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1. Sebuah rumah tinggaldengan rencan denah seperti terlihat pada gambar di atas.Struktur
yang digunakan adalah beton bertulang untuk bagian plat lantai, balok dan kolom. Data-data
penting sebagai dasar perencanaan adalah sebagai berikut:
Tinggi kolom 3 meter 3,5 meter 4 meter
Mutu beton 20 MPa 25 MPa 30 MPa
Jenis tulangan Polos Deformasian
Mutu baja tulangan BJTP 24 BJTP 30 BJTD 30 BJTD35
Rencanakan penulangan struktur plat, balok dan kolom untuk konstruksi bangunan
tersebut. Lengkapi dengan gambar detail penulangan untuk semua hasil perencanaan.

2. Tipe
Ly
(m)
Lx
(m)
Ly/Lx Jenis Kasus
A 4,5 1,5 3 1 arah
B 4,5 3 1,5 2 arah
C 3 2,5 1,2 2 arah
D 2,5 1,5 1,67 2 arah
E 3 3 1 2 arah
F1 3 3 1 2 arah
G 3 1,5 2 1 arah
H 3 2,5 1,2 2 arah
I 3 2,5 1,2 2 arah
J 2,5 1,5 1,67 2 arah
= BalokAnak
= BalokInduk
Plat 2 Arah =
𝐿𝑌
𝐿𝑋
< 2
Plat 1 Arah =
𝐿𝑌
𝐿𝑋
≥ 2
Pink = Pelat Lantai
Hijau = Pelat Atap
Yang terkena silang
maka termasuk balok
eksterior

3. K 3 2,5 1,2 2 arah
L1 3 2,5 1,2 2 arah
M 3 2,5 1,2 2 arah
N 3 3 1 2 arah
F2 3 3 1 2 arah
O 3 2,5 1,2 2 arah
L2 3 2,5 1,2 2 arah
Q 3 2 1,5 2 arah
F3 3 3 1 2 arah
P 3 3 1 2 arah
R 3 1,5 2 1 arah
S 3 1,5 2 1 arah
T 2 1,5 1,3 2 arah
U 3,5 3 1,16 2 arah
V 3,5 3 1,16 2 arah
W 3,5 2 1,75 2 arah
 (Gideon Kusuma Jilid ke-1 halaman 104)
-Tinggi balok : (h) =
1
10
𝑙 −
1
15
𝑙
-Lebar balok : (b) =
1
2
ℎ −
2
3
ℎ
 Perkiraan ukuran balok anak
L = 4,5 m = 450 cm
-Tinggi balok (h) =
1
15
𝑥 𝐿 =
1
15
𝑥 450 = 30𝑐𝑚
-Lebar balok (b) =
2
3
𝑥 ℎ =
2
3
𝑥 30 = 20 𝑐𝑚
 Perkiraan ukuran balok induk
Ly = 4,5m = 450 cm
-Tinggi balok (h) =
1
12
𝑥 𝐿𝑦 =
1
12
𝑥 450 = 37,5 𝑐𝑚 ≈ 40 𝑐𝑚
-Lebar balok (b) =
2
3
𝑥 ℎ =
2
3
𝑥 40 = 26,6 𝑐𝑚 ≈ 30 𝑐𝑚
Bentang bersih terpanjang
Dimensi balok anak
b/h = 20/30 cm
Dimensi balokinduk
b/h = 30/40 cm

4. 𝐿𝑛 = 𝐿 − 2
1
2
𝑥 𝑏 𝑏𝑎𝑙𝑜𝑘 𝑖𝑛𝑑𝑢𝑘 = 450 − 2
1
2
𝑥 30 = 420 𝑐𝑚
Bentang bersih terpendek
𝐿𝑛 = 𝐿 − 2
1
2
𝑥 𝑏 𝑏𝑎𝑙𝑜𝑘 𝑎𝑛𝑎𝑘 = 450 − 2
1
2
𝑥 20 = 430 𝑐𝑚
β =
𝐿𝑛
𝑆𝑛
=
420
430
= 0,977
Cara Perhitungan
 Balok (Induk/Anak) Eksterior
𝑏𝑒 <
1
12
𝐿
𝑏𝑒 < 𝑏𝑤 + 6𝑡
Pilih yang terkecil, be=
*L = yg ditinjau
 Balok (Induk/Anak) Interior
𝑏𝑒 <
1
4
𝐿
𝑏𝑒 < 𝑏𝑤 + 8𝑡
Pilih yang terkecil, be =
*L = yg ditinjau
 Menghitung Momen Inersia Penampang Balok
 Pelat Satu Arah (SNI 2847 2013 tabel 9.5 hal.70) Pilih plat terbesar
 Pelat Dua Arah (SNI 2847 2013 ayat 9.5.3.3 pasal 9)
Untuk plat dengan balok yang membentang diatara tumpuan pada semua
sisinya, tebal minimumnya h, harus memenuhi ketentuan sebagai berikut:

