TUGAS I
MERENCANAKAN BALOK PRATEGANG
Mt = 350 tm
fβc = 47 MPa
Perencanaan Dimensi Balok
Mt = T.z
= T . 0,65 h
T = ππ‘
0,65β
Mencari nilai h
β = π βMt ; Diambil nilai k = 10
β = 10 β350
= 187,08 cm
Diambil 200 cm
π =ππ‘
0,65 β=
350.105
0,65 . 200= 269230,77 ππ
Perhitungan penampang Ab (dengan tafsiran Οbr = 0,40 Οb akhir)
ΟΜ β² Akhir =0.33 bk = 0.33 . 470 = 155,1 kg/cm2
Abhitung = T
Οbr ........................=
269230 ,77
0,40 .155 ,1 = 4339,6 cm2
Dicoba dengan penampang seperti gambar dibawah :
Abrencana = (60 x 20 ) + (155 x 20 ) + ( 40 x 40 )
= 5900 cm2
Abhitung β€ Abrencanaβ¦β¦β¦β¦.( OK)
Menghitung Ta
ππ =π
1ββ Post Tensioning β = 15%
ππ =269230 ,77
1β0,15= 316742.08 πΎπ
ππ = ππ
π΄π ππ = 0,85 .πππ’ ; πππ’ = 18000 πΎπ/ππ2
π΄π = π
ππ=
316742 .08
0,85 . 18000= 21,99 ππ2 = 22 ππ2
80.00
20.00
135.00
40.00
45.00
Menghitung titik berat penampang
a. Menghitung titik berat
π¦1 = ππ’ππ 1 .
1
2β1+ππ’ππ 2 . (
1
2β2+β1)+ ππ’ππ 3 (
1
2β3+β1+β2)
ππ’ππ π‘ππ‘ππ
π¦1 = 1200 .
1
220+3100 . (
1
2155+20)+ 1600 (
1
240+20+155)
5900
= 106,14 cm
π¦2 = β β π¦1
= 200 β 106,14
= 93,86 cm
ππ = 93,86 β 20 = 73,86 ππ
ππ‘ =π .π΄.ππ
π΄π +π.π΄ π =
2.105
32221,58= 6,207 β 7
=7 .21,99 .73,86
5900+7 .21,99= 1,88 ππ
πππ‘ = ππ β ππ‘
= 73,86 β 1,88
y1
y2
et
eat
ea
t2
t1
ta
ya
yb
= 71,98 cm
π¦π = π¦1 + ππ‘
= 106,14 + 1,88
= 108,02 cm
π¦π = β β π¦π
= 200 β 108,02
= 91,98 cm
b. Menghitung momen inersia tampang :
Bagian 1
πΌ =1
12 π1 .β13 + π1 .β1(π¦1 β
1
2β1)2
πΌ =1
12 60 .203 + 60 .20(106,14 β
1
220)2
= 11131479,52 cm4
Bagian 2
πΌ =1
12 π2 .β23 + π2 .β2(π¦1 β (
1
2β2 + β1))2
πΌ =1
12 20 .1553 + 20 .155(106,14 β (
1
2155 + 20))2
= 6437872,09 cm4
Bagian 3
πΌ =1
12 π3 .β33 + π3 .β3(π¦2 β
1
2β3)2
πΌ =1
12 40 .403 + 40 .40(106,14 β
1
2. 40)2
= 12085492,69 cm4
πΌπ‘ππ‘ = Ibagian 1 + Ibagian 2 + Ibagian 3 + Aa. n. eat2
πΌπ‘ππ‘ = 11131479,52 + 6437872,09 + 12085492,69 + 21,99 .7 .73,862
= 30494578,57 cm4
π΄ππ‘ = π΄π + π .π΄π
= 5900 + 7. 21,99
= 6053,93 cm2
a. Menghitung Desain Akhir
Asumsi : desain akhir dengan tidak diijinkan tegangan tarik
1. Pada kondisi awal
ππ =π
π΄ππ‘Β±
ππ.πππ‘.π¦1
πΌπ‘Β±
πππ
πΌπ‘
a. Pada serat atas
ππ =ππ
π΄ππ‘+
ππ.πππ‘.ππ
πΌπ‘β
πππ .π¦π
πΌπ‘
0 =316742 ,08
6053 ,93+
316742 ,08.71,98.108 ,02
30494578,57β
πππ .108 ,02
30494578 ,57
πππ = 4058236913
108,02= 37569310,43 Kgcm = 375,69 tm
π = π1 .ππ
πΌ=
37569310,43kg.cm .108,02 ππ
30494578 ,57 cm4 = 133,08 kg/cm2
b. Pada serat bawah
ππ = βππ
π΄ππ‘β
ππ.πππ‘.π¦π
πΌπ‘+
πππ .π¦π
πΌπ‘
β282 = β316742,08
6053 ,93β
316742,08.71,98.91,98
30494578 ,57+
πππ .91,98
30494578 ,57
πππ = β4906931726
91,98= β53347811,77 πΎπππ = β533,48 π‘π
π = π2 .ππ
πΌ=
β53347811,77kg.cm .91,98 ππ
30494578,57 cm4 = -160,911 kg/cm2
y2=102.01cm
t2=44.81cm
t1=49.15cm
a1=48.00cm
a2=121.50cm
Ta
cb
Tb
Y1 =92.99 cm
fb
fb0
0
Ta
c
Maka, nilai q :
Mbs1 = 375,69 tm
Mbs1 = 1
8 . π1 . πΏ2
375,69 tm = 1
8 . π1 . 352 q1 = 2,453 ton/m = 2453,5 kg/m
Mbs2 = 533,48 tm
Mbs2 = 1
8 . π2 . πΏ2
533,48 tm = 1
8 . π2 . 352 q2 = 3,483 ton/m = 3483,9 kg/m
Q balok = luas balok x berat jenis beton
= 5900 cm2 x 0,0024 kg/cm3
= 14,16 kg/cm
= 1416 kg/m
Catatan :
Tanda negatif (-) berarti tekan, dan tanda positif (+) berarti tarik.
