jembatan beton bertulang

1. 1
PERANCANGAN JEMBATAN BETON BERTULANG DENGAN
TAMPANG BALOK T
Gambar 1. Penampang melintang jembatan
1. Kondisi Jembatan
• Panjang bentang : 17,5 m
• Lebar jembatan : 9 m
• Lebar perkerasan : 7 m
• Tipe jembatan : beton bertulang dengan gelagar balok T
• Jumlah balok gelagar : 6 buah
• Panjang bersih gelagar : 16,5 m
2. Spesifikasi Pembebanan
a. Beban hidup : PPJJR No. 12/1970 (BM 100 %)
• Beban roda T : 100% x 10 t = 10 t
• Beban garis P : 100% x 12 t/m = 12 t/m
• Beban merata q : 100% x 2,2 t/m2 = 2,2 t/m2
b. Beban kejut, 2963,1
5,1750
20
1
50
20
1 =
+
+=
+
+=
L
k
3. Spesifikasi beton dan baja tulangan
a. Beton
• Kuat tekan, fc’ = 25 MPa
• Kuat tekan ijin, fc’ = 10 MPa
• Modulus elastis, Ec = 4700√25 = 23500 MPa
b. Baja tulangan
• Kuat leleh, fy = 400 MPa
• Modulus elastis, Es = 2x105 MPa

2. 2
PERANCANGAN
1. Tiang sandaran
momen lentur, Mu = 1,2×2×100×1,0 = 240 kg-m = 2400 N-m
gaya geser, V = 1,2 × 2 × 100 = 240 kg = 2400 N
Mn = φ bd2k
Mu = Mn
1095,1
1301608,0
102400
2
3
2
=
××
×
=
××
=
db
M
k u
φ
Mpa
3
'
'
108502,2
2585,0
1095,12
11
400
25
85,0
85,0
2
1185,0 −
×=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
−−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
cy
c
perlu
f
k
f
f
ρ
3
min 105,3
400
4,14,1 −
×===
yf
ρ
As = ρ x b x d = 3,5×10-3 ×160×130 = 72,8 mm2
Dipakai tulangan 2∅10 (As = 157,0796 mm2)
Kontrol kapasitas momen balok
Dianggap baja tulangan telah luluh pada saat beton mulai retak (εc = 0,003)
5,18
1602585,0
4000796,157
85,0 '
=
××
×
=
××
×
=
bf
fA
a
c
ys
mm
7647,21
85,0
5,18
1
===
β
a
c mm
7847,2983
7647,21
7647,21130
600600 =⎟
⎠
⎞
⎜
⎝
⎛ −
=⎟
⎠
⎞
⎜
⎝
⎛ −
=
c
cd
fs MPa > fy O K
68,7586944
2
5,18
1304000796,157
2
=⎟
⎠
⎞
⎜
⎝
⎛
−×=⎟
⎠
⎞
⎜
⎝
⎛
−×=
a
dfAM ysn N-mm
=7586,9447 N-m > Mu (2400 N-m) O K
Perencanaan tulangan geser
Vu = 2400 N
3333,1733313016020
6
1
6
1 '
=××=××= dbfV cc N
9999,51993333,173336,0
2
1
2
1
=××=cVφ N > Vu (secara teoritis tidak perlu sengkang)
b=160 mm
h=160 mm d=130 mm

