tugas beton nando
TRANSCRIPT
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Soal :
A. Pembebanan
a. Pembebanan plat Tipe a, c
Bentang arah y = 5 m = mm
1. Data Perencanaan
Tebal Pelat = m = mmfc' = Mpa = Nmm
fy = Mpa = Nmm
Beban Hidup Pelat = = kN/m²
Berat sendiri beton = kg/m³ = kN/m³
2. Pembebanan ( diambil lebar pias plat yang dihitung sebesar : 1 m )
Beban mati :
Berat sendiri = x 1 x =
= kN/m
Beban guna lantai (Hidup) :
Berat beban guna atap = 1 x = kN/m
ql = kN/m
Beban terfaktor qu = 1.2 qd + 1.6 ql = x + x
qu = kN/m
2,501,602,88
2,500
2,500
PERENCANAAN PELAT LANTAI
350
0,12
25
250
0,12 24,00 2,880
2400
120
25
350
2,5
2,880qd
5000
2,50
1,20
7,456
PENYELESAIAN :
24
6 m 5 m 6 m 5 m
6 m
5 m
6 m
a 12
250
13
300
12
250
13
300
13 12 13 12
12 13 12 13
b c d
e f g h
lk ji
300 300
300 300
250 250
250250
y
x
A B C D E
I
H
G
F
Kg/m2
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b. Pembebanan plat Tipe i, k, f, h
Bentang arah y = 6 m = mm
1. Data Perencanaan
Tebal Pelat = m = mm
fc' = Mpa = Nmm
fy = Mpa = Nmm
Beban Hidup Pelat = = kN/m²
Berat sendiri beton = Kg/m³ = kN/m³
2. Pembebanan ( diambil lebar pias plat yang dihitung sebesar : 1 m )
Beban mati :
Berat sendiri = x 1 x =
= kN/m
Beban guna lantai (Hidup) :
Berat beban guna atap = 1 x = kN/m
ql = kN/m
Beban terfaktor qu = 1.2 qd + 1.6 ql = x + x
qu = kN/m
c. Pembebanan plat Tipe e, g, j, l
Bentang arah y = 6 m = mm
1. Data Perencanaan
Tebal Pelat = m = mm
fc' = Mpa = Nmm
fy = Mpa = Nmm
Beban Hidup Pelat = = kN/m²
Berat sendiri beton = Kg/m³ = kN/m³
2. Pembebanan ( diambil lebar pias plat yang dihitung sebesar : 1 m )Beban mati :
Berat sendiri = x 1 x =
= kN/m
Beban guna lantai (Hidup) :
Berat beban guna atap = 1 x = kN/m
ql = kN/m
Beban terfaktor qu = 1.2 qd + 1.6 ql = x + x
qu = kN/m
d. Pembebanan plat Tipe b,dBentang arah y = 5 m = mm
1. Data Perencanaan
Tebal Pelat = m = mm
fc' = Mpa = Nmm
fy = Mpa = Nmm
Beban Hidup Pelat = = kN/m²
Berat sendiri beton = Kg/m³ = kN/m³
3,000
1,20 3,12 1,60 3,00
8,544
2,880
qd 2,880
2,50 2,500
2,500
1,20 2,88 1,60 2,50
7,456
6000
5000
6000
0,13 130
25 25
350 350
300 Kg/m² 3,0
2400 24
0,13 24,00 3,120
qd 3,120
3,00 3,000
350 350
300 Kg/m² 3,0
2400 24
0,12 120
25 25
350 350
250 Kg/m² 2,5
2400 24
0,12 24,00
0,13 130
25 25
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2. Pembebanan ( diambil lebar pias plat yang dihitung sebesar : 1 m )
Beban mati :
Berat sendiri = x 1 x =
= kN/m
Beban guna lantai (Hidup) :
Berat beban guna atap = 1 x = kN/m
ql = kN/m
Beban terfaktor qu = 1.2 qd + 1.6 ql = x + x
qu = kN/m
B. Perhitungan Momen
Dengan menggunakan koefisien momen, didapat :1 Plat i = Plat k
Diketahui : Panjang bentang = m
qu = kN/m
● Momen Tumpuan Kiri
1 2 1 2
24 24
● Momen Lapangan
1 2 1 2
14 14
● Momen Tumpuan Kanan
1 2 1 210 10
2 Plat j = Plat l
Diketahui : Panjang bentang = m
qu = kN/m
● Momen Tumpuan Kiri
1 2 1 2
24 24
● Momen Lapangan
1 2 1 2
14 14
● Momen Tumpuan Kanan
1 2 1 2
10 10
3 Plat e = Plat g
Diketahui : Panjang bentang = m
qu = kN/m
● Momen Tumpuan Kiri
1 2 1 2
11 11
● Momen Lapangan
1 2 1 216 16
● Momen Tumpuan Kanan
= x
0,13 24,00 3,120
qd 3,120
3,00
qu x L
kNm= x 7,456 x 6Mu = x qu x L
x 8,544 x 6
= 26,8416
= 11,184
Mu = x qu x L = x 7,456 x 6
x7,456 6
3,000
3,000
1,20 3,12 1,60 3,00
8,544
= 12,816 kNm
Mu = x qu x L = x 8,544 x 6 = 21,97029 kNm