5. 𝐾 =
1 +
𝑏𝑒
𝑏𝑤
− 1
𝑡
ℎ
[4 − 6
𝑡
ℎ
+ 4
𝑡
ℎ
2
+
𝑏𝑒
𝑏𝑤
− 1
𝑡
ℎ
3
]
1 + (
𝑏𝑒
𝑏𝑤
− 1)(
𝑡
ℎ
)
 Momen Inersia Balok
Ib=
1
12
𝑥 𝑏𝑤 𝑥 ℎ3
𝑥 𝑘
 Momen Inersia Pelat
Is=
1
12
𝑥 𝑙 𝑥 𝑡3
 Mencari αFm
𝛼𝑓 =
𝐸 𝑐𝑏 . 𝐼𝑏
𝐸𝑐𝑠 . 𝐼𝑠
𝛼𝑓 =
𝐼𝑏
𝐼𝑠
 αFm =
𝛼𝐹1 + 𝛼𝐹2 + … + 𝛼𝐹𝑛
𝛴𝛼𝐹
(SNI 2847 2013 Pasal 13.6.1.6 )
 Mencari Tebal Pelat (h)
Tebal plat (t) gunakan dari hmin satu arah
Pelat Lantai B (4,5m x 3m)
𝛽 =
ly
lx
=
4,5
3
= 1,5m
Menentukan lebar efektif balok
(8.12.2 dan 8.12.3 SNI 2847 2013).
Balok eksterior 𝛼𝐹1 (balok induk)
𝑏𝑒 <
1
12
𝐿 →
1
12
𝑥 300 = 25𝑐𝑚
𝑏𝑒 < 𝑏𝑤 + 6𝑡 → 30 + 6(12) = 102 cm
Pilih yang terkecil, maka be = 25 cm
Ecb = Ecs = Ec
 Pelat Satu Arah (SNI 2847 2013 tabel 9.5 hal.70)
Pilih plat plat lantai satu arah terbesar A 4,5m x 1,5m
Pelat masif satu arah pada satu ujung menerus.
hmin =
𝑙
24
0,4 +
𝑓𝑦
700
=
1500
24
0,4 +
300
700
= 51,786 𝑚𝑚 ≈ 120 𝑚𝑚
𝑙 = yg ditinjau

6. 𝐾 =
1 +
25
30
− 1
12
40
[4 − 6
12
40
+ 4
12
40
2
+
25
30
− 1
12
40
3
]
1 + (
25
30 − 1)(
12
40)
= 0,918
Ib=
1
12
𝑥 𝑏𝑤 𝑥 ℎ3
𝑥 𝑘 =
1
12
𝑥30𝑥403
𝑥0,918 = 146901,1 𝑐𝑚4
Is=
1
12
𝑥 𝑙 𝑥 𝑡3
=
1
12
𝑥300𝑥123
= 43200 𝑐𝑚4
αF1 =
146901
43200
= 3,4
Balok interior 𝛼𝐹2(balok induk)
𝑏𝑒 <
1
4
𝐿 →
1
4
𝑥 450 = 112,5 𝑐𝑚
𝑏𝑒 < 𝑏𝑤 + 8𝑡 → 30 + 8(12) = 126𝑐m
Pilih yang terkecil, maka be = 112,5 cm
𝐾 =
1 +
112,5
30 − 1
12
40 [4 − 6
12
40 + 4
12
40
2
+
112,5
30 − 1
12
40
3
]
1 + (
112,5
30
− 1)(
12
40
)
= 1,739
Ib=
1
12
𝑥 𝑏𝑤 𝑥 ℎ3
𝑥 𝑘 =
1
12
𝑥30𝑥403
𝑥1,739 = 278203𝑐𝑚4
Is=
1
12
𝑥 𝑙 𝑥 𝑡3
=
1
12
𝑥 450 𝑥123
= 64800𝑐𝑚4
αF2 =
278203
64800
= 4,293
Balok interior 𝛼𝐹3(balok anak)
𝑏𝑒 <
1
4
𝐿𝑥 →
1
4
𝑥 450 = 112,5 𝑐𝑚
𝑏𝑒 < 𝑏𝑤 + 8𝑡 → 20 + 8(12) = 116 𝑐𝑚
Pilih yang terkecil, maka be = 112,5 cm
𝐾 =
1 +
112,5
20
− 1
12
30
[4 − 6(
12
30
) + 4
12
30
2
+
112,5
20
− 1
12
30
3
]
1 + (
112,5
20
− 1)(
12
30
)
= 1,997
Ib=
1
12
𝑥 𝑏𝑤 𝑥 ℎ3
𝑥 𝑘 =
1
12
𝑥20𝑥303
𝑥1,997 = 89867
Is=
1
12
𝑥 𝑙 𝑥 𝑡3
=
1
12
𝑥450𝑥123
= 64800𝑐𝑚4
αF3 =
89867
64800
= 1,387
αFm =
( 𝛼𝐹1𝑋2)+𝛼𝐹2+𝛼𝐹3
4
=
(2𝑥3,4)+4,293 +1,387
4
= 3,12 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,5)
= 86,06𝑚𝑚 > 90𝑚𝑚 (𝑛𝑜𝑡 𝑜𝑘) ≈ 120 𝑚𝑚
Pelat Lantai C (3m x 2,5m)
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok interior 𝛼𝐹1 (balok induk)