3. 3
walaupun secara teoritis tidak perlu sengkang, tetapi untuk kestabilan struktur dan
peraturan mensyaratkan dipasang tulangan minimum
smaksimum = ½ d = ½ x 130 = 65 mm
luas tulangan geser minimum
3333,43
400
6516025
3
1
3
1 '
min =
××
=
××
=
y
c
v
f
sbf
A mm2
dipakai tulangan ∅8 (As = 100,5310 mm2), maka jarak sengkang
7965,150
16025
3
1
4005310,100
3
1 '
=
×
×
=
×
×
=
bf
fA
s
c
yv
mm
untuk penulangan geser dipakai sengkang ∅8-100
2. Perhitungan plat kantilever
Gambar 2. Pembebanan pada plat kantilever
a. momen lentur (bending moment)
Perhitungan momen lentur
No. Volume (m3) γ
(kg/m3)
W
(kg)
Lengan
(m)
Momen
(kg-m)
1 0,10 × 0,16 × 0,50 = 0,008 2400 19,2 1,8 34,5600
2 0,10×(0,70×0,110)/2 = 0,00385 2400 9,24 1,04 9,6096
3 0,10×0,05×0,50 = 0,0025 2400 6 1,025 6,1500
4 0,10 × (0,15 × 0,50)/2 = 0,00375 2400 9 0,95 8,5500
5 1,00 × 1,00 × 0,20 = 0,2 2400 480 0,5 240,0000
6 1,00 × (1,00 × 0,10)/2 = 0,05 2400 120 0,33 39,6000
7 1,00 × 0,90 × 0,07 = 0,063 2200 138,6 0,375 51,9750
P 2,0 × 100 kg/m 200 1,2 240,0000

4. 4
T 1,2963 × 10000 12963 0,5 6481,5000
Air hujan = 2 × 0,90 × 0,05 = 0,0625 1000 62,5 0,375 23,4375
Railing = 2 × 2m× 6 kg/m = 24 24 1,08 25,9200
Total momen, M 7161,3021
Total momen, M (N-m) 71613,0210
b. Gaya geser (shear force)
Berat tiang sandaran = 1 + 2 + 3 +4 + railing = 67,4400 Kg
Slab kantilever dan perkerasan = 5 + 6 +7 = 738,6000 Kg
Beban roda = 12963,0000 Kg
Beban genangan air hujan = 62,5000 Kg
Toal gaya lintang = 13831,5400 Kg
= 138315,4000 N
c. perhitungan baja tulangan
Mu = 1,2×71613,021 =85935,6252 N-m
Vu = 1,2×138315,400 = 165978,48 N
h = 300 mm d = 300-40 = 260 mm
5890,1
26010008,0
1085935,6252
2
3
2
=
××
×
=
××
=
db
M
k u
φ
MPa
027094,0
200000
400
003,0
003,0
400
85,025
85,0
003,0
003,0
85,0 1
'
=
+
×
×
=
+
×
×
=
s
yy
c
b
E
ff
f β
ρ
ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205
3
'
'
101333,4
2585,0
5890,12
11
400
25
85,0
85,0
2
1185,0 −
×=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
−−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
cy
c
perlu
f
k
f
f
ρ
3
min 105,3
400
4,14,1 −
×===
yf
ρ
As = ρ x b x d = 4,1333x10-3 x 1000 x 260 = 1074,658 mm2
Dipakai tulangan ∅16 (As = 210,0619 mm2), dengan jarak antar tulangan
4686,195
658,1074
10000619,210
=
×
=perlus mm
dipakai tulangan ∅16-125 mm
kontrol terhadap geser beton
7296,0
2601000
165978,48
8
7
8
7
=
××
=
××
=
hb
V
cτ MPa < 0,45 fc = 11,25 MPa O K

5. 5
3. Perhitungan plat bagian dalam (inner slab)
a. Momen lentur akibat beban hidup
Gambar 3. posisi roda
Penyebaran beban hidup (roda) pada slab
P
20cm21 21
6cm
15cm
15cm
50cm21 21
P
Gambar 4. Penyebaran beban hidup pada slab
lx = 1,4 m
ly = ∞
tx = 0,92 m
ty =0,62 m

