5.4 perhitungan struktur asd berdasarkan bms 5.4.1...
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BAB V PERHITUNGAN KONSTRUKSI
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5.4 PERHITUNGAN STRUKTUR ASD BERDASARKAN BMS
5.4.1 Sandaran
5.4.1.1 Pembebanan
Menurut BMS 1992 sandaran untuk pejalan kaki harus direncanakan
untuk dua pembebanan rencana daya layan yaitu q=0,75 kN/m, yang bekerja
secara bersamaan dalam arah menyilang dan vertikal pada sandaran serta tidak
ada ketentuan beban ultimit untuk sandaran.
H = 0,70m + 0,20m +0,32m + 0,9 = 2,12 m
Sandaran menumpu pada rangka induk. Untuk perhitungan sandaran
diambil sandaran yang bawah karena mempunyai bentang yang panjang yaitu :
Dengan perbandingan segitiga
659,130,6
12,230,65,2
=−
=m
mmm
l m, ls = 2*1,659m = 3,318 m
- Beban mati = berat sendiri pipa baja,taksir adalah 10 kg/m
- Beban hidup = qH = qV = 0,75 N/mm =75 kg/m
l ls
Gambar 5.51 Sandaran Pada Jembatan
H = 6,30 m
0,70 m0,20 m
5 m
0,9 m0,32 m hs = 2,12m
ls
5 m
B A 3,318 m
Gambar 5.52 Skema Pembebanan Sandaran
q=75 +10 kg/m
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Sandaran direncanakan menggunakan pipa Ø3” t=3,2mm
5.4.1.2 Data-data Teknis fy
Mpa σijin
Mpa d
mm t
mm A
cm2
Wkg/m
Ix=Iymm4
ix=iy mm
Wx=Wymm3
240 160 76,3 3,2 7,349 5,77 49,2 2,59 12,9
5.4.1.3 Analisa Struktur
- RA = RB = 015,1412
318,3*/852
*==
mmkglq s kg
- Mmax = 972,116318,3*/85*81**
81 22 == mmkglq s kgm
5.4.1.4 Cek Kekuatan dan Kekakuan
a. Terhadap momen
W
M max=σ < σ
67,9012900
11697203 ==
mmNmmσ MPa < 160 Mpa (ok)
b. Terhadap geser
AD
=τ < τ = 0,58*σ
92,19,73415,1410
2 ==mm
Nτ MPa < 0,58*160 = 92,8 MPa (ok)
c. Terhadap lendutan
t
X D
Y
Gambar 5.53 Penampang Pipa Sandaran
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IE
lq**384
**5 4
=∆ <IE
lq**384
**5 4
=∆
436,149200*/10*0,2*3848,331*/085,0*5
426
4
−==∆ Ecmcmkg
cmcmkg cm 922,0360
8,331=
cm
cm (ok)
Jadi pipa Ø 3” dapat dipakai untuk sandaran.
D=76,3mm
Gambar 5.54 Pemasangan Pipa Sandaran
Rangka utama diagonal
Plat landas t=10 mm
begel penjepit U Ø16 mm
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5.4.2 Lantai Kendaraan dan Trotoar
Karena menggunakan metal deck maka beban diarahkan kesatu arah
sehingga termasukdalam sistem pelat satu arah, sehingga bisa diasumsikan
sebagai konstruksi yang terletak menerus diatas beberapa tumpuan.
5.4.2.1 Pembebanan
Dengan menempatkan roda di tengah-tengah pelat diharapkan
mendapatkan momen yang maksimal, dari pada menempatkan 2 roda pada pelat
dengan jarak minimal 1 m.
Untuk tinjauan perhitungan penampang pelat lantai diambil selebar per
segmen metal deck yaitu selebar 400 mm dan sudah dianggap mewakili.
Gelagar memanjang Gelagar melintang
Pelat Lantai Trotoar
5,00 m
0,6 m 1,5 m 1,5 m 1,5 m 1,5 m 0,6 m
Gambar 5.55 Denah Pelat Lantai dan Gelagar
Gelagar memanjang Gelagar melintang
5,00 m Lajur pelat satu arah 0,4 m
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a. Beban mati lantai
- berat sendiri = 0,2m*0,4m*25kN/m3 = 2,0
- beban perkerasan = 0,05m*0,4m*22kN/m3 = 0,44
- beban air hujan = 0,05m*0,4m*9,8kN/m3 = 0,196
Qd2 = 2,636 kN/m
= 2636 N/m
b. Beban hidup lantai
- beban roda 100 kN = 22 /1000
5,0*2,0100 mkN
mkN
=
= 1000 kN/m2*0,2m*(1,3) =260 = 260 kN/m
Ql2 = 260000 N/m
c. Beban mati trotoar
- berat sendiri = 0,5m*0,4m*25kN/m3 = 5,0 = 5,0 kN/m
Qd1 = 5000 N/m
d. Beban hidup trotoar
Luas trotoar A = b* l = 1m*60m = 60 m2
Qd1 Ql1
Qd2 Ql2
Gambar 5.56 Pembebanan Lantai Jembatan
1,5m 1,5m 1,5m 1,5m 0,6m 0,6m
1,5m 1,5m 1,5m 1,5m 0,6m 0,6m
30 cm20 cm
7,2m
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BMS 92 10m2 < A < 100m2 , maka:
- beban pejalan kaki Ql1 = 306033,5
3033,5 −=−
A = 3,33 Kpa = 333 kg/m2
= 333 kg/m2*60m2/60m = 333 kg/m
Ql1 = 3330 N/m
5.4.2.2 Analisa Struktur
Dengan menggunakan bantuan program SAP didapat besarnya:
Momen positif max = 84170,76 Nm = 84170760 Nmm
5.4.2.3 Data-data Teknis MUTU LANTAI DAN TROTOAR MUTU METAL DECK
fc (Mpa) fy (Mpa) fy (Mpa) tdek (mm)
30 400 360 4,5
Panjang total Metal deck per segmen P = 492,4 mm
Luas As = tdek*P = 4,5*492,4 = 2215,8 mm2
Panjang total Metal deck per 1 m lebar P = 1231mm
Berat W = V* γbaja = (0,0045*1,231*1,0)*77 kN/m3 = 0,427 kN/m2
Menentukan titik berat Metal deck :
Y = 54321
)1*5()2*4()3*3()2*2()1*1(AAAAA
YAYAYAYAYA++++
++++
)40*5,4()2,131*5,4()150*5,4()2,131*5,4()40*5,4()25,2*40*5,4()50*2,131*5,4()75,97*150*5,4()50*2,131*5,4()25,2*40*5,4(
++++++++
Y = 8,568,221525,125831
= mm
X = ½*L = ½*400 = 200 mm
Gambar 5.57 Penampang Metal deck Per Segmen
X
Y3Y2
Y1 Y
40 40 15085 85
100 mm 131,2
1
24
3
5
L=400 mm
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5.4.2.4 Cek Momen Nominal Penampang
d = H-Y = 200-56,8 = 143,2 mm
γ = 0,85-0,007*(fc-28) = 0,85-0,007*(30-28) = 0,836 dan 0,65 ≤ γ ≤ 0,85 (ok)
Ku = 261,0836,01*
30360*
2,143*10008,2215*
85,011**
**
85,01
==γc
ys
ff
dbA
a = γ*Ku*d = 0,836*0,261*143,2 = 31,2 mm
z = d-a/2 = 143,2-(31,2/2) = 127,6 mm
Mn = fy*As*d* ⎟⎟⎠
⎞⎜⎜⎝
⎛−
c
ys
ff
dbA **
*6,01
Mn = 7,10150279830360*
2,143*10008,2215*6,012,143*8,2215*360 =⎟
⎠
⎞⎜⎝
⎛− Nmm
Mu ≤ Ø*Mn
84170760 Nmm ≤ 0,9*101502798,7 Nmm
84170760 Nmm ≤ 91352518,82 Nmm
8,417 Tm ≤ 9,135 Tm (ok)
Jadi Metal deck bisa digunakan sebagai tulangan positif searah pelat lantai
jembatan.