7. be = 62,5 cm
be = 126 cm
Pilih be = 62,5 cm
K = 1,39
Ib = 222370,6 cm4
Is = 36000 cm4
αF1 = 6,177
Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 126 cm
Pilih be = 75 cm
K = 1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF2 = 5,543
Balok Eksterior 𝛼𝐹3 (balok induk)
be = 25 cm
be = 102 cm
Pilih be = 25 cm
K = 0,918
Ib = 146901,1 cm4
Is = 43200 cm4
αF3 = 3,4
αFm =
( 𝛼𝐹1𝑋2)+𝛼𝐹2+𝛼𝐹3
4
=
(2𝑥6,177)+5,543+3,4
4
= 5,324 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai D (1,5m x 2,5m)
𝛽 =
ly
lx
=
2,5
1,5
= 1,67 m
Balok Eksterior 𝛼𝐹1 (balok anak)
be = 20,833 cm
be = 92 cm
Pilih be = 20,833 cm
K = 1,02
Ib = 45916,69 cm4
Is = 36000 cm4
αF1 = 1,275
Balok interior 𝛼𝐹2 (balok induk)
be = 37,5 cm
be = 126 cm
Pilih be = 37,5 cm
K = 1,109
Ib = 177489,3 cm4
Is = 21600 cm4
αF2 = 8,217
Balok interior 𝛼𝐹3 (balok induk)
be = 62,5 cm
be = 126 cm
Pilih be = 62,5 cm
K = 1,39
Ib = 222370,6 cm4
Is = 36000 cm4
αF3 = 6,177
αFm =
( 𝛼𝐹2𝑋2)+𝛼𝐹1+𝛼𝐹3
4
=
(2𝑥8,217)+1,275+6,177
4
= 5,972 > 2

8. h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,67 )
= 83,48 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 120 𝑚𝑚
Pelat Lantai E (3m x 3m)
𝛽 =
ly
lx
=
3
3
= 1m
Balok Eksterior 𝛼𝐹1 (balok induk)
be = 25 cm
be = 102 cm
Pilih be = 25 cm
K = 0,918
Ib = 146901,1 cm4
Is = 43200 cm4
αF1 = 3,4
Balok interior 𝛼𝐹2 (balok anak)
be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF2 = 1,814
αFm =
( 𝛼𝐹1𝑋2)+( 𝛼𝐹2𝑋2)
4
=
(2𝑥3,4)+(2𝑥1,814)
4
= 2,607 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1)
= 94,67 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai F1 (3m x 3m)
𝛽 =
ly
lx
=
3
3
= 1m
Balok interior 𝛼𝐹1 (balok anak)
be = 75cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF1 = 1,814
Balok interior 𝛼𝐹2 (balok induk)

9. be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF2 = 5,543
αFm =
( 𝛼𝐹1𝑋2)+( 𝛼𝐹2𝑋2)
4
=
(2𝑥1,814)+(2𝑥5,543)
4
= 3,678 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1)
= 94,67 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai F2 (3m x 3m)
𝛽 =
ly
lx
=
3
3
= 1m
Balok interior 𝛼𝐹1 (balok induk)
be = 75cm
be = 116 cm
Pilih be = 75 cm
K =1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF1 = 5,543
Balok interior 𝛼𝐹2 (balok anak)
be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF2 = 1,814
αFm =
( 𝛼𝐹1𝑋2)+( 𝛼𝐹2𝑋2)
4
=
(2𝑥1,814)+(2𝑥5,543)
4
= 3,678 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1)
= 94,67 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai H (3m x 2,5m)
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok eksterior 𝛼𝐹1 (balok induk)
be = 20,8333 cm
be = 102 cm
Pilih be = 20,8333 cm
K = 0,843
Ib = 134944,2 cm4
Is = 36000 cm4
αF1 = 3,748
Balok eksterior 𝛼𝐹2 (balok induk)

