´
6898.329
´
162.000
´
17.4
=
60015.462 kg
=
145.8 kg
14
Beban hidup garis, P
=
Beban hidup terbagi merata, q
=
1 2 1 2
´
11169.239
´
1583.89
´
17.4
V
g. Perhitungan baja tulangan Pada tumpuan : Mtumpuan = 926581.1 Nm V = 795257.25 N b 500 mm = h = 1000 mm d = 1000 - 60 Rn
=
Mu
=
fbd rperlu
rmin
rperlu As
=
1
=
0.85
´ 18.7 240
=
1.4 fy
=
> rmin
®
rbd
=
=
=
1.4
r
fs Mn
=
´
-
√1
-
1
=
10 3
500
13779.843 kg
=
79525.725 kg 795257.25 N
-
=
0.01201
2.621608 MPa
940 2 2Rn 0.85fc' 1
=
0.00583
=
0.01201
8 D 32
As . fy
a b1
=
=
600
=
As
d-c c fy
=
´
´
0.01201
=
0.85 fc'.b c
=
-
2 0.85
´ ´
2.621608 18.7
940
=
5646.8655 mm2
240
Dipakai baja tulangan a
926581.1
0.85 fc' fy
5584.6195 kg
940 mm
=
0.8
2
=
´
500
´
( As
=
6353.81 mm2 )
6353.81
´ 240 18.7 ´
0.85
´
=
226.0
mm
=
600
940
-
d -
a 2
192.1
=
192.1
mm
500
0.85 226.0
=
1895.182 MPa
>
´
940
fy
OK
226.0
1286928080.46 Nmm
= (
6353.81
240
)´
-
192.1
2 =
1286928.08046 Nm 15
Mn
=
1.39
OK
Mu Perencanaan tulangan geser : Vu = 795257.25 N Vc
=
1 fVc 2
=
√ fc' . bd
1 3 1 2
´
0.65
= ´
1 18.7 3 677028.4
´
500
´
=
220034.24 N
940
=
677028.4 N Vu
<
(Perlu tulangan geser)
SNI 03-2847-2002, pasal 13.5.4. Batas spasi tulangan geser : smaksimum
=
1 2
smaksimum
=
600 mm
=
470 mm dengan luas tulangan geser :
d
=
´
1 2
940
470 mm
=
atau
digunakan spasi Avperlu
=
75 √ fc' . bws fy 1200 1
Avmin
=
18.7
´
< Avmin
s
maka, Vs
=
Æ 12
=
Av
´ 1 √ fc' 3
A v fy d
470
=
264.464 mm2
470
=
1410.476 mm2
1 3
=
18.7
´
500
´
240 ®
Dipakai baja tulangan
´ 240
fy Avperlu
500
1200
√ fc' bw s
3
75
=
fy
Av
=
1410.476 mm2
( Av
=
226.195 mm2 ), maka jarak sengkang : 2
=
b
=
226.195
Vs
=
s
´ 1 3
226.2
´
18.7
´
240 80
=
´
240 500
´
940
=
637868.97 N
637868.97
=
1314897.41 N
150.746 mm
»
80 mm
Kuat geser nominal, Vn : Vn
=
Vc
cek penampang geser : fVn = 0.65
+
´
1314897.41
677028.4
=
+
854683.3
>
Vu
OK 16
Æ 12 -
Jadi dipakai tulangan
80 mm untuk geser, dan
8 D 32
untuk lentur
2 D 32 200 Æ 12 -
80 1000
8 D 32 500 Pada potongan 1 : M1 =
868,981.27 Nm b
hf =
200 d =
940 h=
1000 mm
200 ) =
3700 mm
As ds = 60 mm bw = 500 Lebar efektif balok (b), dipilih yang terkecil diantara : b = 1L = 1 ´ 17400 4 4 bw b = + 16 hf = 500 + ( b
=
jarak p.k.p
=
=
4350 mm
16
´
´
18.7
1400 mm
Kontrol penampang balok- T : Dianggap seluruh flens menerima desakan sepenuhnya hf Mnf = 0.85 fc' .b .hf d = 0.85
´
1400
2 = Mnf
> Rn
M1 =
3,733,506,000 Nmm