5.4.2.5 Penulangan Pelat Lantai dan Trotoar
Tulangan pada serat atas pelat lantai dan trotoar digunakan tulangan
susut, yaitu sebesar:
As = 18%*b*h = 0,18%*1000*200 = 360 mm2
Dipakai tulangan ØD 10-200 = 393 mm2 (arah x dan y)
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5.4.2.6 Gambar Penulangan Pelat dan Trotoar
ØD10-200 ØD10-200
50 cm
Plat 10cm Metal deck 10cm
CL
150 cm 150 cm 60 cm
Gelagar memanjang
500
cm
Gambar 5.58 Penulangan Plat Lantai Kendaraan
ØD10
-200
ØD10-200
ØD10
-200
ØD10-200
Gel
agar
mem
anja
ng
10 c
m
10 c
m
ØD 0
-200
½
.q
ØD10
-200
ØD10-200 ØD 10-200
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5.4.3 Gelagar Memanjang
5.4.3.1 Pembebanan
A. Gelagar Memanjang Tengah
a. Beban mati
- Beban lantai = 0,2m*1,5m*25kN/m3 = 7,5 kN/m
- Beban perkerasan = 0,05m*1,5m*22kN/m3 = 1,65 kN/m
- Beban air hujan = 0,05m*1,5m*9,8kN/m3 = 0,735 kN/m
- beban metal deck = 0,427kN/m2*1,5m = 0,64 kN/m
qd = 10,525 kN/m
= 10,525 N/mm
b. Beban hidup
- Beban D
Beban UDL
untuk L = 50 m, maka:
ql = 4,6155,0*0,8 =⎥⎦⎤
⎢⎣⎡ + Kpa
LKpa = 640 kg/m2 = 6,4 kN/m2
ql = 6,4kN/m2*1,5m = 9,6 kN/m = 9,6N/mm
Gambar 5.60 Pengaruh Beban D pada Gelagar Memanjang
UDL,KEL
½.UDL,KEL ½.UDL,KEL
1,5 m
0,25 m 0,25 m
1,5 m1,5 m1,5m
1,5 m
5,5 m
Arah gelombang Metal deck
Lx
Ly
Gambar 5.59 Penyaluran Beban gelagar memanjang
Gelagar melintang
Gelagar memanjang
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Beban KEL
Beban dinamik (DLA)
Bentang jembatan 20<LE<90 , maka DLA = 0,525-(0,0025* LE) = 0,4
p = 44kN/m*1,5 m*(1,05) = 69,3 kN = 69300 N
- Beban T
qT = 45,67438,005,1*
3,35,1*
2100*
3,3*
2==
bDLAsT = 67,45 kN/m
= 67,45 N/mm
B. Gelagar Memanjang Tepi
a. Beban mati
- Beban lantai = 0,2m*(0,3+0,75)m*25kN/m3 = 5,25 kN/m
- Beban trotoar = 0,3m*0,3m*22t/m3 = 1,98 kN/m
- beban metal deck= 0,427kN/m2*(0,3+0,75)m = 0,448 kN/m
qd = 7,678 kN/m
= 7,678 N/mm
B A
Gambar 5.62 Model Struktur Gelagar Memanjang Tengah
ql = 9,6 N/mm
p = 69300 N
5m
qT = 67,45 N/mm
qd = 10,525 N/mm
Gambar 5.61 Penyebaran Beban Roda Dalam Lantai Baja
α 22,5o
0,2 0,1190,1190,438
α 11,25o0,6
1,5 0,5 T=100kN
Gelagar memanjang
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b. Beban hidup
- Beban UDL
ql = 3kN/m2*0,75m = 2,25 kN/m = 2,25 N/mm
Beban KEL
pl = 22kN/m*0,75m*(0,5) = 8,25 kN = 8250 N
- Beban T
qT = 45,67438,04,0*
3,35,1*
2100*
3,3*
2==
bDLAsT kN/m = 67,45 N/mm
- Beban Hidup Trotoar
pt = 3,33kN/m*0,25m = 0,8325 kN = 83,25 N
5.4.3.2 Menentukan Profil Gelagar Memanjang
Untuk profil gelagar memanjang tengah dan tepi dibuat sama yaitu
dipilih profil IWF 450*200*9*14 dengan data profil sebagai berikut: fy
Mpa Wx cm3
A cm2
Wkg/m
Ixcm3
Hmm
Bmm
tw
mm tf
mm ix
mm
iymm
360 1490 114,2 96,76 33500 450 200 9 14 186 44
H=450mm
Gambar 5.64 Penampang Profil Gelagar Memanjang
Metal deck10 cm10 cm
B= 200mm
t2=14mm
t1=9mm
Plat lantai
r=18mm
B A
Gambar 5.63 Model Struktur Gelagar Memanjang Tepi
ql = 2,25 N/mm
pl = 8250 N
5m
qT = 67,45 N/mm
qd = 7,678 N/mm
pt =83,25 N
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5.4.3.3 Analisa Struktur
Dengan menggunakan bantuan program SAP didapat besarnya:
a. Gelagar Memanjang Tengah
- Momen positif max = 46148036,42 Nmm
- Gaya lintang max = 20403,25 N
b. Gelagar Memanjang Tepi
- Momen positif max = 16844880,33 Nmm
- Gaya lintang max = 9307 N
Dipilih yang terbesar/menentukan yaitu “ a”
5.4.3.4 Cek Kekuatan
a. Tegangan lentur
σbs= σts= 718,3091490
3642,461480max3 ==
cmkgcm
WxM kg/cm2 < σ =1900 kg/cm²..ok
b. Tegangan geser
866,1722,114
325,2040===
cmkg
AwDτ kg/cm2<0,58*1900=1102kg/cm2.....ok
5.4.3.5 Cek Kekakuan/Lendutan
∆= profIE
lPprofIElqu
.**48*
.**384**5 34
+ .....< ∆ = 1500500
500==
L cm
∆= 1293,01283,000106,033500*10*2*48
500*6930033500*10*2*384
500*027575,8*56
3
6
4
=+=+−E cm
<1cm..........ok
σbs = 309,718kg/cm2
h/2 = 22,5 cm
h/2 = 22,5 cm
σts =309,718kg/cm2
Gambar 5.65 Diagram Tegangan Gelagar Memanjang
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5.4.4 Gelagar Melintang
5.4.4.1 Gelagar Melintang Tengah
1. Pembebanan
- Beban Reaksi Gelagar Memanjang
Reaksi gelagar memanjang di sini tanpa beban KEL atau pl
P1 = 2* 38553,25 = 77106,5 N
P2 = 2*9307,83 = 18615,66 N
- Beban D (KEL)
Menurut BMS 92 beban KEL P=44 KN/m
p = 44kN/m*(1,05) = 46,2 kN/m = 46,2 N/mm
p = 22kN/m*(1,05) = 23,1 kN/m = 23,1 N/mm
2. Menentukan Profil Gelagar Melintang
Pilih profil IWF 700*300*13*24 dengan data profil sebagai berikut: fy
Mpa Wx cm3
A cm2
Wkg/m
IxCm3
Hmm
Bmm
tw
mm tf
mm ix
mm
iymm
360 5760 235,5 185 201000 700 300 13 24 293 67,8
H=700mm
Gambar 5.66 Penampang Profil Gelagar Melintang Tengah
10 cm10 cm
B= 300mmt2=24mm
t1=13m
r=28mm
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3. Analisa struktur
Dengan menggunakan bantuan program SAP didapat besarnya:
- Momen positif max = 612196784 Nmm
- Gaya lintang max = 274530,36 N
4. Perhitungan Komposit
a. Untuk aksi komposit sebagian jumlah shear connector direncanakan dulu
yaitu : digunakan Stud (paku) ds = 16 mm dan Hs = 185 mm
Mutu 8.8 fu = 830 Mpa
Syarat = 6,1116185
==s
s
dH ≥ 4 (ok)
Luas 1 paku Asc = 20116*4
*4
22 ==ππ
sd mm2
Kekuatan 1 paku
Q = 0,5*Asc* cc Ef * usc FA *≤ (N)
Q = 0,5*201* 9,25742*30 = 88319,4 N
Q usc FA *≤ = 201*830 = 166830 N (ok)
Posisi gelombang dek baja sejajar dengan penumpu, maka reduksi
kekuatan paku adalah :
Rpa = 0,11**60,0 ≤⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
r
s
r
r
hH
hw
Rpa = 1199,11100185*
100235*60,0 ≥=⎟
⎠⎞
⎜⎝⎛ −⎟
⎠⎞
⎜⎝⎛ maka dipakai reduksi 1
Gambar 5.67 Model Struktur Gelagar Melintang Tengah
0,6 m 1,5 m 1,5 m 1,5 m 0,6 m
KEL
1,5 m
P2 P1 P1 P1 P2
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Jadi kekuatan1 paku = 1*88319,4 = 88319,4 N
Data perencanaan paku, yaitu:
Jarak Bentang (m) Gaya Lintang D (N) Rencana Paku/Baris
0,6 274530,36 2
1,5 187912,82 2
1,5 38553,252 1
44,88319
)36,274530*%10(36,274530)*%10( 111 =
−=
−=
nQDDn buah
Dengan jarak d1 = 220 mm
n2 = 4 buah d2 = 425 mm
n3 = 2 buah d3 = 566 mm
Jumlah paku setengah bentang : 10 buah
b. Cek kekompakan penampang
Untuk penampang komposit hanya ditinjau pada Web saja.