10. be = 25 cm
be = 102 cm
Pilih be = 25 cm
K = 0,918
Ib = 146901,1 cm4
Is = 43200 cm4
αF2 = 3,4
Balok interior 𝛼𝐹3 (balok anak)
be = 62,5 cm
be = 116 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73449,73 cm4
Is = 36000 cm4
αF3 = 2,04
Balok interior 𝛼𝐹4 (balok anak)
be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF4 = 1,814
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
3.748+3.4+2.04+1.814
4
= 2,75 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai I (3m x 2,5m)
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok interior 𝛼𝐹1 (balok anak)
be = 62,5 cm
be = 116 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73449,73 cm4
Is = 36000 cm4
αF1 = 2,04
Balok eksterior 𝛼𝐹2 (balok induk)
be = 25 cm
be = 102 cm
Pilih be = 25 cm
K = 0,918
Ib = 146901,1 cm4
Is = 43200 cm4
αF2 = 3,4
Balok Eksterior 𝛼𝐹3 (balok anak)
be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF3 = 1,814
αFm =
( 𝛼𝐹1𝑋2)+𝛼𝐹2+𝛼𝐹3
4
=
(2𝑥2.04)+3.4+1.814
4
= 2,324 > 2

11. h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai J (1,5m x 2,5m)
𝛽 =
ly
lx
=
2,5
1,5
= 1,67 m
Balok interior 𝛼𝐹1 (balok anak)
be = 62,5 cm
be = 116 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73449,73 cm4
Is = 36000 cm4
αF1 = 2,04
Balok eksterior 𝛼𝐹2 (balok induk)
be = 12,5 cm
be = 102 cm
Pilih be = 12,5 cm
K = 0,672
Ib = 107589,1 cm4
Is = 21600 cm4
αF2 = 4,981
Balok interior 𝛼𝐹3 (balok induk)
be = 62,5cm
be = 126 cm
Pilih be = 62,5 cm
K = 1,39
Ib = 222370,6 cm4
Is = 36000 cm4
αF3 = 6,177
Balok interior 𝛼𝐹4 (balok anak)
be = 37,5 cm
be = 116 cm
Pilih be = 37,5 cm
K = 1,336
Ib = 60120cm4
Is = 21600 cm4
αF4 = 2,783
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
2.04+4.981+6.177+2.783
4
= 3,995 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,67 )
= 83,48 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 120 𝑚𝑚
Pelat Lantai L1 (3m x 2,5m)
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok interior 𝛼𝐹1 (balok induk)
be =62,5 cm
be = 126 cm
Pilih be = 62,5 cm
K = 1,39
Ib = 222370,6 cm4
Is = 36000 cm4
αF1 = 6,177

12. Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 126 cm
Pilih be = 75 cm
K = 1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF2 = 5,543
Balok interior 𝛼𝐹3 (balok anak)
be = 62,5 cm
be = 116 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73449,73 cm4
Is = 36000 cm4
αF3 = 2,04
Balok interior 𝛼𝐹4 (balok anak)
be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF4 = 1,814
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
6.177+5.543+2.04+1.814
4
= 3,894 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai L2 (3m x 2,5m)
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok interior 𝛼𝐹1 (balok induk)
be = 62,5cm
be = 126 cm
Pilih be = 62,5 cm
K = 1,39
Ib = 222370,6 cm4
Is = 36000 cm4
αF1 = 6,177
Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 126 cm
Pilih be = 75 cm
K = 1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF2 = 5,543
Balok interior 𝛼𝐹3 (balok anak)
be = 62,5 cm
be = 116 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73449,73 cm4
Is = 36000 cm4
αF3 = 2,04
αFm =
( 𝛼𝐹1𝑋2)+𝛼𝐹2+𝛼𝐹3
4
=
(2𝑥6,177)+5,543+2,04
4
= 4,984 > 2