=
3,733,506
´
200
940
-
200 2
Nm
, maka balok berperilaku sebagai balok- T Mu1
fbwd
= 2
868981.27 0.8
´
´ 500
=
10 3 ´
2.45864 MPa
940 2 17
rperlu
rmin
rperlu As
=
0.85 fc' fy
=
0.85
´ 18.7 240
=
1.4 fy
=
> rmin
®
rbd
=
=
1
=
r
fs Mn
Mn
=
= =
1
=
0.00583
=
0.01119
-
2 0.85
´ ´
2.458639 18.7
940
=
5259.8193 mm2
8 D 32
As . fy
a b1
=
600
d-c c fy
´
500
´
( As
=
6353.81 mm2 )
0.01119
=
0.85 fc'.b c
-
1
2Rn 0.85fc'
-
0.01119
=
240
Dipakai baja tulangan a
1.4
√1
-
192.1
6353.81 0.85 ´ =
226.0
600
940
´ 240 18.7 ´
=
192.1
mm
500
mm
0.85
As
=
-
226.0
=
1895.182 MPa
>
´
940
fy
OK
226.0
d -
a 2
=
1286928080.46 Nmm
=
1.48
= (
6353.81
240
)´
-
192.1
2 =
1286928.08046 Nm OK
Mu
Cek daktilitas tulangan : As max = 0.0319 hf
=
0.0319
´
=
13388.43 mm2
As min
=
rmin bd
=
As max
>
As pakai
>
b + bw
0.510d hf
200 1400
+
0.00583
´
500
500
-1 0.510
´
´ 200
940
940
=
-
1
2741.6667 mm2
As min
18
Dengan demikian penampang balok memenuhi syarat daktilitas. Pada potongan 2 : M2
=
1,621,932.23 Nm
<
Mnf
3,733,506.00 Nm
=
Perilaku balok sebagai balok- T Rn
=
Mu2
fbwd rperlu
rmin
rperlu As
=
= 2
0.8
0.85 fc' fy
1
=
0.85
´ 18.7 240
=
1.4 fy
=
> rmin
®
rbd
=
=
=
1.4
r
fs Mn
=
Mn Mu
As . fy
=
=
600
=
As
d-c c fy
=
´
-
√1
-
2Rn 0.85fc' 1
0.00583
=
0.02318
=
-
2 0.85
´ ´
940
=
288.2
´
500
´
( As
=
9530.71 mm2 )
9530.71
´ 240 18.7 ´
0.85
´
=
339.1
mm
4.588989 18.7
0.02318
=
600
940
-
=
10896.54 mm2
288.2
mm
500
0.85 =
339.1
=
1063.4546 MPa
>
fy
OK
339.1
d -
a 2
1820523814.99 Nmm 1.12
-
4.58899 MPa
940 2
=
12 D 32
a b1
=
500
1
=
10 3
´
0.02318
0.85 fc'.b c
´
240
Dipakai baja tulangan a
1621932.23
= (
9530.71
´
240
)´
940
-
288.2
2 =
1820523.81499 Nm OK
19
Pada potongan 3 : M3
=
2,258,852.87 Nm
<
Mnf
3,733,506.00 Nm
=
Perilaku balok sebagai balok- T Rn
=
Mu2
fbwd rperlu
rmin
rperlu As
=
= 2
0.8
0.85 fc' fy
1
=
0.85
´ 18.7 240
=
1.4 fy
=
> rmin
®
rbd
=
=
=
1.4
r
fs Mn
=
Mn Mu
As . fy
=
=
600
=
As
d-c c fy
=
´
-
√1
-
2Rn 0.85fc' 1
0.00583
=
0.03695
=
432.3
´
500
( As
=
14296.07
-
2 0.85
´ ´
6.391050 18.7
´
940
=
17367.125 mm2
0.03695
=
14296.07 mm2 )
´ 240 18.7 ´
0.85
´
=
508.6
mm
600
940
-
=
432.3
mm
500
0.85 =
508.6
=
508.96976 MPa
>
fy
OK
508.6
d -
a 2
2483582034.63 Nmm 1.10
-
6.39105 MPa
940 2
=
18 D 32
a b1
=
500
1
=
10 3
´
0.03695
0.85 fc'.b c
´
240
Dipakai baja tulangan a
2258852.87
= (
14296.07
´
240
)´
940
-
432.3
2 =
2483582.03463 Nm OK