Web:
[ ] 185,60250360*
1324*2700
250* =⎟
⎠⎞
⎜⎝⎛ −
=⎟⎠⎞
⎜⎝⎛= yf
tbλ ≤ 82
Dari hasil di atas profil adalah berpenampang “kompak”.
be
a = 5m a = 5m c = 0,2m
Gambar 5.69 Potongan Memanjang Lantai Jembatan
be be
2*2@220 2*2@425 1*2@566
600 1500 1500
700.300.13.24
Gambar 5.68 Pemasangan Stud Gelagar Memanjang Tengah
CL
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c. Lebar efektif plat beton
menurut BMS 92 :
Gelagar tengah
- be ≤ 5l be = 44,1
52,7= m
- be ≤ 12*tmin be = 12*0,10m = 1,2 m
- be ≤ a be = 5 m
Dipilih yang terkecil be = 1,2 m
Tebal beton ekivalen (tbe) dicoba 15 cm, maka:
Angka ekivalensi n = 709,736*4700/10*2
*470010*2 255
≈===MpammN
fcEsEc
Luas beton Fc = 76,21115*7
120* == cmcmtbenbe cm2
Luas profil Fs = 235,5 cm2
Luas total Ft = 235,5cm2 + 211,76cm2 = 467,76 cm2
Ybs = 352
702
==cmh cm
Ybc = 5,8252
15702
=++=++ cmcmcmttbeh cm
Ybkomp = 2
22
76,467)35*5,235()5,82*76,211()*()*(
cmcmcmcmm
FtYbsFsYbcFc +
=+
= 56,46 cm
Gambar 5.70 Penampang Luas Beton Ekivalen Gelagar Melintang Tengah
700 mm
tbe:15cm
t:5 cm
be/n=17,14 cm
b= 300mm
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138
Ytkomp = Htot – Ybkomp = (70cm + 20cm) – 56,46cm = 33,54 cm
Ys = Ybkomp – Ybs = 56,46cm – 35cm = 21,46 cm
Yc = Ytkomp - 2.bet = 33,54 cm – 7,5cm = 26,04 cm
Ikomp = Iprof + (Fs*Ys2) + (Fc*Yc2) + 3**121 tbe
nbe
=201000+(235,5*21,462)+(211,76*26,042)+ 315*14,17*121
=467077,008 cm4
5.4.4.1.7 Cek Kekuatan
a. Tegangan Lentur
7*008,467077
33,54cm*kgcm6121967,84*
*4cmnI
YtM
komp
kompc ==σ
28,62 cmkgc =σ < 0,45*fc = 0,45*360 =162 kg/cm2.......ok
[ ]
komp
kompkomp
komp
ts Yt
tdYtI
YtM−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
=
**
σ
[ ]
207,37454,33
554,33*008,467077
54,33*84,6121967
cmkgts =−⎥
⎦
⎤⎢⎣
⎡
=σ
2740008,467077
46,56*84,6121967*cmkg
IYbM
komp
kompbs ===σ < 1900=σ ....ok
15cm
5 cm
be/n=17,14 cm
σbs = 740kg/cm2
Ys=21,46cm
σc = 62,8 kg/cm2
Gambar 5.71 Diagram Tegangan Gelagar Melintang Tengah Komposit
Ytkomp= 33,54cm
Ybkomp=56,46cm
Yc = 26cm σts = 374,07kg/cm2
Ybs = 35 cm
Ybc = 82,5cm
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139
b. Tegangan Geser
268,3270*2,1
3036,2745 cmkgcmcmkg
AwD
===τ < 211021900*58,0 cmkg==τ
5.4.4.1.8 Cek Kekakuan
- Beban Mati
( )IkompE
alaPIkompElP
IkompElqu
**48*4*3*
**48*
**384**5 2234 −
++=∆ < 500
l=∆
⎢⎣⎡++=∆
*4875*1,23*2
008,467077*10*2*48720*25,30553
008,467077*10*2*384720*2,46*5
6
3
6
4
( ) ( )⎥⎦
⎤⎢⎣
⎡ −+⎥
⎦
⎤−008,467077*10*2*48
210*4720*3*210*66,18615*2008,467077*10*2
60*4720*3*6
22
6
22
= 0,173+ 0,254 + 0,00012 + 0,120 = 0,568 cm < cmcm 44,1500
720==∆
5.4.4.2 Gelagar Melintang Tepi
1. Pembebanan
- Beban Reaksi Gelagar Memanjang
Reaksi gelagar memanjang di sini tanpa beban KEL atau pl
P1 = 30553,25 N
P2 = 9307,83 N
- Beban D (KEL)
Menurut BMS 92 beban KEL P=44 KN/m
p = 44kN/m*(1,05) =46,2 kN/m = 46,2 N/mm
p = 22kN/m*(1,05) = 23,1 kN/m = 23,1N/mm
2. Menentukan Profil Gelagar Memanjang
Pilih profil IWF 588*300*12*20 dengan data profil sebagai berikut: fy
Mpa Wx cm3
A cm2
Wkg/m
Ixcm3
Hmm
Bmm
tw
mm tf
mm ix
mm
iymm
360 4620 192,5 151 118000 588 300 12 20 248 68,5
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140
3. Analisa struktur
Dengan menggunakan bantuan program SAP didapat besarnya:
- Momen positif max = 431593764 Nmm
- Gaya lintang max = 199075 N
4. Perhitungan Komposit
a. Untuk aksi komposit sebagian jumlah shear connector direncanakan dulu
yaitu : digunakan Stud (paku) ds = 16 mm dan Hs = 185 mm
Mutu 8.8 fu = 830 Mpa
Kekuatan1 paku = 1*88319,4 = 88319,4 N
Data perencanaan paku, yaitu: Jarak Bentang (m) Gaya Lintang D (N) Rencana Paku/Baris
0,6 199075,29 2
1,5 119614,92 2
1,5 15276,63 1
44,88319
)29,199075*%10(29,199075)*%10( 111 =
−=
−=
nQDDn buah
Dengan jarak d1 = 275 mm
n2 = 4 buah d2 = 566 mm
n3 = 2 buah d3 = 566 mm
Jumlah paku setengah bentang : 10 buah
2*2@275 2*2@566 1*2@566
600 1500 1500
588.300.12.20
Gambar 5.72 Pemasangan Stud Gelagar Memanjang Tepi
CL
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141
b. Cek kekompakan penampang
Untuk penampang komposit hanya ditinjau pada Web saja.
Web:
[ ] 8,54250360*
1220*2588
250* =⎟
⎠⎞
⎜⎝⎛ −
=⎟⎠⎞
⎜⎝⎛= yf
tbλ ≤ 82
Dari hasil di atas profil adalah berpenampang “kompak”.