6. 6
Beban roda, T = 10000 kg
Bidang kontak = 0,92 m × 0,62 m
Penyebaran beban roda, 1571,22726
62,092,0
2963,110000
=
×
×
=T kg/m2
Dipakai tabel-Bittner (dari Dr. Ing Ernst Bittner)
Dengan lx = 1,4 , ly = ∞ (lantai tidak menumpu pada diafragma)
657,0
4,1
92,0
==
x
x
l
t
fxm = 0,1233
443,0
4,1
62,0
==
x
y
l
t
fym = 0,0661
Mxm = 0,1233 × 22726,1571 × 0,92 × 0,62 = 1598,3379 kgm = 15983,379 Nm
Mym = 0,0661 × 22726,1571 × 0,92 × 0,62 = 856,8543 kgm = 8568,543 Nm
b. momen lentur akibat beban mati
Berat slab = 0,30 × 2400 = 720 kg/m2
Berat perkerasan = 0,06 × 2200 = 132 kg/m2
Berat air hujan = 0,05 × 1000 = 50 kg/m2
Total qDL = 902 kg/m2
7920,1764,1902
10
1
10
1 22
=××=××= xDLxm lqM kgm = 1767,920 Nm
9307,587920,176
3
1
3
1
=×=×= xmym MM kgm = 589,307 Nm
c. momen total
Mx = 15983,379 + 1767,920 = 17661,299 Nm
My =8568,543 + 589,307 = 9157,85 Nm
d. perhitungan baja tulangan
arah melintang lx
M = 17661,299 Nm
h = 300 mm d = 300-40 = 260 mm
3267,0
26010008,0
1017661,299
2
3
2
=
××
×
=
××
=
db
M
k
φ
MPa
027094,0
200000
400
003,0
003,0
400
85,025
85,0
003,0
003,0
85,0 1
'
=
+
×
×
=
+
×
×
=
s
yy
c
b
E
ff
f β
ρ
ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205

7. 7
4
'
'
102313,8
2585,0
3267,02
11
400
25
85,0
85,0
2
1185,0 −
×=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
−−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
cy
c
perlu
f
k
f
f
ρ
3
min 105,3
400
4,14,1 −
×===
yf
ρ
As = ρ x b x d = 3,5 x10-3 x 1000 x 260 = 910 mm2
Dipakai tulangan ∅16 (As = 210,0619 mm2), dengan jarak antar tulangan
8373,230
910
10000619,210
=
×
=perlus mm
dipakai tulangan ∅16-125 mm
arah memanjang ly
M = 9157,85 Nm
h = 300 mm d = 300-40 = 260 mm
1693,0
26010008,0
109157,85
2
3
2
=
××
×
=
××
=
db
M
k
φ
MPa
027094,0
200000
400
003,0
003,0
400
85,025
85,0
003,0
003,0
85,0 1
'
=
+
×
×
=
+
×
×
=
s
yy
c
b
E
ff
f β
ρ
ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205
4
'
'
102495,4
2585,0
1693,02
11
400
25
85,0
85,0
2
1185,0 −
×=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
−−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
cy
c
perlu
f
k
f
f
ρ
3
min 105,3
400
4,14,1 −
×===
yf
ρ
As = ρ x b x d = 3,5 x10-3 x 1000 x 260 = 910 mm2
Dipakai tulangan ∅16 (As = 210,0619 mm2), dengan jarak antar tulangan
8373,230
910
10000619,210
=
×
=perlus mm
dipakai tulangan ∅16-125 mm
4. Perhitungan Gelagar
a. beban mati (dead load)
Hand rail = {(0,10 × 0,16 × 1,00 × 2400)/2} × 1,1871 = 22,7923 kg/m
Railing = 2 × 1,00 × 6 × 1,1871 = 14,2452 kg/m
Perkerasan = 0,06 × 2200 × 4,5716 = 603,4512 kg/m
Air hujan = 0,05 × 1000 × 4,5716 = 228,5800 kg/m
Pelat lantai = 0,30 × 2400 × 4,5716 = 3291,5520 kg/m

8. 8
Gelagar = 1,00 × 0,50 × 2400 × 1,00 = 1200,0000 kg/m
Total = 5360,6207 kg/m
Balok melintang (diafragma), Tb = 0,30 × 0,60 × 2400 × 0,9 = 388,8 kg
Gambar 5. Garis pengaruh momen
Gambar 6. Potongan memanjang balok pada perhitungan momen lentur
b. momen lentur akibat beban mati
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−×=→
L
x
L
x
LqMM DLxqDL 1
2
1 2
Momen pada potongan 1, x = 2,0 m (M1 DL)
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
2
1
5,16
2
5,166207,5360
2
1 2
qDLM = 77729,0002 kgm
MTb= ½ × 388,8 × 2 = 388,8000 kgm
M1 DL = 78117,8002 kgm
781178,0020 Nm