20
Pada potongan 4 : M4
=
2,779,743.20 Nm
<
Mnf
3,733,506 Nm
=
Perilaku balok sebagai balok- T Rn
=
Mu2
fbwd rperlu
rmin
rperlu As
=
= 2
0.8
0.85 fc' fy
1
=
0.85
´ 18.7 240
=
1.4 fy
=
> rmin
®
rbd
=
=
=
1.4
r
fs Mn
=
Mn Mu
As . fy
=
=
600
=
As
d-c c fy
=
´
-
√1
-
2Rn 0.85fc' 1
0.00583
=
0.05984
=
528.4
´
500
( As
=
17472.97
-
2 0.85
´ ´
7.864823 18.7
´
940
=
28124.244 mm2
0.05984
=
17472.97 mm2 )
´ 240 18.7 ´
0.85
´
=
621.6
mm
600
940
-
=
528.4
mm
500
0.85 =
621.6
=
307.339 MPa
>
fy
OK
621.6
d -
a 2
2834063926.31 Nmm 1.02
-
7.86482 MPa
940 2
=
22 D 32
a b1
=
500
1
=
10 3
´
0.05984
0.85 fc'.b c
´
240
Dipakai baja tulangan a
2779743.20
= (
17472.97
´
240
)´
940
-
528.4
2 =
2834063.92631 Nm OK
21
BAB III PERENCANAAN BANGUNAN BAWAH 3.1. Tinjauan Stabilitas Abutment di Daerah Titik A (kondisi muka air tinggi maksimum 1.50 m dibawah titik A)
0.50
1.00
RA = 795257.25 N = 79.5 Ton 1.20
A
wc1
0.50
Pe1
0.40
2.67
wc2
1.50
we2
0.60
MAT -1.50 M
we1 we3
7.70
wc3
4.00
Pe2 Pw2
0.70
5.00
ww1 wc4
Pe3
0.30
6.50
Pw1
we4wc5
wc6
1.00
1.50
wc7 ww2 wc8 1.50
1
hw = 5.00 Pu 22
Keterangan : RA = gaya vertikal pada tumpuan A Wc = gaya berat beton (weight of concrete) We = gaya berat tanah timbunan (weight of earthfill) Ww = gaya berat air (weight of water) Pe = gaya tekan tanah (pressure of earth) Pw = gaya tekan air (pressure of water) Pu = gaya uplift Gaya yang bekerja pada abutment terhadap titik (1) pada kondisi muka air tinggi 1,50 m dibawah garis A
Tabel 3.1. Perhitungan reaksi & momen yang terjadi pada abutment Nota si RA
V
Uraian
H
x
Mx=V.x
y
My=H.y
( ton ) 79.526
( ton ) -
(m) 2
( tm ) 159.051
(m) -
( tm ) -
2.040
-
2.75
5.610
-
-
0.240 13.200 2.760
-
2.67 2 2.67
0.641 26.400 7.369
-
-
Wc1 1.70
×
0.50
×
1.00
×
2.40
Wc2 0.50 Wc3 5.50 Wc4 0.50
× × ×
0.50 1.00 0.50
× × ×
0.40 1.00 4.60
× × ×
1.00 2.40 1.00
×
2.40
×
2.40
Wc5 0.50 Wc6 1.50
× ×
1.00 0.30
× ×
0.30 1.00
× ×
1.00 2.40
×
2.40
0.360 1.080
-
3.33 2.25
1.199 2.430
-
-
Wc7 0.50 Wc8 4.00
× ×
1.50 0.70
× ×
0.30 1.00
× ×
1.00 2.40
×
2.40
0.540 6.720
-
1 2
0.540 13.440
-
-
We1 6.70
×
1.00
×
1.00
×
1.80
12.060
-
3.5
42.210
-
-
We2 0.50 We3 0.50
× ×
0.50 0.50
× ×
0.40 4.60
× ×
1.00 1.00
× ×
1.80 1.80
0.180 2.070
-
2.83 2.83
0.509 5.858
-
-
We4 0.50
×
1.00
×
0.30
×
1.00
×
1.80
0.270
-
3.67
0.991
-
-
Ww1 4.00
×
1.50
×
1.00
×
1.00
6.000
-
0.75
4.500
-
-
Ww2 0.50
×
1.50
×
0.30
×
1.00
