This document summarizes the planning and design calculations for a pre-stressed concrete beam with the following parameters: 1. The required bending moment (Mt) is 350 ton-meters. The concrete compressive strength (f'c) is 47 MPa. 2. The initial dimensions of the beam are calculated as 200 cm height (h) and 4339.6 cm^2 cross-sectional area (Ab). 3. The final design meets the required bending moment of 350 ton-meters with a uniform prestress force (q) of 2285.71 kg/m distributed over the beam length. Stresses in the concrete are calculated to remain below the allowable limits.

1. TUGAS I
MERENCANAKAN BALOK PRATEGANG
Mt = 350 tm
f’c = 47 MPa
Perencanaan Dimensi Balok
Mt = T.z
= T . 0,65 h
T =
𝑀𝑡
0,65ℎ
Mencari nilai h
ℎ = 𝑘 √Mt ; Diambil nilai k = 10
ℎ = 10 √350
= 187,08 cm
Diambil 200 cm
𝑇 =
𝑀𝑡
0,65 ℎ
=
350.105
0,65 . 200
= 269230,77 𝑘𝑔
Perhitungan penampang Ab (dengan tafsiran σbr = 0,40 σb akhir)
σ̅′
Akhir =0.33 bk = 0.33 . 470 = 155,1 kg/cm2
Abhitung =
T
σbr
........................=
269230 ,77
0,40 .155 ,1
= 4339,6 cm2
Dicoba dengan penampang seperti gambar dibawah :

2. Abrencana = (60 x 20 ) + (155 x 20 ) + ( 40 x 40 )
= 5900 cm2
Abhitung ≤ Abrencana………….( OK)
Menghitung Ta
𝑇𝑎 =
𝑇
1−∆
Post Tensioning ∆ = 15%
𝑇𝑎 =
269230 ,77
1−0,15
= 316742.08 𝐾𝑔
𝑓𝑎 =
𝑇𝑎
𝐴𝑎
𝑓𝑎 = 0,85 . 𝑓𝑎𝑢 ; 𝑓𝑎𝑢 = 18000 𝐾𝑔/𝑐𝑚2
𝐴𝑎 =
𝑇
𝑓𝑎
=
316742 .08
0,85 . 18000
= 21,99 𝑐𝑚2
= 22 𝑐𝑚2
80.00
20.00
135.00
40.00
45.00

3. Menghitung titik berat penampang
a. Menghitung titik berat
𝑦1 =
𝑙𝑢𝑎𝑠 1 .
1
2
ℎ1+𝑙𝑢𝑎𝑠 2 . (
1
2
ℎ2+ℎ1)+ 𝑙𝑢𝑎𝑠 3 (
1
2
ℎ3+ℎ1+ℎ2)
𝑙𝑢𝑎𝑠 𝑡𝑜𝑡𝑎𝑙
𝑦1 =
1200 .
1
2
20+3100 . (
1
2
155+20)+ 1600 (
1
2
40+20+155)
5900
= 106,14 cm
𝑦2 = ℎ − 𝑦1
= 200 – 106,14
= 93,86 cm
𝑒𝑎 = 93,86 − 20 = 73,86 𝑐𝑚
𝑒𝑡 =
𝑛 .𝐴.𝑒𝑎
𝐴𝑏+𝑛.𝐴
𝑛 =
2.105
32221,58
= 6,207 ≈ 7
=
7 .21,99 .73,86
5900+7 .21,99
= 1,88 𝑐𝑚
𝑒𝑎𝑡 = 𝑒𝑎 − 𝑒𝑡
= 73,86 – 1,88
y1
y2
et
eat
ea
t2
t1
ta
ya
yb