6
7,456
6,00
8,544
Mu = x qu x L =
kNm
= 19,17257 kNm
Mu = x
x 8,544 x
6 = 30,7584 kNm
6,00
8,544
Mu = x qu x L = x 8,544 x 6 = 27,96218 kNm
Mu = x qu x L = x 8,544 x
6 = 19,224 kNmMu = x qu x L =
I H G F
1/24
1/14
1/10
1/16 1/14
1/241/11 1/101/11
6 m 5 m6 m
http://newkidjoy.blogspot.com/2011/03/perencanaan-struktur-beton-pelat-lantai.htmlhttp://newkidjoy.blogspot.com/2011/03/perencanaan-struktur-beton-pelat-lantai.html
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1 2 1 2
11 11
4 Plat f = Plat h
Diketahui : Panjang bentang = m
qu = kN/m
● Momen Tumpuan Kiri
1 2 1 2
11 11
● Momen Lapangan
1 2 1 2
16 16
● Momen Tumpuan Kanan
1 2 1 2
11 11
5 Plat a = Plat c
Diketahui : Panjang bentang = m
qu = kN/m● Momen Tumpuan Kiri
1 2 1 2
10 10
● Momen Lapangan
1 2 1 2
14 14
● Momen Tumpuan Kanan
1 2 1 2
24 24
6 Plat b = Plat dDiketahui : Panjang bentang = m
qu = kN/m
● Momen Tumpuan Kiri
1 2 1 2
10 10
● Momen Lapangan
1 2 1 2
14 14
● Momen Tumpuan Kanan
1 2 1 2
24 24
6,00
7,456
Mu = x qu x L = x
Mu = x qu x L = x 8,544 x 6 = 27,96218 kNm
7,456 x 6 = 24,40145 kNm
Mu = x qu x L = x 7,456 x 6 = 16,776 kNm
x 7,456 x
6 = 24,40145 kNm
5,00
7,456
Mu = x qu x L = x 7,456 x 5 = 18,64 kNm
Mu = x qu x L = x 7,456 x
5,00
8,544
Mu = x qu x L = x
5 = 13,31429 kNm
Mu = x qu x L = x 7,456 x 5 = 7,766667 kNm
Mu = x qu x L =
8,544 x 5 = 21,36 kNm
Mu = x qu x L = x 8,544 x 5 = 15,25714 kNm
5 = 8,9 kNmMu = x qu x L = x 8,544 x
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Berikut rekapan momen dapat diperhatikan pada tabel berikut :
Plat a
Plat b
Plat c
Plat d
Plat e
Plat f
Plat g
Plat h
Plat i
Plat j
Plat k
Plat l
C. Penulangan Plat1. Penulangan Plat a dan c
a) Tumpuan Kiri
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpaβ1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 94 ²
fc x 25
1 2 m Rnm
=0,85 0,85
0,85=
fy
600
600
0,0244
= 0,0326
0,75
0,75 0,0326
25
350x
600
600
350
x
25
350
0,85
12
120 20 12 94
Rn =Mn
= =1000b . d²
18,640 18,64
Mn =Mu
ɸ=
18,64
0,8
ρ = 1 -fy
=m0,85
fy=
0,85
350
Tumpuan Kiri Lapangan Tumpuan Kanan
MOMEN (kNm)
19,173 26,842
21,970 30,758
19,173
18,640
21,360
18,640
21,360
27,962
24,401
27,962
24,401
11,184
ρ b
fy
1,4=
1,4
350=
26,842
21,970 30,758
PLAT
12,816
11,184
12,816
13,314 7,767
15,257 8,900
13,314 7,767
15,257 8,900
19,224 27,962
16,776 24,401
19,224 27,962
16,776 24,401
0,004
10 ⁶
10 ⁶23300000,000 Nmm
2,636939792 N/mm²23300000,00
120
20
16,47059=
=
M terbesar 30,75821,97027,962
1
ρ
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2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi tumpuan kiri
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
=mm
2
> As bagiAs ada > As perlu OK
→ dipakai tulangan 8 - mm200
200
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
200
251,327
758,627 151,725
1000 240120
8
0,002
240
1/4 . π . Ф2 .b
= ρ1
16,47059-
As ada x x 12 x1000
120
10000,008071 758,6273
120
= = 942,9 mm
116,47059 2,63694
1350
0,008071
0,008071dipakai
0,004 0,0080705 0,0244
94
120
As bagi
0,25 3,142 8 1000
240
209,44
120 600
450
209,440
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b) Lapangan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 94 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perluOK
Maka, Dipakai tulangan Ø 12 -
530,5197
200
As ada
=0,85 0,85 25
x600
350 600 350
0,75
0,75 0,0326