c. Lebar efektif plat beton
menurut BMS 92 :
Gelagar tepi
- be ≤ (L/10)+c be = (7,2/10)+0,2 = 0,92 m
- be ≤ 6*tmin be = 6*0,10 = 0,6 m
- be ≤ (a/2)+c be = (5/2)+0,2 = 2,7 m
Dipilih yang terkecil be = 0,6 m
Tebal beton ekivalen (tbe) dicoba 15 cm, maka:
Angka ekivalensi n = 709,736*4700/10*2
*470010*2 255
≈===MpammN
fcEsEc
Luas beton Fc = 57,12815*7
60* == cmcmtbenbe cm2
Luas profil Fs = 192,5 cm2
Luas total Ft = 192,5cm2 + 128,57cm2 = 321,07 cm2
Ybs = 4,2928,58
2==
cmh cm
Gambar 5.73 Penampang Luas Beton Ekivalen Gelagar Melintang Tengah
588 mm
tbe:15cm
t:5 cm
be/n=8,57 cm
b= 300mm
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BAB V PERHITUNGAN KONSTRUKSI
142
Ybc = 3,7152
158,582
=++=++ cmcmcmttbeh cm
Ybkomp = 2
22
07,321)4,29*5,192()3,71*57,128()*()*(
cmcmcmcmm
FtYbsFsYbcFc +
=+
= 46,178 cm
Ytkomp = Htot – Ybkomp = (58,8cm + 20cm) – 46,178cm = 32,622 cm
Ys = Ybkomp – Ybs = 46,178cm – 29,4 cm = 16,778 cm
Yc = Ytkomp - 2.bet = 32,622 cm – 7,5cm = 25,122 cm
Ikomp = Iprof + (Fs*Ys2) + (Fc*Yc2) + 3**121 tbe
nbe
=118000+(192,5*16,7782)+(128,57*25,1222)+ 315*57,8*121
=255754,670 cm4
5.4.4.2.1 Cek Kekuatan
a. Tegangan Lentur
7*67,255754
32,622cm*kgcm4315937,64*
*4cmnI
YtM
komp
kompc ==σ
264,78 cmkgc =σ < 0,45*fc = 0,45*360 =162 kg/cm2.......ok
[ ]
komp
kompkomp
komp
ts Yt
tdYtI
YtM−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
=
**
σ
[ ]
2129,466622,32
5622,32*67,255754
622,32*64,4315937
cmkgts =−⎥
⎦
⎤⎢⎣
⎡
=σ
226,77967,255754
178,46*64,4315937*cmkg
IYbM
komp
kompbs ===σ <
1900=σ ....ok
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BAB V PERHITUNGAN KONSTRUKSI
143
b. Tegangan Geser
2272,5648,58*6,0
529,19907 cmkgcmcmkg
AwD
===τ < 211021900*58,0 cmkg==τ
5.4.4.2.2 Cek Kekakuan
- Beban Mati
( )IkompE
alaPIkompElP
IkompElqu
**48*4*3*
**48*
**384**5 2234 −
++=∆ < 500
l=∆
⎢⎣⎡++=∆
*4875*1,23*2
67,255754*10*2*48720*25,30553
67,255754*10*2*384720*2,46*5
6
3
6
4
( ) ( )⎥⎦
⎤⎢⎣
⎡ −+⎥
⎦
⎤−67,255754*10*2*48
210*4720*3*210*83,9307*267,255754*10*2
60*4720*3*6
22
6
22
= 0,316+ 0,464 + 0,0002 + 0,11 = 0,889 cm < cmcm 44,1500
720==∆
15cm
5 cm
be/n=8,57 cm
σbs = 779,26kg/cm2
Ys=16,78cm
σc = 78,64 kg/cm2
Gambar 5.71 Diagram Tegangan Gelagar Melintang Tengah Komposit
Ytkomp= 32,62cm
Ybkomp=46,18cm
Yc = 25,12cm σts = 466,129kg/cm2
Ybs = 29,4 cm
Ybc = 71,3cm
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BAB V PERHITUNGAN KONSTRUKSI
144
5.4.4.3 Sambungan Gelagar Memanjang dengan Gelagar Melintang
a. Beban yang bekerja :
- Beban reaksi gelagar memanjang = 30553,25 N
- Beban KEL = 69300 N
Pu = 99853,25 N
b. Material penyambung
- Plat penyambung = 2L 120.120.15 fup = 520 Mpa
fy
Mpa
A cm2
w mm
bmm
Wx=Wycm3
Ix=Iy
cm4
ix= iymm
iη
mm iξ
mm
360 33,9 55 120 52,5 446 36,3 23,4 45,6
- Baut mutu 8.8 = d 20 mm fuf = 830 Mpa
Ac = 225 mm2 As = 245 mm2
c. Syarat jarak baut
- Tengah 2,5*d ≤ S ≤ 7*d 60 – 168 diambil 80 mm
- Tepi 1,5*d ≤ S1≤ 3*d 36 – 72 diambil 40 mm
1. Baut Geser
a. Cek pola baut
Lj = 320mm (300 < Lj < 1300) kr = 1,075 – Lj/4000 = 0,995
Ngeser = Ø*Vf Ntumpu = Ø*Vb
= Ø*0,62* fuf*kr*nc*Ac = Ø*3,2*df*tp* fup
= 0,7*0,62*400*0,995*2*225 = 0,7*3,2*20*15*520
= 161288,505 N = 349440 N
Ntumpu = Ø*Vb Ntumpu = Ø*Vb
Gambar 5.72 Sambungan Gelagar Memanjang Terhadap Gelagar Melintang dengan Baut Geser
Pu e1
IWF700*300*13*24
IWF 450*200*9*14
4080
80
40
320 450 80
Y2Y1
Baut geser
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BAB V PERHITUNGAN KONSTRUKSI
145
= Ø*ae*tp* fup = Ø*ae*tp* fup
= 0,7*40*15*520 = 0,7*80*15*520
= 218400 N = 436800 N
Pilih terkecil N = 161288,505 N
M = Pu*e1 = 99853,25 *(13/2+55) = 6140974,875 Nmm
Gaya yang bekerja pada baut:
Akibat gaya geser KVP = 3125,249634
25,99853==
nPu N
Akibat momen KHM = 655,23028120*240*2
120*56140974,872*21*2
*2222 =
+=
+ YYYM N
K = 22HMVP KK + = 051,33765655,230283125,24693 22 =+ ≤N=161288,505N
(ok)
Pola baut geser bisa digunakan!
b. Cek pelat penyambung
a. Kapasitas pelat hubung terhadap kombinasi geser dan momen
Adalah konservatif untuk hanya mempertimbangkan leleh umum dari
pelat seperti yang diuraikan dalam BMS 1992 yaitu:
Ixy = 2*446*104 = 8920000 mm4
ixy = 272,363390*2
8920000= mm
567,10250360*1*
968,265148
250** === y
fxy
eny
fK
iLλ
αc = 0,999
Akibat geser τ = 39,10320*15*2
999,0*25,99853*==
p
cu
AP α
Mpa
Akibat momen σ = 99,11320*15*6/1*2
875,61409742 ==
pWM Mpa
22 *3 τσσ +=i ≤ Ø*fy
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BAB V PERHITUNGAN KONSTRUKSI
146
= 21,62 Mpa ≤ 0,7*360 = 252 Mpa (ok)
b. Kekuatan pelat hubung dalam tumpu
Menurut BMS 1992 “kekuatan pelat dalam tumpu adalah memadai bila
melampaui gaya tumpuan rencana akibat baut dalam geser”
Ø*Vb tumpu = 349440 N
Ø*Vb ujung = 218400 N
Ø*Vb antar lobang = 436800 N
Pilih terkecil N = 218400 N
Vbu ≤ Ø*Vbn
nPu ≤ 218400 N
24963,3125 ≤ 218400 N (ok)
Pelat penyambung bisa digunakan!
2. Baut Geser dan Tarik
e2
Pu
IWF700*300*13*24
IWF 450*200*9*14
4080
80
40
450
80
Titik putar
240160
80
Baut geser+tarik
Pu
IWF 700*300*13*24 IWF 450*200*9*14
4080
80
40
320
700 80
Baut tarik+geser
Gambar 5.73 Sambungan Gelagar Memanjang Terhadap Gelagar Melintang Dengan Baut Geser dan Tarik
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BAB V PERHITUNGAN KONSTRUKSI
147
“Metode pendekatan”
Ntarik = Ø*Ntf Ngeser = Ø*Vf
= Ø*As*fuf = 161288,505 N
= 0,7*245*830 Ntumpu = Ø*Vb
= 142315 N = 218400 N
= 436800 N
M = Pu*e2 = 99853,25*(13/2) = 694046,125 Nmm
ΣY2 = 2*(802+1602+2402) = 179200 mm2
Akibat gaya geser KV = 3,249634
25,99853==
nPu N
Akibat momen KH = 526,929179200
240*694046,125*2 ==
ΣYYM N
1023,0142315
526,929505,1612883,24963
**
2222
≤=⎟⎠⎞
⎜⎝⎛+⎟
⎠
⎞⎜⎝
⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛
tf
H
f
V
NK
VK
φφ (ok)
Pola baut geser dan tarik bisa digunakan!