4. = 71,98 cm
𝑦𝑎 = 𝑦1 + 𝑒𝑡
= 106,14 + 1,88
= 108,02 cm
𝑦𝑏 = ℎ − 𝑦𝑎
= 200 – 108,02
= 91,98 cm
b. Menghitung momen inersia tampang :
Bagian 1
𝐼 =
1
12
𝑏1 .ℎ13
+ 𝑏1 .ℎ1(𝑦1 −
1
2
ℎ1)2
𝐼 =
1
12
60 .203
+ 60 .20(106,14−
1
2
20)2
= 11131479,52 cm4
Bagian 2
𝐼 =
1
12
𝑏2 .ℎ23
+ 𝑏2 .ℎ2(𝑦1 − (
1
2
ℎ2 + ℎ1))2
𝐼 =
1
12
20 .1553
+ 20 .155(106,14− (
1
2
155 + 20))2
= 6437872,09 cm4
Bagian 3
𝐼 =
1
12
𝑏3 .ℎ33
+ 𝑏3 .ℎ3(𝑦2 −
1
2
ℎ3)2
𝐼 =
1
12
40 .403
+ 40 .40(106,14−
1
2
. 40)2
= 12085492,69 cm4
𝐼𝑡𝑜𝑡 = Ibagian 1 + Ibagian 2 + Ibagian 3 + Aa. n. eat2
𝐼𝑡𝑜𝑡 = 11131479,52 + 6437872,09 + 12085492,69 + 21,99 .7 .73,862
= 30494578,57 cm4

5. 𝐴𝑏𝑡 = 𝐴𝑏 + 𝑛 . 𝐴𝑎
= 5900 + 7. 21,99
= 6053,93 cm2
a. Menghitung Desain Akhir
Asumsi : desain akhir dengan tidak diijinkan tegangan tarik
1. Pada kondisi awal
𝑓𝑏 =
𝑇
𝐴𝑏𝑡
±
𝑇𝑎.𝑒𝑎𝑡.𝑦1
𝐼𝑡
±
𝑀𝑏𝑠
𝐼𝑡
a. Pada serat atas
𝑓𝑏 =
𝑇𝑎
𝐴𝑏𝑡
+
𝑇𝑎.𝑒𝑎𝑡.𝑌𝑎
𝐼𝑡
−
𝑀𝑏𝑠.𝑦𝑎
𝐼𝑡
0 =
316742 ,08
6053 ,93
+
316742 ,08.71,98.108 ,02
30494578,57
−
𝑀𝑏𝑠.108 ,02
30494578 ,57
𝑀𝑏𝑠 =
4058236913
108,02
= 37569310,43 Kgcm = 375,69 tm
𝜎 =
𝑀1.𝑌𝑎
𝐼
=
37569310,43kg.cm .108,02 𝑐𝑚
30494578 ,57 cm4
= 133,08 kg/cm2
b. Pada serat bawah
𝑓𝑏 = −
𝑇𝑎
𝐴𝑏𝑡
−
𝑇𝑎.𝑒𝑎𝑡.𝑦𝑏
𝐼𝑡
+
𝑀𝑏𝑠.𝑦𝑏
𝐼𝑡
−282 = −
316742,08
6053 ,93
−
316742,08.71,98.91,98
30494578 ,57
+
𝑀𝑏𝑠.91,98
30494578 ,57
𝑀𝑏𝑠 = −
4906931726
91,98
= −53347811,77 𝐾𝑔𝑐𝑚 = −533,48 𝑡𝑚
𝜎 =
𝑀2.𝑌𝑏
𝐼
=
−53347811,77kg.cm .91,98 𝑐𝑚
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

6. Maka, nilai q :
 Mbs1 = 375,69 tm
Mbs1 =
1
8
. 𝑞1 . 𝐿2
375,69 tm =
1
8
. 𝑞1 .352 q1 = 2,453 ton/m = 2453,5 kg/m
 Mbs2 = 533,48 tm
Mbs2 =
1
8
. 𝑞2 . 𝐿2
533,48 tm =
1
8
. 𝑞2 . 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
𝑓𝑏 = −
𝑇𝑎
𝐴𝑏𝑡
±
𝑇𝑎.𝑒𝑎𝑡.𝑦
𝐼𝑡
±
𝑀𝑏𝑠.𝑦
𝐼𝑡
a. Pada serat atas
𝑓𝑎𝑘ℎ𝑖𝑟 =
𝑇𝑎
𝐴𝑏𝑡
+
𝑇𝑎.𝑒𝑎𝑡.𝑦𝑎
𝐼𝑡
−
𝑀𝑡.𝑦𝑎
𝐼𝑡
−188 =
316742 ,08
6053,93
+
316742 ,08.71,98.108 ,02
30494578 ,57
−
𝑀𝑡.108 ,02
30494578,57
𝑀𝑡 = −
9791217684
108,02
= −90642637,33 𝐾𝑔𝑐𝑚 = −906,43 𝑡𝑚
𝜎 =
𝑀1.𝑌𝑎
𝐼
=
−90642637,33 kg.cm .108,02 𝑐𝑚
30494578,57 cm4
= -321,081 kg/cm2