0,0244
1,4
16,47059 1,883528
= 0,0326
=1,4
=
350
0,005644
0,004 0,0056438 0,0244
dipakai 0,005644
0,005644 1000 94
120
20
12
0,004fy 350
13,314 13,314286 10 ⁶
Mn =Mu
=13,314 10 ⁶
= 16642857,143
120 20 12 94
25 0,85
350
ρ b =0,85
x600
fy 600
m =fy
=350
= 16,470590,85 0,85
Nmmɸ 0,8
Rn =Mn
=16642857,14
= 1,883528423 N/mm²b . d² 1000
12 x1000
= 565,7 mm200
ρ = 1 -fy
ρ =1
1 1 -16,47059
200
ρb
= x x
1
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Menentukan tulangan bagi Lapangan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
c) Tumpuan Kanan
Tebal pelat (h) = mm
Tebal penutup (p) = mmDitentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
20 100
450
209,440 200
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
200
251,327
200
8
530,520 106,104
0,002 1000 120 240
240
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
240
209,44
7,767 7,7666667 10 ⁶
0,0244
1,4=
1,4= 0,004
fy 350
0,85
ρ b =0,85
x600
fy 600
=0,85 25
x600
94
25 0,85
350
Mn =Mu
=7,767 10 ⁶
= 9708333,333 Nmmɸ 0,8
= 0,0326350 600 350
0,75 ρb
0,75 0,0326
120
20
12
120 20 12
1
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x 94 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi tumpuan kanan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
0,002 1000 120 240
240
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
240
209,44
20 100
450
209,440 200
8
303,137 60,627
350
1000
16,47059 1,098725
m =fy
=350
= 16,470590,85 0,85
Rn =Mn
=9708333,33
= 1,098724913 N/mm²b . d²
200
As ada = x x 12 x1000
=
0,003225
0,004 0,0032249 0,0244
dipakai 0,003225
0,003225 1000 94 303,1367
ρ = 1 -fy
ρ =1
1
565,7 mm200
200
1 -16,47059
0,25
1/4 . π .Ф2 .b
s
3,142 8 1000
200
251,327
200
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2. Penulangan Plat b dan d
a) Tumpuan Kiri
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
= ρ min =
Mu = = x Nmm
x
x 104 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
As ada x x 12 x1000
= 942,9 mm120
120
1 -16,47059 2,468565
16,47059 350
0,007519
0,004 0,0075186 0,0244
dipakai 0,007519
0,007519 1000 104 781,9319
120
10 ⁶
Mn =Mu
=21,36 10 ⁶
= 26700000,000 Nmmɸ 0,8
Rn =Mn
=26700000,00
= 2,468565089 N/mm²b . d² 1000
kNm
x600
fy 600
=0,85 0,85 25
x600
= 0,0326350 600 350
0,75
0,75 0,0326
ρb
130
20
12
130 20 12 104
25 0,85
350
ρ b =0,85
0,02441,4
=1,4
= 0,004fy 350
21,360 21,36
m =fy
=350
=
= 16,470590,85 0,85
ρ = 1 -fy
ρ =1
1
1
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Menentukan tulangan bagi tumpuan kiri
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
b) Lapangan
Tebal pelat (h) = mm
Tebal penutup (p) = mmDitentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 104 ²
fc x 25
Nmmɸ 0,8
Rn =Mn
=19071428,57
= 1,763260778 N/mm²
b . d² 1000
m =fy
=350
= 16,470590,85
0,75 ρb
0,75 0,03260,0244
1,4= = 0,004
fy
15,257 15,257143 10 ⁶
Mn =Mu
=15,257 10 ⁶
= 19071428,571
=0,85
x600
fy 600
=0,85 0,85 25
x600
= 0,0326350 600 350
193,329 180
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
180
279,253
180
130
20
12
130 20 12
8
781,932 156,386
0,002 1000 130 260
260
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
260
193,33
130 650
450
1,4
350
104
25 0,85
350
ρ b
0,85
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi Lapangan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
260
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