9. 9
Momen pada potongan 2, x = 4,0 m (M2 DL)
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
4
1
5,16
4
5,166207,5360
2
1 2
qDLM = 134015,5175 kgm
MTb= ½ × 388,8 × 4 = 777,6000 kgm
M2 DL = 134793,1175 kgm
1347931,1750 Nm
Momen pada potongan 3, x = 6,0 m (M3 DL)
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
6
1
5,16
6
5,166207,5360
2
1 2
qDLM = 168859,5521 kgm
MTb= ½ × 388,8 × 6 = 1166,4000 kgm
M3 DL 170025,9521 kgm
1700259,5210 Nm
Momen pada potongan 4, x = 8,25 m (M4 DL)
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
25,8
1
5,16
25,8
5,166207,5360
2
1 2
qDLM = 182428,6232 kgm
MTb= ½ × 388,8 × 8,25 = 1603,8000 kgm
M4 DL 184032,4232 kgm
1840324,2320 Nm
c. Beban hidup (live load)
koefisien kejut = 1,2963
beban garis, 6294,258595716,4
75,2
12000
2963,1 =××=P kg
beban terbagi merata, 28,36575716,4
75,2
2200
=×=q kg/m
d. Momen lentur akibat beban hidup
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−×=
L
x
L
x
LPPM x 1
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−×=
L
x
L
x
LqqM x 1
2
1 2
Momen pada potongan 1, x = 2,0 m (M1 LL)
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−×=
5,16
2
1
5,16
2
5,166294,25859PM x = 45450,2577 kgm
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
2
1
5,16
2
5,1628,3657
2
1 2
qM x = 53030,5600 kgm
M1 LL = 98480,8177 kgm
984808,1770 Nm

10. 10
Momen pada potongan 2, x = 4,0 m (M2 LL)
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−×=
5,16
4
1
5,16
4
5,166294,25859PM x = 78362,5133 kgm
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
4
1
5,16
4
5,1628,3657
2
1 2
qM x = 91432,0000 kgm
M2 LL = 169794,5133 kgm
1697945,1330 Nm
Momen pada potongan 3, x = 6,0 m (M3 LL)
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−×=
5,16
6
1
5,16
6
5,166294,25859PM x = 98736,7668 kgm
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
6
1
5,16
6
5,1628,3657
2
1 2
qM x = 115204,3200 kgm
M3 LL 213941,0868 kgm
2139410,8680 Nm
Momen pada potongan 4, x = 8,25 m (M4 LL)
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−×=
5,16
25,8
1
5,16
25,8
5,166294,25859PM x = 106670,9713 kgm
( )
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−××=
5,16
25,8
1
5,16
25,8
5,1628,3657
2
1 2
qM x = 124461,8100 kgm
M4 LL 231132,7813 kgm
2311327,8130 Nm
Tabel. Momen lentur total
Pembebanan M.1 M.2 M.3 M.4
Beban mati, DL
Beban hidup, LL
781178,0020
984808,1770
1347931,1750
1697945,1330
1700259,5210
2139410,8680
1840324,2320
2311327,8130
Total, Mu
(1,2MD+1,6ML) 2513106,6856 4334229,6228 5463368,8140 5906513,5792
e. Gaya geser (shearing force)
Beban mati terbagi merata = 0,5 × 5360,6207 × 16,5 44225,1208 kg
Balok melintang = 1,4 × 388,8 544,3200 kg
Beban hidup garis P = 0,5 × 6294,25859 12928,8147 kg
Beban hidup terbagi merata q = 0,5 × 28,3657 × 16,5 30172,5600 kg
Total V 87870,8155 kg
878708,1550 N