0.225
-
0.5
0.113
-
-
×
1.00
23
Pe1 0.50
×
0.36
× ( 2.70 )2 ×
1.00
×
1.60
-
2.100
-
-
5.9
12.387
Pe2 0.36
×
2.70
×
×
1.00
×
1.60
-
7.776
-
-
2.47
19.207
Pe3 0.50
×
0.36
× ( 5.00 ) ×
1.00
×
1.60
-
7.200
-
-
1.63
11.736
Pw1 0.50
× ( 5.00 )2 ×
1.00
×
1.00
-
-12.500
-
-
1.63
-20.375
Pw2 0.50
× ( 5.00 )2 ×
1.00
×
1.00
-
12.500
-
-
1.63
20.375
×
4.00
×
1.00
-
2
-24.000
-
-
-
246.861
Pu
0.60
5.00
5.00 2
×
-12.000
Jumlah
115.271
17.076
43.330
3.2. Perhitungan Stabilitas Abutment (kondisi muka air tinggi maksimum 1.50 m dibawah titik A) a. Stabilitas terhadap guling SF
=
ΣMx ΣMy
=
246.861 43.330
=
5.7
>
1.5
…..(aman)
= ( 0.5
×
115.271 ) + ( 20 17.076
=
8.1
>
=
203.531 tm
b. Stabilitas terhadap geser SF
=
dimana : f = C = ΣV = ΣH =
f .
ΣV + ΣH
C.A
×
4.00
4
×
1.00 )
…..(aman)
koefisien geseran kohesi jumlah gaya vertikal jumlah gaya horisontal
c. Stabilitas daya dukung tanah M
=
e
=
eo
=
246.861 M ΣV
-
1 × b 6 ( berarti R dalam inti )
b 2
43.330 =
203.531 115.271
=
1 6
×
-
4.00 2
=
-0.234
4.00
=
0.67
>
e = -0.234
24
s12
smax
=
ΣV A
×
=
4.80
× (
=
3.11 ton/m2
1
1
±
6 .e b
=
115.271 6
+
-0.4 )
=
0.31 kg/cm2
3.3. Perencanaan Tiang Pancang Data : • Beban yang dipikul abutment (V) • Diameter tiang pancang • Jumlah tiang pancang yang diperlukan • Jarak antar tiang pancang • Kedalaman tiang pancang • •
Tekanan konus Total koefisien
×
×
4
<
s'1
=
= = = = =
115.271 ton 40 cm 10 batang 1.20 m 10 m
= =
20 kg/cm2 450 kg/cm
1
±
3.20 kg/cm2
6
(0.234) 4
OK
6.00
1.00
2.00
4.00
1.00
0.60
1.20
1.20
1.20
1.20
0.60
Beban yang dipikul oleh 1 tiang : = 115.271 = 11.53 ton 10 Daya dukung yang dibangkitkan P, yaitu : q1 = A P × 3 A
=
P q1
= =
0.25
×
3.14
20 kg/cm2 1256.00 ×
× ( 40 )2 =
20
=
1256.00 cm2
8373.33 kg 25
3 Daya dukung yang dibangkitkan C, yaitu : q2 = K C × K C q2
= = =
5 3.14 × 40 450 kg/cm 125.60 5
×
Total daya dukung (Q)
= 450
=
125.60 cm =
11304.00 kg
8373.33
+
11304.00
=
19677.33 kg
=
1)
5-
19.68
ton
Kontrol : (menurut LOS ENGELES GROUP) N = 1 d × {m.(n-1)+(m-1)+ 2.(m-1)+(n-1)} S×M
=
1-
=
1-
=
0.64
0.4 2 0.02
× { 5(
2-
1)
+ ( 5-
+
2(
1)
+ (
2-
10 × ( 18.0 ) ×
Jadi daya dukung yang diberikan oleh 1 tiang pancang : = Q N × = 19.68 0.64 × = 12.59 ton > 11.53 ton
OK
26
Daftar Tulangan Baja (Bulat)
Daftar Tulangan Baja (Ulir)
Sumber : TABEL PROFIL KONSTRUKSI BAJA
Sumber : TABEL PROFIL KONSTRUKSI BAJA