260
193,33
20 100
450
193,329 180
1/4 . π .Ф2 .b
s
0,25 3,142 8
0,005266 1000 104 547,6935
200
As ada = x x 12 x1000
= 565,7 mm200
ρ =1
1
- ρ = 1fy
1 -16,47059 1,763261
35016,47059
0,005266
0,004 0,0052663 0,0244
dipakai 0,005266
200
8
547,693 109,539
0,002 1000 130 260
1000
180
279,253
180
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c) Tumpuan Kanan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 104 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perluOK
Maka, Dipakai tulangan Ø 12 -
1,028569
16,47059 350
0,003014
0,004 0,0030136 0,0244
dipakai 0,003014
0,003014 1000 104 313,4099
200
As ada = x x 12 x1000
= 565,7 mm200
Rn =Mn
=11125000,00
= 1,028568787 N/mm²b . d² 1000
m =fy
=350
= 16,470590,85 0,85
0,0326350 600 350
0,75 ρb
0,75 0,0326
0,0244
1,4 1,4= 0,004
fy 350
8,9 10 ⁶
20
12
130 12 104
25 0,85
350
ρ b = x600
fy 600
=25
x600
130
0,85
20
=
0,85 0,85=
ɸ
8,900
Mn =Mu
=8,900 10 ⁶
= 11125000,000 Nmm0,8
ρ = -1fy
ρ =1
1 1 -16,47059
200
1
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Menentukan tulangan bagi tumpuan kanan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x 2 x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
3. Penulangan Plat e dan g
a) Tumpuan Kiri
Tebal pelat (h) = mm
Tebal penutup (p) = mmDitentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = = x Nmm
x
x 104 ²
fc x 25
= 34952727,273 Nmmɸ 0,8
Rn =Mn
=34952727,27
= 3,231576116 N/mm²
b . d² 1000
m =fy
=350
= 16,470590,85
350
0,75 ρb
0,75 0,03260,0244
1,4=
1,4= 0,004
fy 350
27,962 kNm 27,962182 10 ⁶
130
20
12
130 20 12 104
25 0,85
350
ρ b =0,85
x600
fy 600
1000 130 260
260
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
260
193,33
20 100
450
193,329 180
8
313,410 62,682
0,002
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
180
279,253
180
=0,85 0,85 25
x600
= 0,0326350 600
0,85
10 ⁶Mn =
Mu=
27,96218
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi tumpuan kiri
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
180
279,253
180
2601/4 . π . Ф
2 .b
As bagi
0,25 3,142 8 1000
260
193,33
130 650
450
193,329 180
1/4 . π .Ф2
.b
s
0,25 3,142 8
As ada = x x 12 x1000
= 1131 mm
100
8
1047,05 209,410
0,002 1000 130 260
ρ = 1 -fy
ρ =
1
1 1 -
16,47059 3,231576
16,47059 350
0,010068
0,004 0,0100678 0,0244
dipakai 0,010068
100
0,010068 1000 104 1047,052
100
1000
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b) Lapangan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 104 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perluOK
Maka, Dipakai tulangan Ø 12 - 140
0,00672
0,004 0,0067196 0,0244
dipakai 0,00672
0,00672 1000 104 698,8369
140
As ada = x x 12 x1000
= 808,2 mm140
24030000,000 Nmmɸ 0,8
Rn =Mn
=24030000,00
= 2,22170858 N/mm²b . d² 1000
m =fy
=350
= 16,470590,85 0,85
= 0,0326350 600 350
0,75 ρb
0,75 0,0326
0,0244
1,4=
1,4= 0,004
fy 350
130
20
130 20 12 104
25 0,85
350
ρ b =0,85
x600
fy 600
12
0,85=
0,85 25x
600
19,224 19,224 10 ⁶
Mn =Mu
=19,224 10 ⁶
=
ρ = 1 -fy
ρ =1
1 1 -16,47059 2,221709
16,47059 350
1
-
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Menentukan tulangan bagi Lapangan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x 2 x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
c) Tumpuan Kanan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mmTinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 104 ²