5.4.5 Pertambatan/Ikatan Angin
5.4.5.1 Pembebanan
Gambar 5.75. Pengaruh Angin dan Beban hidup Terhadap Rangka Utama
Tew1=30%
Tew2
6,17
m
3,08
5 m
3,
085
m
5,22
m
0,95
m
45 m
50 m
6,17 m
Gambar 5.74. Bidang Sisi Rangka Utama dan Beban Hidup
Beban hidup pada lantai kendaraan
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BAB V PERHITUNGAN KONSTRUKSI
148
Luas bidang rangka Ab = (50m+45m)*0,5*6,17m = 293,075 m2
Beban angin yang timbul:
a. Rangka induk
Menurut BMS 92 “luas ekivalen diambil sebagai luas padat jembatan
dalam elevasi proyek tegak lurus, untuk jembatan rangka diambil
sebesar 30% luas yang dibatasi oleh unsur rangka terluar”.
Tew1 = 30%*[0,0006*Cw*(Vw)2*Ab]*(1,2)
= 30%*0,0006*1,2*302*293,075*(1,2)
= 68,368 kN
b. Beban hidup
Menurut BS 92 “apabila ada kendaraan di atas jembatan,maka beban
garis merata tambahan arah horizontal harus diterapkan pada
permukaan lantai jembatan”.
Tew2 = [0,0012*Cw*(Vw)2]*L*(1,2)
= 0,0012*1,2*302*50*(1,2)
= 77,76 kN
1. Tekanan angin atas
RA= 307,4615,6
)95,0*76,77()085,3*368,68(15,6
)95,0*2()085,3*1(=
+=
+ ewew tTkN
P1= 145,59307,46
9==AR kN
P2= 572,22145,5
2==
P kN
RA
RB
Tew1 6,17 m
3,085 m
3,085 m
5,22 m
0,95 m
Tew2
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BAB V PERHITUNGAN KONSTRUKSI
149
2. Tekanan angin bawah
RB = 296,10015,6
)22,5*76,77()085,3*368,68(15,6
)22,5*2()085,3*1(=
+=
+ ewew tT
P1= 029,1010
296,10010
==BR kN
P2= 014,52029,10
2==
P kN
5.4.5.2 Pendimensian Ikatan Angin Atas
1. Batang Vertikal (1-2)
L= 7,2 m, P = -43,448 kN (tekan)
Dicoba profil IWF 200*200*8*12, dengan data profil :
fy
Mpa
A cm2
Wkg/m
Hmm
Bmm
tw
mm tf
mm ix
mm
iy mm
250 63,53 49,9 200 200 8 12 86,2 50,2
Koefisien tekuk (sendi-sendi) K=1
Le = Ke*L = 1*7200 = 7200 mm
a. Tentukan kapasitas penampang (Nn)
Kelangsingan penampang
168250250*
12*28200
=≤=⎟⎠⎞
⎜⎝⎛ −
=− eyflense λλ
4522250250*
8)12*2(200
=≤=⎟⎠⎞
⎜⎝⎛ −
=− eywebe λλ
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BAB V PERHITUNGAN KONSTRUKSI
150
Karena semua elemen pelat adalah kurang dari batas kelangsingan, lebar
efektif dari tiap elemen pelat adalah lebar aktual (Kf = 1)
Jadi Nn = Kf*An*fy
= 1*6353*250
= 1588250 N
b. Tentukan kapasitas tekan unsur (Nc)
Kelangsingan unsur
20043,1432,50
7200
min
≤===iLeλ (batas kelangsingan batang tekan)
4,143250250*1*
2,507200
250**
min
=== yf
eny
fK
iL
λ
αc = 0,212 (interpolasi)
Nc = αc * Nn
Nc = 0,212*1588,25 kN
Nc = 336,709 kN
Nu ≤ Ø*Nc
43,448 kN ≤ 0,9*336,709 kN
4,3448 T ≤ 30,304 T (ok)
2. Batang Diagonal (21-22,55-56)
L= 4,383 m, P = -56,021 kN (tekan)
Dicoba profil IWF 169*125*5,5*8, dengan data profil:
fy
Mpa
A cm2
W kg/m
Hmm
Bmm
tw
mm tf
mm ix
mm
iy mm
Iy
cm4
250 29,65 23,3 169 125 5,5 8 71,8 29,7 1530
Koefisien tekuk (sendi-sendi) didapat K=1
Le = Ke*L = 1*4383 = 4383 mm
a. Tentukan kapasitas penampang (Nn)
Kelangsingan penampang
169,8250360*
8*25,5125
=≤=⎟⎠⎞
⎜⎝⎛ −
=− eyflense λλ
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454,33250360*
5,5)8*2(169
=≤=⎟⎠
⎞⎜⎝
⎛ −=− eywebe λλ
Karena semua elemen pelat adalah kurang dari batas kelangsingan, lebar
efektif dari tiap elemen pelat adalah lebar aktual (Kf = 1)
Jadi Nn = Kf*An*fy
= 1*2965*250
= 741200 N
b. Tentukan kapasitas tekan unsur (Nc)
Kelangsingan unsur
20057,1477,29
4383
min
≤===iLeλ (batas kelangsingan batang tekan)
57,147250250*1*
7,294383
250**
min
=== yf
eny
fK
iLλ
αc = 0,229 (interpolasi)
Nc = 0,229*741,2 = 169,735 kN
Nu ≤ Ø*Nc
56,021 kN ≤ 0,9*169,735 kN
5,6021 T ≤ 15,276 T (ok)
3. Batang Diagonal (37-56)
L= 4,383 m, P = 39,893 kN (tarik)
Dipakai IWF 169*125*5,5*8
a. Menentukan kuat tarik rencana
- Leleh unsur
Nt = Ag*fy = 2965*250 = 741200 N
- Patahan unsur
Nt = 0,85*Kt*An*fu = 0,85*0,85*75%*2965*410 = 658730,344 N
Ambil terkecil Nt = 658730,344 N
Nu ≤ Ø*Nt
39893 N ≤ 0,9*658730,344 N
3,8893 T ≤ 59,286 T (ok)
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BAB V PERHITUNGAN KONSTRUKSI
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b. Kelangsingan unsur
57,1477,29
4383
min
===iLλ ≤ 300 (konstruksi sekunder) ok
5.4.5.3 Sambungan
Untuk profil I sambungan harus pada kedua sayapnya (jumlah minimal baut
adalah 4 buah/2 per sayap).
a. Material penyambung
- Plat penyambung = tp 10 mm fup = 410 Mpa
- Baut mutu 4.6 = d 16 mm fuf = 400 Mpa
Ac = 144 mm2 As = 157 mm2
b. Syarat jarak baut
- Tengah 2,5*d ≤ S ≤ 7*d 40 – 112 diambil 70 mm
- Tepi 1,5*d ≤ S1≤ 3*d 24 – 48 diambil 35 mm
1. Batang diagonal dengan plat buhul ikatan angin
A. Batang tarik
Besar gaya batang Pu = 39,893 kN
- Kekuatan nominal penyambung
Ngeser = Vf
= 0,62* fuf*kr*nc*Ac
= 0,62*400*1*1*144
= 35712 N
Ntumpu = Vb Ntumpu = Vb
= ae*tp* fup = 3,2*df*tp* fup
= 35*10*410 = 3,2*16*10*410
= 143500 N = 209920 N
Pilih terkecil N = 35712 N
- Jumlah baut
24,135712*9,0
39893*
===N
Pn u
φ~ 4 buah (min)
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BAB V PERHITUNGAN KONSTRUKSI
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- Cek kekuatan tarik profil
Ag = 2965 mm2
An = Ag – (n*dl*tf) = 2965-(4*18*8) = 2389 mm2
Leleh unsur Nt = Ag *fy = 2965*250 = 741250 N
Patahan unsur Nt = 0,85*Kt*An* fup = 0,85*0,85*2389*410 = 707681,5 N
Pilih terkecil Nt = 707681,5 N
Pu ≤ Ø*Nt
39893 N ≤ 0,9*707681,5 N
3,9893 T ≤ 63,691 T (ok)
B. Batang tekan
Besar gaya batang: Pu = 24,119 kN
- Kekuatan nominal penyambung
Ngeser = Vf Ntumpu = Vb
= 0,62* fuf*kr*nc*Ac = 3,2*df*tp* fup
= 0,62*400*1*1*144 = 3,2*16*10*410
= 35712 N = 209920 N
Pilih terkecil N = 35712 N
- Jumlah baut
75,035712*9,0
24119*
===N
Pn u
φ~ 4 buah (min)
- Cek kekuatan
Pu ≤ Ø*N
24119 N ≤ 0,9*35712 N
2,4119 T ≤ 2,49 T (ok)
2. Plat buhul ikatan angin dengan rangka utama
Pu = 2*39,893*sin 55,20 = 32,76 kN (tarik)
- Kekuatan nominal penyambung
Ntarik = Ntf Ngeser = Vf
= As*fuf = 35712 N
= 157*400
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BAB V PERHITUNGAN KONSTRUKSI
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= 62800 N
Ntumpu = Vb
= 209920 N
= 143500 N
Pilih yang terkecil Nt = 35712 N
- Jumlah baut
019,135712*9,0
32766*
===N
Pn u
φ atau pembautan mengikuti pola dari sambungan
rangka utama!