11. 11
f. Perhitungan baja tulangan
Pada tumpuan
V = 878708,1550 N h = 1300 mm
b = 500 mm d = 1300 - 60 = 1240 mm
Perencanaan tulangan geser
Vu = 878708,1550 N
6667,5072911217,550025
6
1
6
1 '
=××=××= dbfV cc N
5,1521876667,5072916,0
2
1
2
1
=××=cVφ N < Vu (perlu sengkang)
Gambar 7. Diagram gaya geser (SFD)
Hasil perhitungan dapat dilihat pada tabel berikut
No. Penampang titik 1 titik 2 titik 3 titik 4
kritis 0 - 2 m 2 - 4 m 4 - 6 m 6 - 8,25 m
1 Vu (N) 878708.1550 698350.141 517992.127 337089.793
2 Vc (N) 507291.6667 507291.6667 507291.6667 507291.6667
3 ½ φ Vc (N) 152187.5 152187.5 152187.5 152187.5
Perlu sengkang Perlu sengkang Perlu sengkang Perlu sengkang
4 Vs (N) 957221.925 656625.235 356028.545 54524.655
5 s (mm) 79.91645314 116.5014334 214.8641793 1402.994317
6 s mak (mm) 608.75 608.75 608.75 608.75
7 Dipakai D10 - 75 D10 - 110 D10 - 200 D10 - 500
Potongan I-I (8,25 m dari tumpuan)
lebar efektif, diambil nilai terkecil dari :
375,45,174
1
4
1
=×== LbE m
( ) 53003001650016 =×+=+= fwE hbb mm
1400== gelagarjarakbE mm
CL
878708,1550
698350.141 517992.127
517447.807
337089.793
111642.2755
bw = 500 mm
bE = 1400 mm
hf = 300 mm
h = 1300 mm

12. 12
Mu = 5906513,5792 N-m
s
yb
b
E
fd
c
+
=
003,0
003,0
b
s
y
b d
E
f
a
+
=
003,0
003,0
85,0
ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm
dalam keadaan setimbang (ΣH = 0)
( ){ }tbbbaffA wfwbcyb ×−+×××=× '
85,0
( ){ } ( ){ } 34425
400
30050014005007562585,085,0 '
=
×−+×××
=
×−+×××
=
y
wfwbc
b
f
tbbbaf
A mm2
kemampuan sayap mendukung momen
9906750000
2
300
12602585,03001400
2
85,0 '
=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
t
dftbM cf Nmm
M = 9906750 Nm > 5906513,5792 Nm → blok beton a ada di dalam sayap
Letak garis netral, c
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
85,0 ' a
dfabM cf
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
12602585,01400,25906513579
a
a
a2 – 2520a + 397076,5431 = 0
a = 168,8889 mm, c = 168,8889/0,85 = 198,6928 mm
luas tulangan yang diperlukan
1114,4486
400
8889,1685002585,085,0 '
=
×××
=
×××
=
y
wc
f
abf
A mm2 < 0,75×Ab = 0,75×34425
Dipakai tulangan ∅30 (As = 706,8583 mm2), jumlah tulangan yang dibutuhkan
3,6
8583,706
1114,4486
==n dipakai 8∅30 (As = 5654,8664 mm2)
Potongan II-II (6 m dari tumpuan)
Mu = 5463368,8140 N-m
s
yb
b
E
fd
c
+
=
003,0
003,0
b
s
y
b d
E
f
a
+
=
003,0
003,0
85,0
ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm

13. 13
dalam keadaan setimbang (ΣH = 0)
( ){ }tbbbaffA wfwbcyb ×−+×××=× '
85,0
( ){ } ( ){ } 34425
400
30050014005007562585,085,0 '
=
×−+×××
=
×−+×××
=
y
wfwbc
b
f
tbbbaf
A mm2
kemampuan sayap mendukung momen
9906750000
2
300
12602585,03001400
2
85,0 '
=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
t
dftbM cf Nmm
M = 9906750 Nm > 5463368,8140 Nm → blok beton a ada di dalam sayap
Letak garis netral, c
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
85,0 ' a
dfabM cf
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
12602585,01400,05463368814
a
a
a2 – 2520a + 367285,2984 = 0
a = 155,3214 mm, c = 155,3214/0,85 = 182,7311 mm
luas tulangan yang diperlukan
7247,4125
400
155,32145002585,085,0 '
=
×××
=
×××
=
y
wc
f
abf
A mm2 < 0,75×Ab = 0,75×34425
Dipakai tulangan ∅30 (As = 706,8583 mm2), jumlah tulangan yang dibutuhkan
8,5
8583,706
7247,4125
==n dipakai 6∅30 (As = 4241,1501 mm2)
Potongan III-III (4 m dari tumpuan)
Mu = 4334229,6228 N-m
s
yb
b
E
fd
c
+
=
003,0
003,0
b
s
y
b d
E
f
a
+
=
003,0
003,0
85,0
ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm
dalam keadaan setimbang (ΣH = 0)
( ){ }tbbbaffA wfwbcyb ×−+×××=× '
85,0
( ){ } ( ){ } 34425
400
30050014005007562585,085,0 '
=
×−+×××
=
×−+×××
=
y
wfwbc
b
f
tbbbaf
A mm2