fc x 25m =
fy=
350= 16,47059
0,85 0,85
350
0,75 ρb
0,75 0,0326
0,02441,4
=1,4
= 0,004fy 350
27,962 27,962182 10 ⁶
Mn =Mu
=27,962 10 ⁶
= 34952727,273ɸ 0,8
0,25 3,142 8 1000
180
279,253
180
130
20
12
130 20 12 104
25 0,85
350
8
139,767
0,002 130 260
260
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
260
193,33
20 100
698,837
1000
450
193,329 180
1/4 . π .Ф2 .b
s
ρ b =0,85
x600
fy 600
=0,85 0,85 25
x600
= 0,0326350 600
Mn
Nmm
Rn = =34952727,27
= 3,231576116 N/mm²b . d² 1000
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi tumpuan kanan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
180
279,253
180
dipakai 0,010068
0,010068 1000 104 1047,052
100
As ada = x x 12 x1000
= 1131 mm100
ρ = 1 -fy
ρ =
1
1 1 -
16,47059
16,47059 350
3,231576
0,010068
0,004 0,0100678 0,0244
100
8
1047,05 209,410
0,002 1000 130 260
2601/4 . π . Ф
2 .b
As bagi
0,25 3,142 8 1000
260
193,33
3,142
20 100
450
193,329 180
1/4 . π .Ф2
.b
s
0,25 8 1000
-
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4. Penulangan Plat f dan h
a) Tumpuan Kiri
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = = x Nmm
x
x 94 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perluOK
Maka, Dipakai tulangan Ø 12 -
As ada = x x 12 x1000
= 1131 mm100
100
m =fy
=350
= 16,470590,85
ρ = 1 -fy
ρ =1
1 1 -16,47059
16,47059 350
24,401 kNm 24,401455 10 ⁶
Mn =Mu
=24,40145 10 ⁶
= 30501818,182 Nmmɸ 0,8
Rn =Mn
=30501818,18
= 3,451993909 N/mm²b . d² 1000
= 0,0326350 600 350
0,75 ρb
0,75 0,0326
0,0244
1,4 = 1,4 = 0,004fy 350
120
20
12
120 20 12 94
25 0,85
350
ρ b =0,85
x600
fy 600
0,85
25=
0,85 0,85x
600
3,451994
0,010828
0,004 0,0108285 0,0244
dipakai 0,010828
0,010828 1000 94 1017,877
100
1
-
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Menentukan tulangan bagi tumpuan kiri
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x 2 x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
b) Lapangan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mmTinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 94 ²
fc x 25
600
450
209,440 200
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
200
251,327
200
8
1017,88 203,575
0,002 1000 120 240
240
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
240
209,44
120
120
20
12
120 20 12 94
25 0,85
600= 0,0326
350 600 350
0,75 ρb
0,75 0,0326
350
ρ b =0,85
x600
fy 600
=0,85 0,85 25
x
10 ⁶
Mn =Mu
=19,224 10 ⁶
= 24030000,000
0,02441,4
=1,4
= 0,004fy 350
19,224 19,224
m =fy
=350
= 16,470590,85 0,85
Nmmɸ 0,8
Rn =Mn
=24030000,00
= 2,71955636 N/mm²b . d² 1000
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi Lapangan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
3,142 8 1000
240
209,44
20 100
450
209,440
1/4 . π .Ф2
.b
s
0,25 3,142 8 1000
200
251,327
8
784,284 156,857
0,002 1000 120
16,47059 2,719556
16,47059 350
0,008343
0,004 0,0083434 0,0244
dipakai 0,008343
ρ = 1 -fy
ρ =
1
1 1 -
140
0,008343 1000 94 784,2838
140
As ada = x x 12 x1000
= 808,2 mm140
240
2401/4 . π . Ф
2 .b
As bagi
0,25
200
200
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c) Tumpuan Kanan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 94 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perluOK
Maka, Dipakai tulangan Ø 12 -
As ada = x x 12 x1000
= 1616 mm70
70
ρ =1
1 1 -16,47059 3,955718
16,47059 350
0,012612
0,004 0,012612 0,0244