- Cek kekuatan tarik pelat
Ag = 2*10*300 = 6000 mm2
An = 6000-[(2*18*10*2)+(45*4
10*25*2*22
)] = 5141,111 mm2
Pilih terkecil An = 5141,111 mm2
Leleh unsur Nt = Ag *fy = 6000*250 = 1500000 N
Patahan unsur Nt = 0,85*Kt*An* fup = 0,85*0,85*5141,111*410=1522925,606 N
Pilih terkecil Nt = 1500000 N
Pu ≤ Ø*Nt
32,766 kN ≤ 0,9*1500 kN
21,383 kN ≤ 1350 kN (ok)
Gambar 5.76 Hubungan Batang Diagonal dengan Buhul
35
3555
3555
125 125
35 35
3535
Rangka utama
35169
125
55 35 35
P P
55,2o45
25
300
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BAB V PERHITUNGAN KONSTRUKSI
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3. Batang vertikal dengan rangka utama
Gaya tekan tidak pengaruh, sehingga yang diperhitungkan adalah beban reaksi
batang. Reaksi berat profil : R = (0,499*7,2*0,6)/2 = 2,15 kN
- Kekuatan nominal penyambung
Ngeser = Vf
= 35712 N
Ntumpu = Vb Ntumpu = Vb
= ae*tp* fup = 3,2*df*tp* fup
= 50*10*410 = 3,2*16*10*410
= 205000 N = 209920 N
Pilih yang terkecil N = 35712 N
- Cek kekuatan
R ≤ Ø*N
2155 N ≤ 0,9*35712 N
0,215 T ≤ 2,49 T (ok)
- Jumlah baut
067,035712*9,0
2155*
===N
Rnφ
~ 4 buah
P
Gambar 5.77 Hubungan Batang Vertikal dengan Rangka Utama
45110
200
200
200
11045 45
45Rangka utama
100
R
50
50
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5.4.5.4 Pendimensian Ikatan Angin Bawah
1. Batang Tekan
L= 8,766/2 = 4,383 m, P = -55,992 kN (tekan)
Dicoba profil 2L 90*90*11, dengan data profil L:
fy
Mpa
A cm2
Wkg/m
bmm
dmm
Ix=Iy
cm4
ix= iymm
iη
mm iξ
mm
250 18,7 14,7 90 11 138 27,2 17,5 34,1
Koefisien tekuk (sendi-sendi) K=1
Le = Ke*L = 1*4383 = 4383 mm
a. Tentukan kapasitas penampang (Nn)
Kelangsingan penampang
1618,7250250*
111190
=≤=⎟⎠⎞
⎜⎝⎛ −
= eye λλ
Karena elemen pelat adalah kurang dari batas kelangsingan, lebar efektif
dari tiap elemen pelat adalah lebar aktual (Kf = 1)
Jadi Nn = Kf*An*fy
= 1*2*1870*250
= 935000 N
b. Tentukan kapasitas tekan unsur (Nc)
Kelangsingan unsur
Ixy = 2*138*104 = 2720000 mm4
ixy = 968,261870*2
2720000= mm
20052,162968,26
4383≤===
xy
exy i
Lλ (batas kelangsingan batang tekan)
52,162250250*1*
968,264383
250** === y
fxy
eny
fK
iL
λ
αc = 0,192 (interpolasi)
Nc = αc * Nn = 0,192*935 = 179,52 kN
Nu ≤ Ø*Nc
55,992 kN ≤ 0,9*179,52 kN
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BAB V PERHITUNGAN KONSTRUKSI
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55,992 kN ≤ 161,568 kN (ok)
2. Batang Tarik
P = 47,844 kN (tarik)
Dipakai profil L 90*90*11
a. Menentukan kuat tarik rencana
- Leleh unsur
Nt = Ag*fy = 1780*250 = 445000 N
- Patahan unsur
Nt = 0,85*Kt*An*fu = 0,85*0,85*85%*1780*410 = 470849,638 N
Ambil terkecil Nt = 445000 N
Nu ≤ Ø*Nt
47,844 kN ≤ 0,9*445 kN
47,844 kN ≤ 400,5 kN (ok)
b. Kelangsingan unsur
2505,17
4383
min
===iLλ ≤ 300 (konstruksi skunder) ok
3. Kopel
Batang tekan dihubungkan pelat kopel yang direncanakan berjumlah 5
buah sehingga:
5,73015
43831 =
+==
nL
L e mm
5074,415,175,730
min
11 ≤===⎟⎠⎞
⎜⎝⎛
iL
iL
c
(ok)
758,16774,413,162 22212 =+=⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
cm
e
bn
e
iL
iL
iL
αc = 0,181 (interpolasi)
Nc = αc * Nn = 0,181*935 kN = 169,235 kN
Nu ≤ Ø*Nc
55,992 kN ≤ 0,9*169,235 kN
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BAB V PERHITUNGAN KONSTRUKSI
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55,992 kN ≤ 152,312 kN (ok)
Menurut “Centunion” ikatan angin bawah merupakan ikatan angin
sementara (temporary bracing), karena fungsi utamanya adalah sebagai penjaga
kestabilan konstruksi jembatan rangka bentang satu, saat proses perakitan bentang
yang lain (sistim kantilever), sehingga pada akhirnya ikatan angin ini bisa dicopot.
Selain alasan diatas bahwa gaya akibat tekanan angin bawah sudah
dianggap bisa ditahan oleh gelagar-gelagar jembatan.