diameter (mm) berat (kg/m) 5 6 6.35 7 8 9 9.52 10 12 12.7 13 14 15 16 18 19 20 22 25 25.4 26 28 30 32 34 35 36 38
0.153 0.220 0.251 0.300 0.393 0.499 0.557 0.620 0.887 0.997 1.040 1.209 1.337 1.580 1.994 2.230 2.465 2.980 3.850 3.980 4.136 4.830 5.510 6.310 7.130 7.600 7.990 8.900
luas tampang nominal (mm2) 19.63 28.27 31.67 38.48 50.27 63.62 71.18 78.54 113.10 126.68 132.73 153.94 176.71 201.06 254.47 283.53 314.16 380.13 490.87 506.71 530.93 615.75 706.86 804.25 907.92 962.11 1017.88 1134.11
kode 10 13 16 19 22 25 29 32 35 38 41
berat (kg/m) diameter (mm) 0.560 0.995 1.560 2.250 3.040 3.980 5.040 6.230 7.510 8.950 10.500
9.53 12.70 15.90 19.10 22.20 25.40 28.60 31.80 34.90 38.10 41.30
luas tampang nominal (mm2) 71.33 126.68 198.56 286.52 387.08 506.71 642.42 794.23 956.62 1140.09 1339.65
Mutu Baja Tulangan
Mutu
Sebutan
U - 22 U - 24 U - 32 U - 39 U - 48
baja lunak baja lunak baja sedang baja keras baja keras
Tegangan leleh karakteristik (σau) atau tegangan karakteristik yang memberikan regangan tetap 0.2% (σ0.2) (kg/cm²)
Mpa
2200 2400 3200 3900 4800
220 240 320 390 480
Catatan : Mpa = Mega Pascal, satuan sistem Internasional 1 Mpa = 10 kg/cm²
Kelas dan Mutu Beton σ'bk
σ'bm
Kelas
Mutu
(kg/cm²)
dg.s=46 (kg/cm²)
Mpa
I
B0
-
-
-
II
B1 K125 K175 K225
125 175 225
200 250 300
12.5 17.5 22.5
III
K>225
>225
>300
>30
K250 K300
250 300
325 375
32.5 37.5
Catatan : Mpa = Mega Pascal, satuan sistem Internasional 1 Mpa = 10 kg/cm²
Satuan Amerika Serikat
Satuan metrik konvensional
Satuan SI
Panjang 1 ft 1 in
0.3048 m 2.540 cm
0.3048 m 25.40 mm
Luas 1 ft² 1 in²
929.0 cm² 6.452 cm²
0.09290 m² 645.2 mm²
Berat jenis 1 lb/ft²
16.018 kg/m³
1 lb
0.4536 kg
16.018 kg/m³
Gaya 4.448 N
Momen 1 ft.lb 1 in.lb
0.1383 kg.m 0.01152 kg.m
1 lb/ft² (psf) 1 lb/in² (psi)
4.882 kg/m² 0.07031 kg/cm²
1.356 N.m 0.1130 N.m
Tegangan
Modulus penampang 1 in³ Momen inersia 1 in4
47.88 N/m² 6895 N/m²
16.39 cm³
16387.06 mm³
41.62 cm4
416231.43 mm4
= = = =
47.88 6.895 0.006895 0.006895
Pa kPa N/mm² MPa
Satuan gaya pada sistem SI adalan newton (N). 1 newton adalah gaya yang diperlukan agar massa 1 kg mengalami percepatan 1 m/det². Karena berat merupakan gaya, maka ukurannya adalah newton. Karena percepatan akibat gravitasi bumi adalah 9,807m/det², maka berat 1 kg adalah 9,807 N. Perlu dicatat pula bahwa : 1 1 1
Pa = kPa = MPa =
1 1,000 1,000,000
N/m² N/m² N/m²
= =
0.001 N/mm² 1 N/mm²
N/mm²
