MERENCANAKAN BALOK BETON PRATEGANG DENGAN METODE BALANCING

1. TUGAS II
MERENCANAKAN BALOK PRATEGANG DENGAN METODE BALANCING
Sebuah balok sederhana dengan gambar seperti di bawah ini. Tentukan gaya prategang, P pada
bentang AB dan BC ! (fc’ = 50 MPa)
Penyelesaian :
h = (300 + ½ ei) mm
PAB = PBC
h
LWb
.8
.
2
1
=
ei
LWb
.2
.
2
2
)300.(8
.
2
1
2
1
ei
LWb

=
ei
LWb
.2
.
2
2
(Wb.L1
2
) ( 2ei) = (Wb.L2
2
) (2400 + 4 ei)
2 ei L1
2
= L2
2
(2400 + 4 ei)
2 ei (16)2
= (5)2
(2400 + 4 ei)
2 ei 256 = 25 (2400 + 4 ei)
512 ei = 60000 + 100 ei
412 ei = 60000
ei = 145,63 mm = 0,1450 m
h = (300 + ½ ei) mm
h = [300 + ½ (145,63)]
h = 372,8 mm = 0,372 m
16.00 5.00
q = 2,5 t/m
50
50
0.30
ei

2. Gaya prategang yang terjadi :
PAB =
h
LWb
.8
.
2
1
=
)372,0(8
)16)(5,2( 2
= 215,05 ton
PBC =
ei
LWb
.2
.
2
2
=
)145,0(2
)5)(5,2( 2
= 215,5 ton
Penyelesaian :
Dipakai kondisi akhir :
Kontrol nilai fb :
fb =
Ab
PAB
= 2
)3001000(
05,215
mmx
ton
= 7,168 x 10-4
ton/mm2
fb = 7,168 MPa < fb (ijin)
fb = 7,168 MPa < 27 MPa . . . . . . . . (Ok)
b = 30 cm
h
d' = 20 cm
d = 80 cm
Aa

3. fya =
Aa
PP efef 
=
Aa
PP efef 0
020
=
Aa
Pef)2,01( 
=
Aa
Pef2,1
Aa =
Mpa
Nx
1600
)1005,215(2,1 4
= 1612,87 mm2
Fc’ = 50 Mpa, maka :
β1 = 0,85 – 0,008 (fc’ – 30) dan β1 ≥ 0,65 MPa (untuk fc’ > 30 MPa)
β1 = 0,85 – 0,008 (50 – 30)
β1 = 0,69
ΣH = 0
T – Cc = 0
[ Aa fy ] – [ 0,85 fc’ a b ] = 0 ; dimana a = β1 c
[ Aa fy ] – [ 0,85 fc’ β1 c b ] = 0
c =
bfc
fyAa
.'..85,0
.
1
=
)300)(69,0)(50(85,0
)1600)(87,1612( 2
mmMpa
Mpamm
= 293,33 mm
a = β1 c = 0,69 (293,33mm) = 202,39 mm
b = 30 cm
h
d' = 20 cm
d = 80 cm
Aa
c
a
T
Cc℄
0,85 f'c0,003
eaΔea
Δea Tot
Za

4. Kontrol regangan baja :
a =
EaAa
Pef
.
=
)101,2)(87,1612(
1005,215
52
4
Mpaxmm
Nx
= 0,0063
Δa didapat dari perbandingan segitiga sebagai berikut :
cd
a


=
c
003,0
; maka Δa =
c
cd )(003,0 
=
33,293
)33,293800(003,0 
= 0,0051
Maka :
a total = a + Δa > y =
Ea
fya
=
MPax
MPa
5
101,2
1600
= 0,0076
a total = 0,0063 + 0,0051
a total = 0,011 > 0,0076 . . . . . (baja leleh) → fa = fya
Menentukan Za :
Za = d – ½ a = 800 – ½ (202,39) =698,805 mm
Menentukan momen nominal, Mn :
Mn = T.Za
Mn = (Aa.fy) Za
Mn = 1612,87 mm2
. 1600 MPa . 698,805 mm
Mn = 1.803.330.593 Nmm
Mn = 1.803,33 kNm
Menentukan momen ultimit, Mu :
Mu =  Mn
Mu = 0,8 . 1.803.330.593 Nmm
Mu = 1.442.664.474 Nmm
Mu = 1.442,66 kNm
Jadi, momen ultimit = 1.442,66 kNm