5.4.6 Rangka Utama
5.4.6.1 Pembebanan
1. Beban Mati
a. Berat sendiri profil rangka baja
b. Ikatan Angin
P2 = (1/2*7,2*0,499) + (1/2*8,496*0,233) = 2,786 kN
P1 = 2*(1/2*8,496*0,233) = 1,979 kN
c. Beban reaksi gelagar melintang
P3 = 274,263 kN (melintang tengah)
P4 = 199,352 kN (melintang tepi)
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BAB V PERHITUNGAN KONSTRUKSI
159
2. Beban Hidup Berjalan
a. Beban UDL
q = 6155,0*0,8 =⎥⎦⎤
⎢⎣⎡ + Kpa
L,4Kpa = 640 kg/m2 = 6,4 kN/m2
q = (6 kN/m2*5,5m)/2+(3,2kN/m2*0,25m) = 16,9 kN/m
b. Beban KEL
p = (44kN/m*5,5m/2)+(22kN/m*0,25m)*(0,5) =123,75 kN
3. Beban Tekanan Angin
a. Angin atas
b. Angin bawah
Beban-beban diatas diselesaikan dengan program SAP (lampiran)
5.4.6.2 Kombinasi Gaya Batang
Setelah gaya-gaya batang dari beban-beban diatas dikombinasikan maka
didapat kelompok-kelompok gaya batang yang menentukan yaitu:
- Batang horisontal (1,10) P = 2487,486 kN (tarik)
- Batang horisontal (2,9) P = 6712,098 kN (tarik)
- Batang horisontal (3,8) P = 9935,865 kN (tarik)
- Batang horisontal (4,7) P = 12108,053 kN (tarik)
- Batang horisontal (5,6) P = 13229,580 kN (tarik)
- Batang horizontal (11,19) P = - 4764,348 kN (tekan)
- Batang horizontal (12,18) P = -8496,994 kN (tekan)
- Batang horizontal (13,17) P = -11173,420 kN (tekan)
- Batang horizontal (14,16) P = -12823,949 kN (tekan)
- Batang horizontal (15) P = -13399,628 kN (tekan)
- Batang diagonal (20,39) P = -6381,761 kN (tekan)
- Batang diagonal (21,38) P = 6205,888 kN (tarik)
- Batang diagonal (22,37) P = -4912,664 kN (tekan)
- Batang diagonal (23,36) P = 4854,772 kN (tarik)
- Batang diagonal (24,35) P = -3534,115 kN (tekan)
- Batang diagonal (25,34) P = 3464,701 kN (tarik)
- Batang diagonal (26,33) P = -2202,875 kN (tekan)
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BAB V PERHITUNGAN KONSTRUKSI
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- Batang diagonal (27,32) P = 2123,122 kN (tarik)
- Batang diagonal (28,31) P = -817,142 kN (tekan)
- Batang diagonal (29,30) P = 722,503 kN (tarik)
5.4.6.3 Pendimensian
1. Batang Horisontal (15)
L= 5 m, P = -13399,628 kN (tekan)
Dicoba profil IWF 458*417*30*50, dengan data profil :
fy
Mpa
A cm2
W kg/m
HMm
Bmm
tw
mm tf
mm ix
mm
iy mm
Iy
cm4
360 528,6 415 458 417 30 50 188 107 60500
Koefisien tekuk (sendi-sendi) Ke=1
Le = Ke*L = 1*5000 = 5000 mm
a. Tentukan kapasitas penampang (Nn)
Kelangsingan penampang
166,4250360*
50*230417
=≤=⎟⎠⎞
⎜⎝⎛ −
=− eyflense λλ
4532,14250360*
30)50*2(458
=≤=⎟⎠⎞
⎜⎝⎛ −
=− eywebe λλ 4
Karena semua elemen pelat adalah kurang dari batas kelangsingan, lebar
efektif dari tiap elemen pelat adalah lebar aktual (Kf = 1)
Jadi Nn = Kf*An*fy
= 1*52860*360
= 19029600 N
b. Tentukan kapasitas tekan unsur (Nc)
Kelangsingan unsur
200471075000
min
≤===iLeλ (batas kelangsingan batang tekan)
4,56250360*1*
1075000
250**
min
===fyK
iL
fe
nyλ
αc = 0,815 (interpolasi)
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BAB V PERHITUNGAN KONSTRUKSI
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Nc = 0,815*19029,6= 15509,124 kN
Nu ≤ Ø*Nc
13399 kN ≤ 0,9*15509,124 kN
1339,9 T ≤ 1395,82 T (ok)
2. Batang Horisontal (5,6)
L= 5 m, P = 13229,580 kN (tarik)
Dipakai profil IWF 458*417*30*50
a. Menentukan kuat tarik rencana
- Leleh unsur
Nt = Ag*fy = 52860*360 = 19029600 N
- Patahan unsur
Nt = 0,85*Kt*An*fu = 0,85*0,85*85%*52860*520 = 16880576,7 N
Ambil terkecil Nt = 16880576,7 N
Nu ≤ Ø*Nt
13229,58 kN ≤ 0,9*16880,58 kN
1322,96 T ≤ 1519,25 T (ok)
b. Kelangsingan unsur
471075000
min
===iLλ ≤ 240 (konstruksi utama) ok
Dengan cara perhitungan yang sama maka pendimensian rangka utama dapat
ditabelkan sebagai berikut:
Tabel 5.4 Hasil Perhitungan Dimensi Rangka Utama
NO BATANG
POSISI BATANG
PANJANG BATANG
m
GAYA BATANG Nu (kN)
PROFIL TERPILIH
αc
KAPASI TAS
TEKAN UNSUR Nc (kN)
LELEH UNSUR Nt (kN)
PATAHAN UNSUR Nt
kN Nu ≤ Ø*Nn
1,2,9,10 Horisontal 5 +6712,098 400*408*
21*21 - - 8005,98 7205,38 6712,098
3,8 Horisontal 5 +9935,865 428*407*
20*35 - - 11518,77 10366,89 9935,865
4,7 Horisontal 5 +12108,05 458*417*
30*50 - - 19029,6 16880,6 12108,05
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BAB V PERHITUNGAN KONSTRUKSI
162
11,12,18,
19 Horisontal 5 -8496,994
428*407*
20*35 0,792 9255,85 - - 8496,994
14,16 Horisontal 5 -12823,949 458*417*
30*50 0,815 13958,2 - - 12823,949
20,21,38,
39 Diagonal 6,77 -6381,761
400*408*
21*21 0,783 8122,68 - - 6381,76
22,23,33,
34 Diagonal 6,77 -4912,664
394*405*
18*18 0,779 5411,37 - - 4912,664
24,25,34,
35 Diagonal 6,77 +3464,701
390*300*
10*16 - - 4896 3908,78 3464,701
26-29,30-
33 Diagonal 6,77 -2202,0875
386*299*
9*14 0,579 2253,03 - - 2202,0875
A. Hubungan Gelagar Melintang Tengah dengan Rangka Utama
Baut mengalami geser dan tarik secara bersamaan!
Pu = 969231,78 N
a. Material penyambung
- Plat penyambung = tp 20 mm fup = 520 Mpa
- Baut mutu 8.8 = d 24 mm fuf = 830 Mpa
Ac = 324 mm2 As = 353 mm2
b. Syarat jarak baut
Baut dipasang lurus
- Tengah 2,5*d ≤ S ≤ 7*d 60 – 168 diambil 140 mm
- Tepi 1,5*d ≤ S1≤ 3*d 36 – 72 diambil 64 mm
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BAB V PERHITUNGAN KONSTRUKSI
163
“Metode pendekatan”
Kapasitas nominal penyambung
- Tegak lurus sumbu baut:
Lj = 548 mm (300 < Lj < 1300) kr = 1,075 – Lj/4000 = 0,938
Ngeser = Vf
= 0,62* fuf*kr*nc*Ac
= 0,62*830*0,938*2*324
= 312786,23 N
Ntumpu = Vb Ntumpu = Vb
= ae*tp* fup = 3,2*df*tp* fup
= 64*10*520 = 3,2*24*10*520
= 332800 N = 399360 N
Ambil terkecil NVf = 312786,23 N
- Searah sumbu baut:
420
e Pu
Titik putar
Baut geser+tarik
140
64
700 750
548
202,5
IWF 458*417*30*50
DIA
FRA
GM
A
IWF
458*
417*
30*5
0
140
280
64
IWF 700*300*13*24
Gambar 5.35 Hubungan Gelagar Melintang Tengah Terhadap Rangka Utama
Pelat sambung t=20 mmPelat simpul t=20 mm
140
140
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BAB V PERHITUNGAN KONSTRUKSI
164
Ntarik = Ntf
= As*fuf
= 353*830
= 292990 N
ΣY2 = 2*(1402+2802+4202) = 548800 mm2
M = Pu*e = 969231,78*( ½*414+ 20) = 220015614,1 Nmm
Akibat gaya geser KV = 973,1211538
78,969231==
nPu N
Akibat momen KH = 297,168379548800
420*1220015614,*2 ==
ΣYYM N
1674,0292990*7,0
297,16837923,312786*7,0
973,121153**
2222
≤=⎟⎠
⎞⎜⎝
⎛+⎟
⎠
⎞⎜⎝
⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛
tf
H
f
V
NK
VK
φφ (ok)
Pola baut geser dan tarik bisa digunakan!
B. Hubungan Gelagar Melintang Tepi dengan Rangka Utama
Baut mengalami geser dan tarik secara bersamaan!
Pu = 685390,01 N
a. Material penyambung (idem)
b. Syarat jarak baut
Baut dipasang lurus
- Tengah 2,5*d ≤ S ≤ 7*d 60 – 168 diambil 150 mm
- Tepi 1,5*d ≤ S1≤ 3*d 36 – 72 diambil 69 mm
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BAB V PERHITUNGAN KONSTRUKSI
165
“Metode pendekatan”
Kapasitas nominal penyambung
- Tegak lurus sumbu baut:
Lj = 588 mm (300 < Lj < 1300) kr = 1,075 – Lj/4000 = 0,928
Ngeser = Vf
= 0,62* fuf*kr*nc*Ac
= 0,62*830*0,928*2*324
= 309451,622 N
Ntumpu = Vb Ntumpu = Vb
= 3,2*df*tp* fup = ae*tp* fup
= 3,2*24*10*520 = 69*10*520
= 399360 N = 358800 N
Ambil terkecil NVf = 309451,622 N
- Searah sumbu baut:
450
e Pu
Titik putar
69
Baut geser+tarik
150
69
600
750
588
150
IWF 390*300*10*16
DIA
FRA
GM
A
IWF
390*
300*
10*1
6
150300
IWF 588*300*12*20
Gambar 5.36 Hubungan Gelagar Melintang Tepi Terhadap Rangka Utama
Pelat sambung t=20 mm
Pelat simpul t=20 mm
150
150
150 800
Jacking force Ru=141,8 TonPelat pengaku badan t=10*144
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BAB V PERHITUNGAN KONSTRUKSI
166
Ntarik = Ntf
= 292990 N
ΣY2 = 2*(1502+3002+4502) = 630000 mm2
M = Pu*e = 685390,01*( ½*390+20) = 147358852,2 Nmm
Akibat gaya geser KV = 751,856738
01,685390==
nPu N
Akibat momen KH = 323,105256630000
450*2,147358852*2 ==
∑YYM N
1416,0292990*7,0
323,105256622,309451*7,0
751,85673**
2222
≤=⎟⎠
⎞⎜⎝
⎛+⎟
⎠
⎞⎜⎝
⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛
tf
H
f
V
NK
VK
φφ (ok)
Pola baut geser dan tarik bisa digunakan!