dipakai 0,012612
= 3,955718342 N/mm²b . d² 1000
m =fy
=350
= 16,470590,85 0,85
ρ = 1 -fy
0,004fy 350
27,962 27,962182 10 ⁶
Mn =Mu
=27,962 10 ⁶
= 34952727,273 Nmmɸ 0,8
25 0,85
350
ρ b =0,85
x600
fy 600
=0,85 0,85 25
x600
=350 600 350
120
20
12
120 20 12 94
0,0326
0,75 ρb
0,75 0,0326
0,0244
1,4=
1,4=
Rn =Mn
=34952727,27
0,012612 1000 94 1185,526
70
1
-
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Menentukan tulangan bagi tumpuan kanan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x 2 x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
5. Penulangan Plat i dan k
a) Tumpuan Kiri
Tebal pelat (h) = mm
Tebal penutup (p) = mmDitentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = = x Nmm
x
x 94 ²
fc x 25
b . d² 1000
m =fy
=350
= 16,470590,85
0,75
0,75 0,03260,0244
1,4=
1,4= 0,004
fy 350
11,184 kNm 11,184 10 ⁶
Mn =Mu
=11,184 10 ⁶
= 13980000,000ɸ 0,8
0,85
350
ρ b =0,85
x600
fy 600
=0,85 0,85 25
x600
= 0,0326350 600 350
20 100
450
209,440 200
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
200
251,327
200
120
20
12
8
1185,53 237,105
0,002 1000 120 240
240
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
240
209,44
ρb
120 20 12 94
25
Nmm
Rn =Mn
=13980000,00
= 1,582163875 N/mm²
0,85
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi tumpuan kiri
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
0,25 8 1000
200
251,327
200
442,043 88,409
0,002 1000 120 240
2401/4 . π . Ф
2 .b
As bagi
0,25 3,142 8 1000
240
209,44
120 600
dipakai 0,004703
0,004703 1000 94 442,0431
150
As ada = x x 12 x1000
= 754,3 mm150
ρ = 1 -fy
ρ =
1
1 1 -
16,47059
35016,47059
1,582164
0,004703
0,004 0,0047026 0,0244
150
8
3,142
450
209,440 200
1/4 . π .Ф2
.b
s
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b) Lapangan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 94 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perluOK
Maka, Dipakai tulangan Ø 12 -
As ada = x x 12 x1000
= 808,2 mm140
140
1 -16,47059 2,712281
16,47059 350
0,008319
0,004 0,0083194 0,0244
dipakai 0,008319
0,008319 1000 94 782,0191
140
350
19,173 19,172571 10 ⁶
Mn =Mu
=19,173 10 ⁶
= 23965714,286 Nmmɸ 0,8
Rn =Mn
=23965714,29
= 2,712280929 N/mm²b . d² 1000
120
12
120 20 12 94
25 0,85
350
ρ b =0,85
x600
fy 600
20
=0,85 0,85 25
x600
=350 600 350
0,75 ρb
0,0326
0,75 0,0326
0,0244
1,4=
1,4= 0,004
fy
m =fy
=350
= 16,470590,85 0,85
ρ = 1 -fy
ρ =1
1
1
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Menentukan tulangan bagi Lapangan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x 2 x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
c) Tumpuan Kanan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mmTinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 94 ²
fc x 25
33552000,00= 3,7971933 N/mm²
b . d² 1000
m =fy
=350
= 16,470590,85
350
0,75 ρb
0,75 0,0326
0,02441,4
=1,4
= 0,004fy 350
26,842 26,8416 10 ⁶
Mn =Mu
=26,842 10 ⁶
= 33552000,000ɸ 0,8
120
20
12
120 20 12 94
25 0,85
350
ρ b =0,85
x600
fy 600
8
782,019 156,404
0,002 1000 120 240
240
1/4 . π . Ф2 .b
As bagi
0,25 3,142 8 1000
240
20
209,44
100
450
209,440 180
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
180
279,253
180
=0,85 0,85 25
x600
= 0,0326350 600
0,85
Nmm
Rn =Mn
=
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi tumpuan kanan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
s
0,25 3,142 8 1000