5. Prategang Parsial
0,85. f’c.β1.c.b = Ta
0,85. 50 . 0,69 . c . 300 = Aa. Fy
0,85. 50 . 0,69 . c . 300 = ( 87,1612 .68%)) . 1600
8797,5 . c = 1754802,56
c = 199,46 mm
∆𝜀𝑎
𝜀𝑐𝑢
=
ℎ𝑎−𝑐
𝑐
∆𝜀𝑎
0,003
=
800−199,46
199,46
∆𝜀𝑎 =
0,003(800−199,46)
199,46
= 0,009
Δtot = ∆𝜀𝑎 + 𝜀𝑎
𝜀𝑎 =
𝑃 𝑒𝑓
68%.1252,125.2,1𝑥105
=
215,05𝑥104
68%.1612,87.2,1𝑥105
= 0,0093
Δtot = 0,009 + 0,0093 =0,018 > 𝜀𝑦 = 0,0076 (leleh)
𝑀𝑛𝑝 = 𝑇𝑝. 𝑍𝑝
32%. 𝑀𝑛 = 𝐴𝑝. 𝑓𝑦𝑝. 𝑍𝑝 𝑍𝑝 = ℎ𝑝 −
𝑎
2
32%. 1.803.330.593 = 𝐴𝑝. 𝑓𝑦𝑝. (ℎ𝑝 −
𝑎
2
)
32%. 1.803.330.593 = 𝐴𝑝. 400. (940 −
137,62
2
)
577065789,8 = 𝐴𝑝. 348476
𝐴𝑝 = 1655,97𝑚𝑚
𝑀𝑛 = 𝐴𝑎. 𝑓𝑦𝑎. 𝑍𝑎 + 𝐴𝑝. 𝑓𝑦𝑝. 𝑍𝑝
𝑀𝑛 = 0,68.1612,87.1600.731,18 + 1655,97.400.871,18
= 1860135714 nmm = 1860,13 KNm
b = 30 cm
h
d' = 20 cm
d = 80 cm
Aa
c a = ß1.c
Ta
Cc
0,85 f'c0,003
eaΔea
Za
Tp
Zp
ha
hp
ep

6.  Menghitung tegangan yang terjadi
Gambar awal:
Momen inersia penampang = 1/12 b. h3
= 1/12. 300. 10003
= 2,5 x 1010
mm4
= 2,5x106
cm4
 ea = h/2 = 100 cm/2 = 50 cm
 Aa =
Mpa
Nx
1600
)1005,215(2,1 4
= 1612,87 mm2
 𝑒𝑡 =
𝑛.𝐴𝑎.𝑒𝑎
𝐴𝑏+𝑛.𝐴𝑎
; dimana n adalah perbandingan antara modulus elastisitas beton dan baja :
n =
𝐸𝑎
𝐸𝑐
=
𝐸𝑎
4700√𝑓′𝑐
=
200000
4700√50
= 6,01  6
𝑒𝑡 =
6.16,12.50
3000+6.16,12
= 1,56 cm
 eat = ea – et = 50 cm – 1,56 cm = 48,43 cm
 Ya = (y1) + et = (100/2) cm + 1,56 cm = 51,56 cm
 Yb = h – Ya = 100 cm – 51,56 cm = 48,44 cm
 Abtotal = (30 . 100) + (6. 16,12) = 3096,72 cm2
 𝑇𝑎 =
𝑇
(1−15%)
=
𝐴𝑎.𝑓𝑦𝑎+𝐴𝑝.𝑓𝑦𝑝
1−15%
=
16,12.16000+16,55.4000
1−15%
=
334120
1−15%
= 393082,35 kg
Cb
Ta
Tb
Cas
Ar
Y1
Y2
10.00
3.00
2.00
eat
et
ea
Yb
Ya