C. Hubungan Antar Rangka Utama
a. Material penyambung
- Plat penyambung sayap = tpf 20 mm fup = 410 Mpa
Plat penyambung badan = tpw 15 mm fup = 410 Mpa
- Baut mutu 8.8 = d 20 mm fuf = 830 Mpa
Ac = 225 mm2 As = 245 mm2
b. Syarat jarak baut
Baut dipasang lurus
- Tengah 2,5*d ≤ S ≤ 7*d 50 – 140 diambil 100 mm
- Tepi 1,5*d ≤ S1≤ 3*d 30 – 60 diambil 50 mm
Baut dipasang selang-seling
- Tengah V 2,5*d ≤ sg ≤ 7*d 50 – 140 diambil 100 mm
- Tengah H sp ≤ 7*d- ½*sg 90 diambil 80 mm
c. Kekuatan nominal penyambung
- Untuk sambungan irisan 2 Ngeser = Vf
= 0,62* fuf*kr*nc*Ac
= 0,62*830*1*2*225
= 231570 N
Jika dengan pelat pengisi, Ngeser = 85%* Vf = 196834,5 N
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BAB V PERHITUNGAN KONSTRUKSI
167
- Untuk sambungan irisan 1 Ngeser = Vf
= 0,62* fuf*kr*nc*Ac
= 0,62*830*1*1*225
= 115785 N
Jika dengan pelat pengisi, Ngeser = 85%* Vf = 98417,25 N
Ntumpu = Vb Ntumpu = Vb
= ae*tp* fup = 3,2*df*tp* fup
= 50*20*410 = 3,2*20*20*410
= 410000 N = 524800 N
Ntumpu = Vb
= ae*tp* fup
= 100*20*410
= 820000 N
d. Contoh perhitungan jumlah baut pada batang dengan Pu terbesar.
Sambungan batang-batang horisontal direncanakan menggunakan
irisan 2, ini dimaksudkan untuk memperkecil jumlah baut agar kebutuhan
luasan pelat simpul menjadi lebih kecil.
1. Batang horisontal tarik (6)
Besar gaya batang Pu = 7160,929 kN
- Kekuatan nominal penyambung
Untuk sambungan irisan 2 Ngeser = Vf = 231570 N
Ntumpu = Vb = 410000 N
= 524800 N
= 820000 N
Pilih terkecil N = 231570 N
- Jumlah baut
45231570*7,0
7160929*
===N
Pn u
φ ~ 48 buah
Pada sayap = 16+16 buah
Pada badan = 16 buah
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BAB V PERHITUNGAN KONSTRUKSI
168
- Cek kekuatan tarik profil
Ag = 29540 mm2
An1 = Ag – (n*dl*tf)-(n*dl*tw) = 29540-(4*22*28)-(2*22*18) = 26284 mm2
An2 = Ag – [(n*dl*tf)+ (g
fp
sts
*4*2
Σ )]-(n*dl*tw)
= 29540-[(8*22*28)+(100*4
28*80*2*22
)]-(2*22*18) = 22028 mm2
Pilih terkecil An2 = 22028 mm2
Leleh unsur Nt = Ag *fy = 29540*360 = 9570960 N
Patahan unsur Nt = 0,85*Kt*An* fup = 0,85*0,85*22028*520 = 8275919 N
Pilih terkecil Nt = 8275919 N
Pu ≤ Ø*Nt
7160,929 kN ≤ 0,9*8275,919 kN
7160,929 kN ≤ 7448,328 kN (ok)
2. Batang horisontal tekan (18)
Besar gaya batang Pu = -7355 kN
- Kekuatan nominal penyambung
Untuk sambungan irisan 2 Ngeser = Vf = 231570 N
Ntumpu = Vb = 820000 N
= 524800 N
Pilih terkecil N = 231570 N
- Jumlah baut
46231570*7,0
7355000*
===N
Pn u
φ ~ 48 buah
Pada sayap = 16+16 buah
Pada badan = 16 buah
Berikut tabel jumlah baut rangka utama :
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169
Tabel 5.2 Jumlah Baut Rangka Utama
SIMPUL BATANG IRISAN SAMBUNGAN
TEREDUKSI
GAYA BATANG
(kN)
JUMLAH BAUT
(n)
TERPASANG JUMLAH BAUT
TERPASANG (n)
SAYAP SAYAP BADAN
1 H-1 1 - 1119,024 14 8 8 - 16
D-24 1 - -3020,966 38 20 20 - 40
2
D-24 1 - -3020,966 38 20 20 - 40
H-13 1 - -2230,898 28 14 14 - 28
D-36 1 - 2991,843 37 20 20 - 40
3
H-1 2 - 1119,024 7 8 8 4 20
D-36 1 - 2991,843 37 20 20 - 40
D-25 1 - -2507,554 31 20 20 - 40
H-2 2 - 3115,38 20 8 8 8 24
4
D-25 1 - -2507,554 31 16 16 - 32
H-13 2 - -2230,898 14 10 10 6 26
H-14 2 - -4070,94 26 10 10 10 30
D-37 1 - 2486,201 31 16 16 - 32
5
H-2 2 √ 3115,38 23 10 10 8 28
D-37 1 √ 2486,201 37 20 20 - 40
D-26 1 √ -2029,047 30 20 20 - 40
H-3 2 - 4669,811 29 10 10 10 30
6
D-26 1 √ -2029,047 30 16 16 - 32
H-14 2 √ -4070,94 30 12 12 10 34
H-15 2 - -5513,735 35 12 12 12 36
D-38 1 √ 2006,882 30 16 16 - 32
7
H-3 2 - 4669,811 29 14 14 10 38
D-38 1 √ 2006,882 30 16 16 - 32
D-27 1 √ -1567,916 23 16 16 - 32
H-4 2 - 5957,774 37 14 14 14 42
8
D-27 1 √ -1567,916 23 12 12 - 24
H-15 2 √ -5513,735 40 14 14 14 42
H-16 2 - -6535,412 41 14 14 14 42
D-39 1 √ 1547,16 23 12 12 - 24
9
H-4 2 √ 5957,774 44 16 16 16 48
D-39 1 √ 1547,16 23 12 12 - 24
D-28 1 √ -1116,649 17 12 12 - 24
H-5 2 - 6765,158 42 16 16 14 46
10 D-28 1 √ -1116,649 17 10 10 - 20
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BAB V PERHITUNGAN KONSTRUKSI
170
H-16 2 - -6535,412 41 16 16 14 46
H-17 2 - -7156,632 45 16 16 16 48
D-40 1 √ 1093,32 16 10 10 - 20
11
H-5 2 - 6765,158 42 16 16 14 46
D-40 1 √ 1093,32 16 8 8 - 16
D-29 1 √ -681,717 10 8 8 - 16
H-6 2 - 7160,929 45 16 16 16 48
12
D-29 1 √ -681,717 10 6 6 - 12
H-17 2 - -7156,632 45 16 16 16 48
H-18 2 - -7355 46 16 16 16 48
D-41 1 √ 658,801 10 6 6 - 12
13
H-6 2 - 7160,929 45 16 16 16 48
D-41 1 √ 658,801 10 6 6 - 12
D-30 1 √ 658,801 10 6 6 - 12
H-7 2 - 7160,929 45 16 16 16 48
Catatan :
1. Pada jumlah baut terpasang terlihat jumlah baut yang lebih besar dan seragam (terutama
pada sayap), ini dimaksudkan untuk mempermudah perhitungan luas pelat simpulnya dan
penataan pembautannya.
2. H = batang horisontal
D = batang diagonal
√ = tereduksi 15% akibat penggunaan pelat pengisi
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