200
251,327
200
8
1132,10 226,421
0,002 1000 120 240
2401/4 . π . Ф
2 .b
As bagi
0,25 3,142 8 1000
240
209,44
20 100
16,47059 350
0,012044
0,0120436 0,0244
dipakai 0,012044
0,012044 1000 94 1132,103
70
As ada = x x 12 x1000
= 1616 mm70
ρ = 1 -fy
ρ =
1
1 1 -
16,47059 3,797193
0,004
70
450
209,440 200
1/4 . π .Ф2
.b
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6. Penulangan Plat j dan i
a) Tumpuan Kiri
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = = x Nmm
x
x 104 ²
fc x 25
1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perluOK
Maka, Dipakai tulangan Ø 12 -
150
As ada = x x 12 x1000
= 754,3 mm150
Nmm0,8
16020000,001,481139053 N/mm²
1000
350= 16,470590,85 0,85
ρ = 1 -fy
ρ =1
1 1 -16,47059 1,481139
16,47059 350
600
fy
=0,85 0,85 25
x600
= 0,0326350 600 350
0,75 ρb
0,0326
0,0244
130
20
12
130 20 12 104
25 0,85
600
350
ρ b =0,85
x
1,4 1,4fy
0,75
= = 0,004350
Rn =Mn
= =
ɸ
12,816 kNm 12,816 10 ⁶
Mn =Mu
=12,816 10 ⁶
= 16020000,000
b . d²
m =fy
=
0,004391
0,004 0,0043906 0,0244
dipakai 0,004391
0,004391 1000 104 456,6202
150
1
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Menentukan tulangan bagi tumpuan kiri
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x 2 x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
b) Lapangan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mmTinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 104 ²
fc x 25
= =27462857,14
= 2,53909552 N/mm²b . d² 1000
m =fy
=350
= 16,470590,85 0,85
0,85 25x
600= 0,0326
350 600 350
0,75
0,75 0,0326
0,02441,4 1,4
= 0,004fy 350
8 1000
260
193,33
650
450
193,329 180
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
180
279,253
180
8
456,620 91,324
0,002 1000 130 260
130
260
1/4 . π . Ф2 .b
As bagi
0,25 3,142
130
20
12
130 20 12 104
25 0,85
ρb
350
ρ b =0,85
x600
fy 600
=0,85
ɸ
Mn
=
Mn =Mu
=
21,970 21,970286 10 ⁶
21,970 10 ⁶= 27462857,143 Nmm
0,8
Rn
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi Lapangan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
s
0,25 3,142 8 1000
180
279,253
180
0,007749 1000 104 805,9036
140
As ada = x x 12 x1000
= 808,2 mm140
ρ =
1
1 1 -
16,47059 2,539096
16,47059 350
0,007749
0,004 0,0077491 0,0244
dipakai 0,007749
ρ = 1 -fy
140
8
805,904 161,181
0,002 1000 130 260
2601/4 . π . Ф
2 .b
As bagi
0,25 3,142 8 1000
260
193,33
20 100
450
193,329 180
1/4 . π .Ф2
.b
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c) Tumpuan Kanan
Tebal pelat (h) = mm
Tebal penutup (p) = mm
Ditentukan diameter tulangan f p = mm
Tinggi efektif d = h – p – ½ f p
= - - ½ = mm
fc' = Mpa = , untuk fc' < 30 Mpa
fy = Mpa
β1 fc'
+ fy
x x
+
ρ maks = x
= x
=
ρ min =
Mu = kNm = x Nmm
x
x 104 ²
fc x 25
Nmmɸ 0,8
Rn =Mn
=38448000,00
= 3,554733728 N/mm²b . d² 1000
m =fy
=350
= 16,470590,85 0,85
0,0244
1,4=
1,4= 0,004
fy 350
30,758 30,7584 10 ⁶
Mn =Mu
=30,758
= 38448000,000
600
fy 600
0,85 0,85 25x
600= 0,0326
350 600 350
0,75 ρb
0,75 0,0326
130
20
12
130 20 12 104
25 0,85
10 ⁶
=
350
ρ b =0,85
x
1
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1 2 m Rn
m
2 x x
=
ρ min = < ρ = < ρ maks=
ρ =
Asperlu = ρ b d = x x = mm
Diperlukan tulangan Ø 12 -
1 22 2 2
4 7
As ada > As perlu OK
Maka, Dipakai tulangan Ø 12 -
Menentukan tulangan bagi tumpuan kanan
Ditentukan diameter tulangan f p = mm
As bagi = 20% As perlu 20% x = mm2
As bagi = 0,002 . b . h = x x = mm2