14. 14
kemampuan sayap mendukung momen
9906750000
2
300
12602585,03001400
2
85,0 '
=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
t
dftbM cf Nmm
M = 9906750 Nm > 4334229,6228 Nm → blok beton a ada di dalam sayap
Letak garis netral, c
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
85,0 ' a
dfabM cf
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
12602585,01400,84334229622
a
a
a2 – 2520a + 291376,7813 = 0
a = 121,4820 mm, c = 121,4820/0,85 = 142,92 mm
luas tulangan yang diperlukan
8656,3226
400
121,48205002585,085,0 '
=
×××
=
×××
=
y
wc
f
abf
A mm2 < 0,75×Ab = 0,75×34425
Dipakai tulangan ∅30 (As = 706,8583 mm2), jumlah tulangan yang dibutuhkan
6,4
8583,706
8656,3226
==n dipakai 6∅30 (As = 4241,1501 mm2)
Potongan IV- IV (2 m dari tumpuan)
Mu = 2513106,6856 N-m
s
yb
b
E
fd
c
+
=
003,0
003,0
b
s
y
b d
E
f
a
+
=
003,0
003,0
85,0
ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm
dalam keadaan setimbang (ΣH = 0)
( ){ }tbbbaffA wfwbcyb ×−+×××=× '
85,0
( ){ } ( ){ } 34425
400
30050014005007562585,085,0 '
=
×−+×××
=
×−+×××
=
y
wfwbc
b
f
tbbbaf
A mm2
kemampuan sayap mendukung momen
9906750000
2
300
12602585,03001400
2
85,0 '
=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
t
dftbM cf Nmm
M = 9906750 Nm > 2513106,6856 Nm → blok beton a ada di dalam sayap

15. 15
Letak garis netral, c
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
85,0 ' a
dfabM cf
⎭
⎬
⎫
⎩
⎨
⎧
−××××=
2
12602585,01400,62513106685
a
a
a2 – 2520a + 168948,3486 = 0
a = 68,9284 mm, c = 68,9284/0,85 = 81,0922 mm
luas tulangan yang diperlukan
9106,1830
400
68,92845002585,085,0 '
=
×××
=
×××
=
y
wc
f
abf
A mm2 < 0,75×Ab = 0,75×34425
Dipakai tulangan ∅30 (As = 706,8583 mm2), jumlah tulangan yang dibutuhkan
6,2
8583,706
9106,1830
==n dipakai 3∅30 (As = 2120,5750 mm2)
Tabel Penulangan balok
Pembebanan M.1 M.2 M.3 M.4
Beban mati, DL
Beban hidup, LL
781178,0020
984808,1770
1347931,1750
1697945,1330
1700259,5210
2139410,8680
1840324,2320
2311327,8130
Total, Mu
(1,2MD+1,6ML) 2513106,6856 4334229,6228 5463368,8140 5906513,5792
tulangan 3∅30 6∅30 6∅30 8∅30

16. 16
DAFTAR PUSTAKA
Agus Iqbal Manu, Ir.,Dipl. Heng., 1995, Dasar-Dasar Perencanaan Jembatan Beton
Bertulang, Cetakan I,P.T. Mediatana Saptakarya, Jakarta
Bambang Supriyadi, DR.,Ir., CES.,DEA., 2000, Jembatan, Edisi pertama, Beta Offset,
Jogjakarta
Departemen Pekerjaan Umum, Standar Bangunan Atas Jembatan Gelagar Beton
Bertulang Tipe T, 1993, Departemen Pekerjaan Umum Ditjen Bina Marga
Dit. Bina Program Jalan Subdit. Perencanaan Teknik Jembatan