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 :

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

Menghitung titik berat penampang

a. Menghitung titik berat

𝑦1 = 𝑙𝑢𝑎𝑠 1 .

1

2ℎ1+𝑙𝑢𝑎𝑠 2 . (

1

2ℎ2+ℎ1)+ 𝑙𝑢𝑎𝑠 3 (

1

2ℎ3+ℎ1+ℎ2)

𝑙𝑢𝑎𝑠 𝑡𝑜𝑡𝑎𝑙

𝑦1 = 1200 .

1

220+3100 . (

1

2155+20)+ 1600 (

1

240+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

= 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

2ℎ1)2

𝐼 =1

12 60 .203 + 60 .20(106,14 −

1

220)2

= 11131479,52 cm4

Bagian 2

𝐼 =1

12 𝑏2 .ℎ23 + 𝑏2 .ℎ2(𝑦1 − (

1

2ℎ2 + ℎ1))2

𝐼 =1

12 20 .1553 + 20 .155(106,14 − (

1

2155 + 20))2

= 6437872,09 cm4

Bagian 3

𝐼 =1

12 𝑏3 .ℎ33 + 𝑏3 .ℎ3(𝑦2 −

1

2ℎ3)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

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

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

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

+ =

Daerah aman kabel

𝑎1 =𝑀𝑚𝑖𝑛

𝑇𝑎=

𝑀𝑏𝑠

𝑇𝑎

𝑞 = 0,59𝑥2400 = 1416 𝐾𝑔/𝑚

𝑀𝑏𝑠 =1

8𝑞𝐿2 =

1

81416.352 = 216825 𝐾𝑔𝑚

𝑎1 =𝑀𝑏𝑠

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

𝜏 =𝑉

7

8𝑏.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

820.200

= 11,43 𝐾𝑔/𝑐𝑚2

𝜌 = √𝜏 2 + (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 = 24,55𝑘𝑔

𝑐𝑚2 = 2,455𝑀𝑝𝑎

𝜌𝑖𝑗𝑖𝑛 = 0,43√47 = 2,95 𝑀𝑝𝑎

Cek :

= 2,455 𝑀𝑝𝑎 > ijin = 2,95𝑀𝑝𝑎

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