→ Dipilih yang terbesar, As bagi mm2
s =
= x x2
x
= mm
s ≤ 5 h = 5 x =
s ≤ mm
→ Dipilih yang terkeci ≈ mm
As ada =
= x x2
x
= mm2
> As bagi
As ada > As perlu OK
→ dipakai tulangan 8 - mm
180
279,253
180
0,25 3,142 8 1000
260
193,33
20 100
450
193,329 180
1/4 . π .Ф2 .b
s
0,25 3,142 8 1000
70
As ada = x x 12 x1000
= 1616 mm70
70
8
1163,45 232,690
fy ρ = 1 -
0,011187
16,47059 ρ =
11 1 -
16,47059 3,554734
350
0,004 0,011187 0,0244
dipakai 0,011187
0,011187 1000 104 1163,451
0,002 1000 130 260260
1/4 . π . Ф2 .b
As bagi
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Berikut Rekapan Tulangan Pokok masing-masing plat :
Plat a Ø 12 - Ø 12 - Ø 12 -
Plat b Ø 12 - Ø 12 - Ø 12 -
Plat c Ø 12 - Ø 12 - Ø 12 -
Plat d Ø 12 - Ø 12 - Ø 12 -
Plat e Ø 12 - Ø 12 - Ø 12 -
Plat f Ø 12 - Ø 12 - Ø 12 -
Plat g Ø 12 - Ø 12 - Ø 12 -
Plat h Ø 12 - Ø 12 - Ø 12 -
Plat i Ø 12 - Ø 12 - Ø 12 -
Plat j Ø 12 - Ø 12 - Ø 12 -
Plat k Ø 12 - Ø 12 - Ø 12 -
Plat l Ø 12 - Ø 12 - Ø 12 -
Berikut Rekapan Tulangan bagi masing-masing plat :
Plat a Ø 8 - Ø 8 - Ø 8 -
Plat b Ø 8 - Ø 8 - Ø 8 -
Plat c Ø 8 - Ø 8 - Ø 8 -
Plat d Ø 8 - Ø 8 - Ø 8 -
Plat e Ø 8 - Ø 8 - Ø 8 -
Plat f Ø 8 - Ø 8 - Ø 8 -
Plat g Ø 8 - Ø 8 - Ø 8 -
Plat h Ø 8 - Ø 8 - Ø 8 -
Plat i Ø 8 - Ø 8 - Ø 8 -Plat j Ø 8 - Ø 8 - Ø 8 -
Plat k Ø 8 - Ø 8 - Ø 8 -
Plat l Ø 8 - Ø 8 - Ø 8 -
Dikarenakan jarak tulangan masing-masing pelat berbeda-beda, untuk mempermudah pemasangannya
dilapangan maka semua jarak tulangan yang dipasang berdasarkan portal arah melintang (Y),
kemudian diambil jarak terkecil
Tulangan pokok yang di pasang searah melintang atau sumbu Y:
Plat a
Plat e
Plat i
Plat b
Plat f
Plat j
Plat c
Plat g
Plat k
Plat d
Plat h
Plat l
12 - 70 1616,327
Ø 12 - 140 808,1633
Ø 12 - 140 808,1633
Ø 12 - 140 808,1633
Ø 12 - 70 1616,327 Ø 12 - 140 808,1633
Ø 12 - 70 1616,327
Ø 12 - 70 1616,327
Ø
PlatTumpuan Kiri Lapangan
Tulangan As Tulangan As
180 279,253 180 279,253 180 279,253
200 251,3274 180 279,2527 200 251,3274
180 279,2527 180 279,2527 180 279,2527
180 279,2527 180 279,2527 180 279,2527
200 251,3274 200 251,3274 200 251,3274
200 251,327 180 279,253 200 251,327
180 279,253 180 279,253 180 279,253
180 279,253 180 279,253 180 279,253
200 251,327 200 251,327 200 251,327
200 251,327 200 251,327 200 251,327
180 279,253 180 279,253 180 279,253
200 251,327 200 251,327 200 251,327
150 754,2857 140 808,1633 70 1616,327
150 754,2857 140 808,1633 70 1616,327
PlatTumpuan Kiri Lapangan Tumpuan Kanan
Tulangan As Tulangan As Tulangan As
100 1131,429 140 808,1633 70 1616,327
150 754,2857 140 808,1633 70 1616,327
150 754,2857 140 808,1633 70 1616,327
100 1131,429 140 808,1633 100 1131,429
100 1131,429 140 808,1633 70 1616,327
100 1131,429 140 808,1633 100 1131,429
120 942,8571 200 565,7143 200 565,7143
120 942,8571 200 565,7143 200 565,7143
120 942,8571 200 565,7143 200 565,7143
PlatTumpuan Kiri Lapangan Tumpuan Kanan
Tulangan As Tulangan As Tulangan As
120 942,8571 200 565,7143 200 565,7143
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Tulangan bagi yang di pasang :
Plat a
Plat e
Plat i
Plat b
Plat f
Plat j
Plat c
Plat g
Plat k
Plat d
Plat h
Plat l
Ø 12 - 180 279,253 Ø 12 - 180 279,253
Ø 12 - 180 279,253 Ø 12 - 180 279,253
Ø 12 - 180 279,253 Ø 12 - 180 279,253
PlatTumpuan Kiri Lapangan
Tulangan As Tulangan As
Ø 12 - 180 279,253 Ø 12 - 180 279,253