7. b. Pada serat bawah
𝑓𝑎𝑘ℎ𝑖𝑟 = −
𝑇𝑎
𝐴𝑏𝑡
−
𝑇𝑎.𝑒𝑎𝑡.𝑦𝑏
𝐼𝑡
+
𝑀𝑡.𝑦𝑏
𝐼𝑡
0 = −
316742 ,08
6053,93
−
316742 ,08.71,98.91,98
30494578,57
+
𝑀𝑡.91,98
30494578 ,57
𝑀𝑡 =
501582070 ,6
91,98
= −5453164,49 𝐾𝑔𝑐𝑚 = −545,316 𝑡𝑚
𝜎 =
𝑀2.𝑌𝑏
𝐼
=
−5453164 ,49 kg.cm .91,98 𝑐𝑚
30494578,57 cm4
= -164,482 kg/cm2
Maka, nilai q :
 Mt1 = 906,43 tm
Mt1 =
1
8
. 𝑞1 . 𝐿2
906,43 tm =
1
8
. 𝑞1 .352 q1 = 5,9195 ton = 5919,5 kg/m
 Mt2 = 545,316 tm
Mt2 =
1
8
. 𝑞2 . 𝐿2
545,316 tm =
1
8
. 𝑞2 .352 q2 = 3,561 ton = 3561,2 kg/m
 Mt = 350 tm
Mt =
1
8
. 𝑞′ . 𝐿2
350 tm =
1
8
. 𝑞′ .352 q’ = 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
+ =

8. Daerah aman kabel
𝑎1 =
𝑀 𝑚𝑖𝑛
𝑇𝑎
=
𝑀 𝑏𝑠
𝑇𝑎
𝑞 = 0,59𝑥2400 = 1416 𝐾𝑔/𝑚
𝑀 𝑏𝑠 =
1
8
𝑞𝐿2
=
1
8
1416.352
= 216825 𝐾𝑔𝑚
𝑎1 =
𝑀 𝑏𝑠
𝑇𝑎
=
216825
316742 ,08
= 0,68𝑚 = 68𝑐𝑚
𝑡1 =
𝐼𝑏2
𝑦1
𝐼𝑏2
=
𝐼𝑡
𝐴𝑏𝑡
=
30494578,57
6053,93
= 5037,15
=
5037 ,15
106,14
= 47,46 𝑐𝑚
𝑡2 =
𝐼𝑏2
𝑦2
=
5037 ,15
92,86
= 54,24𝑐𝑚
𝑎2 =
𝑀𝑡
𝑇
=
350000
269230,77
= 129,9 𝑐𝑚
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

9. 𝜏 =
𝑉
7
8
𝑏.ℎ
𝑀𝑡 = 350𝑡𝑚
𝑀𝑡 =
1
8
. 𝑞. 𝐿2
350000 =
1
8
. 𝑞. 352
𝑞 = 2285,71 𝐾𝑔/𝑚
𝑉 =
1
2
. 𝑞. 𝐿
=
1
2
. 2285,71.35 = 40000 𝐾𝑔
𝜏 =
40000
7
8
20.200
= 11,43 𝐾𝑔/𝑐𝑚2
𝜌 = √ 𝜏2 + (
𝜎𝑏
2
)
2
−
𝜎𝑏
2
𝜎𝑏 =
𝑇
𝐴𝑏𝑡
=
269230 ,77
6053,93
= 44,47𝐾𝑔/𝑐𝑚2
𝜌 = √(11,43)2 + (
44,47
2
)
2
−
44,47
2
𝜌 = √130,645 + 494,395 − 22,235 = 24,55
𝑘𝑔
𝑐𝑚2 = 2,455𝑀𝑝𝑎
𝜌𝑖𝑗𝑖𝑛 = 0,43√47 = 2,95 𝑀𝑝𝑎
Cek :
 = 2,455 𝑀𝑝𝑎 > ijin = 2,95𝑀𝑝𝑎
Jadi diperlukan tulangan geser