7. Tegangan yang Terjadi
Berdasarkan diagram tegangan-regangan awal, untuk kondisi awal :
 Pada serat atas :
𝑓𝑏 𝑎𝑤𝑎𝑙 =
𝑇𝑎
𝐴𝑏 𝑡𝑜𝑡
+
𝑇𝑎.𝑒𝑎𝑡.𝑌𝑎
𝐼𝑡𝑜𝑡
−
𝑀𝑏𝑠1.𝑌𝑎
𝐼𝑡𝑜𝑡
= 0
0 =
393082,35 kg
3096,72 cm2
+
393082,35 kg . 48,43 cm . 51,56 𝑐𝑚
2,5x106 cm4
−
𝑀1 . 51,56 𝑐𝑚
2,5x106 cm4
Didapat nilai M1 = 25191703,95 kg.cm = 251,9 tm
𝜎 =
𝑀1.𝑌𝑎
𝐼
=
25191703,95 kg.cm .51,56 𝑐𝑚
2,5x106 cm4
= 519,55 kg/cm2
 Pada serat bawah :
𝑓𝑏 𝑎𝑤𝑎𝑙 = −
𝑇𝑎
𝐴𝑏 𝑡𝑜𝑡
−
𝑇𝑎.𝑒𝑎𝑡.𝑌𝑏
𝐼𝑡𝑜𝑡
+
𝑀𝑏𝑠2.𝑌𝑏
𝐼𝑡𝑜𝑡
= 0,6 𝑓′𝑐
− 300 = −
393082,35 kg
3096,72 cm2
−
393082,35 kg . 48,43 cm . 48,44 𝑐𝑚
2,5x106 cm4
+
𝑀1 . 48,44 𝑐𝑚
2,5x106 cm4
Didapat nilai M2 = -25588127,24 kg.cm
M2 = -255,88 tm
𝜎 =
𝑀2.𝑌𝑏
𝐼
=
−25588127,24 kg.cm .48,44𝑐𝑚
2,5x106 cm4
= -495,79 kg/cm2
Cb
Ta
Tb
Cas
Ar
Y1
Y2
10.00
3.00
2.00
eat
et
ea
Yb
Ya
C
Ta
Tp
fb
fb
0

8. Berdasarkan diagram tegangan-regangan awal, untuk kondisi akhir :
 Pada serat atas :
𝑓𝑏 𝑎𝑘ℎ𝑖𝑟 =
𝑇
𝐴𝑏 𝑡𝑜𝑡
+
𝑇.𝑒𝑎𝑡.𝑌𝑎
𝐼𝑡𝑜𝑡
−
𝑀𝑡 . 𝑌𝑎
𝐼𝑡𝑜𝑡
= −0,4 𝑓′𝑐
−200 =
334120 kg
3096,72 cm2
+
334120kg . 48,43 cm . 51,56 𝑐𝑚
2,5x106cm4
−
𝑀1 . 51,56 𝑐𝑚
2,5x106 cm4
= −0,4 .500 𝑘𝑔/𝑐𝑚2
Didapat nilai Mt1 = -31110388,39 kg.cm = -311,10 tm
𝜎 =
𝑀1.𝑌𝑎
𝐼
=
−31110388,39 kg.cm .51,56 𝑐𝑚
2,5x106 cm4
= -641,62 kg/cm2
 Pada serat bawah :
𝑓𝑏 𝑎𝑘ℎ𝑖𝑟 = −
𝑇
𝐴𝑏 𝑡𝑜𝑡
−
𝑇.𝑒𝑎𝑡.𝑌𝑏
𝐼𝑡𝑜𝑡
+
𝑀𝑡.𝑌𝑏
𝐼𝑡𝑜𝑡
= 0
= −
334120 kg
3096,72 cm2
−
334120kg . 48,43 cm . 48,44 𝑐𝑚
2,5x106 cm4
+
𝑀2 . 48,44 𝑐𝑚
2,5x106 cm4
= 0 𝑘𝑔/𝑐𝑚2
Didapat nilai Mt2 = 21945196,61 kg.cm = 219,45 tm
𝜎 =
𝑀2.𝑌𝑏
𝐼
=
21945196,61 kg.cm .48,44 𝑐𝑚
2,5x106 cm4
= 425,21 kg/cm2
Catatan :
Tanda negatif (-) berarti tekan, dan tanda positif (+) berarti tarik.

9. 10.00
3.00
-495,79 kg/cm2 425,21 kg/cm2 -70,58 kg/cm2
+ =