Kontrol :
q1 > Q balok
2453,5 kg/m > 1416 kg/m.........................................................( OK !!! )
q2 > Q balok
3483,9 kg/m > 1464 kg/m.........................................................( OK !!! )
2. Pada kondisi akhir
ππ = βππ
π΄ππ‘Β±
ππ.πππ‘.π¦
πΌπ‘Β±
πππ .π¦
πΌπ‘
a. Pada serat atas
πππβππ =ππ
π΄ππ‘+
ππ.πππ‘ .π¦π
πΌπ‘β
ππ‘.π¦π
πΌπ‘
β188 =316742 ,08
6053,93+
316742 ,08.71,98.108 ,02
30494578 ,57β
ππ‘.108 ,02
30494578,57
ππ‘ = β 9791217684
108,02= β90642637,33 πΎπππ = β906,43 π‘π
π = π1 .ππ
πΌ=
β90642637,33 kg.cm .108,02 ππ
30494578,57 cm4 = -321,081 kg/cm2
b. Pada serat bawah
πππβππ = βππ
π΄ππ‘β
ππ.πππ‘ .π¦π
πΌπ‘+
ππ‘.π¦π
πΌπ‘
0 = β316742 ,08
6053,93β
316742 ,08.71,98.91,98
30494578,57+
ππ‘.91,98
30494578 ,57
ππ‘ = 501582070 ,6
91,98= β5453164,49 πΎπππ = β545,316 π‘π
π = π2 .ππ
πΌ=
β5453164 ,49 kg.cm .91,98 ππ
30494578,57 cm4 = -164,482 kg/cm2
Maka, nilai q :
Mt1 = 906,43 tm
Mt1 = 1
8 . π1 . πΏ2
906,43 tm = 1
8 . π1 . 352 q1 = 5,9195 ton = 5919,5 kg/m
Mt2 = 545,316 tm
Mt2 = 1
8 . π2 . πΏ2
545,316 tm = 1
8 . π2 . 352 q2 = 3,561 ton = 3561,2 kg/m
Mt = 350 tm
Mt = 1
8 . πβ² . πΏ2
350 tm = 1
8 . πβ² . 352 qβ = 2,285 ton = 2285,7 kg/m
Dipilih q terkecil = 2285,7 kg/m
Q balok = luas balok x berat jenis beton
= 5900 cm2 x 0,0024 kg/cm3
= 14,16 kg/cm
= 1416 kg/m
q terkecil > Q balok
2285,7 kg/m > 1416 kg/m....................................................( OK !!! )
y2=102.01cm
t2=44.81cm
t1=49.15cm
a1=48.00cm
a2=121.50cm
Ta
cb
Tb
Y1 =92.99 cm
140,35 kg/cm2
-160,35 kg/cm2
-278,95 kg/cm2
152,74 kg/cm2
-138,6 kg/cm2
-7,61 kg/cm2
+ =
Daerah aman kabel
π1 =ππππ
ππ=
πππ
ππ
π = 0,59π₯2400 = 1416 πΎπ/π
πππ =1
8ππΏ2 =
1
81416.352 = 216825 πΎππ
π1 =πππ
ππ=
216825
316742 ,08= 0,68π = 68ππ
π‘1 =πΌπ2
π¦1 πΌπ2 =
πΌπ‘
π΄ππ‘=
30494578,57
6053,93= 5037,15
=5037 ,15
106,14= 47,46 ππ
π‘2 =πΌπ2
π¦2=
5037 ,15
92,86= 54,24ππ
π2 =ππ‘
π=
350000
269230,77= 129,9 ππ
y2=102.01cm
t2=44.81cm
t1=49.15cm
a1=48.00cm
a2=121.50cm
Ta
cb
Tb
Y1 =92.99 cm
Ta
Tb
Ta
Tb
y1=86.56cm
y2=108,43cm
π =π
7
8π.β
ππ‘ = 350π‘π
ππ‘ =1
8. π. πΏ2
350000 =1
8. π. 352 π = 2285,71 πΎπ/π
π =1
2. π. πΏ
=1
2. 2285,71.35 = 40000 πΎπ
π =40000
7
820.200
= 11,43 πΎπ/ππ2
π = βπ 2 + (ππ
2)
2
βππ
2 ππ =
π
π΄ππ‘=
269230 ,77
6053,93= 44,47πΎπ/ππ2
π = β(11,43)2 + (44,47
2)
2
β44,47
2
π = β130,645 + 494,395 β 22,235 = 24,55ππ
ππ2 = 2,455πππ
πππππ = 0,43β47 = 2,95 πππ
Cek :
= 2,455 πππ > ijin = 2,95πππ
Jadi diperlukan tulangan geser