13. h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai O (3m x 2,5m)
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok interior 𝛼𝐹1 (balok anak)
be = 62,5 cm
be = 116 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73449,73 cm4
Is = 36000 cm4
αF1 = 2,04
Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 126 cm
Pilih be = 75 cm
K = 1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF2 = 5,543
Balok Eksterior 𝛼𝐹3 (balok induk)
be = 20,8333 cm
be = 102 cm
Pilih be = 20,8333 cm
K = 0,843
Ib = 134944,1 cm4
Is = 36000 cm4
αF3 = 3,748
αFm =
( 𝛼𝐹2𝑋2)+𝛼𝐹1+𝛼𝐹3
4
=
(2𝑥5,543)+2,04+3,748
4
= 4,219 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai M (3m x 2,5m)
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok interior 𝛼𝐹1 (balok anak)
be = 62,5 cm
be = 126 cm
Pilih be = 62,5 cm
K = 1,39
Ib = 222370,6 cm4
Is = 36000 cm4
αF1 = 6,177
Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 126 cm
Pilih be = 75 cm
K = 1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF2 = 5,543
Balok eksterior 𝛼𝐹3 (balok induk)

14. be = 20,8333 cm
be = 102 cm
Pilih be = 20,8333 cm
K = 0,843
Ib = 134944,1 cm4
Is = 36000 cm4
αF3 = 3,748
Balok interior 𝛼𝐹4 (balok anak)
be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF4 = 1,814
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
6,177+5,543+3,748 +1,814
4
= 4,321 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai N (3m x 3m)
𝛽 =
ly
lx
=
3
3
= 1m
Balok interior 𝛼𝐹1 (balok anak)
be = 75 cm
be = 116 cm
Pilih be = 75 cm
K = 1,742
Ib = 78377,14 cm4
Is = 43200 cm4
αF4 = 1,814
Balok eksterior 𝛼𝐹2 (balok induk)
be = 25 cm
be = 102 cm
Pilih be = 25 cm
K = 0,918
Ib = 146901,1 cm4
Is = 43200 cm4
αF2 = 3,4
Balok interior 𝛼𝐹3 (balok induk)
be = 75 cm
be = 126 cm
Pilih be = 75 cm
K = 1,497
Ib = 239473,1 cm4
Is = 43200 cm4
αF3 = 5,543
αFm =
( 𝛼𝐹1𝑋2)+𝛼𝐹2+𝛼𝐹3
4
=
(2𝑥1,814)+3.4+5,543
4
= 3,143 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
No. Pelat αFm Tebal (mm)

15. 1 B 3,12 120
2 C 5,324 100
3 D 5,972 120
4 E 2,607 100
5 F1 3,678 100
6 F2 3,678 100
7 H 2,75 100
8 I 2,324 100
9 J 3,995 120
10 L1 3,894 100
11 L2 4,984 100
12 O 4,219 100
13 M 4,321 100
14 N 3,143 100
Tebal plat (t) gunakan dari hmin satu arah
Pelat Lantai V
𝛽 =
ly
lx
=
3,5
3
= 1,167 m
Balok interior 𝛼𝐹1(balok anak)
𝑏𝑒 <
1
4
𝐿𝑦 →
1
4
𝑥 350 = 87,5 𝑐𝑚
𝑏𝑒 < 𝑏𝑤 + 8𝑡 → 20 + 8(12) = 110 𝑐𝑚
Pilih yang terkecil, maka be = 87,5cm
K=
1+
87,5
20
−1
10
30
[4−6
12
30
+4
10
30
2
+
87,5
20
−1
10
30
3
]
1+(
87,5
20
−1)(
10
30
)
= 1,831
Ib=
1
12
x bw x h3
x k =
1
12
x20x303
x1,831 = 82390 cm4
Is=
1
12
x l x t3
=
1
12
x350x103
= 29166,67 cm4
αF1 =
82390
29166,67
= 2,825
 Pelat Satu Arah (SNI 2847 2013 tabel 9.5 hal.70)
Pilih plat terbesar R 3m x 1,5m
Pelat masif satu arah pada dua ujung menerus.
hmin =
𝑙
28
0,4 +
𝑓𝑦
700
=
1500
28
0,4 +
300
700
= 44,39 𝑚𝑚 ≈ 100 𝑚𝑚

16. Balok interior αF2 (balok induk)
be <
1
4
Lx →
1
4
x 300 = 75 cm
be < bw + 8t → 30 + 8(10) = 110cm
Pilih yang terkecil, maka be = 75 cm
K=
1+
75
30
−1
10
40
[4−6
10
40
+4
10
40
2
+
75
30
−1
10
40
3
]
1+(
75
30
−1)(
10
40
)
= 1,484
Ib=
1
12
x bw x h3
x k =
1
12
x30x403
x1,484 = 237386cm4
Is=
1
12
x l x t3
=
1
12
x300x103
= 25000cm4
αF2 =
237386
25000
= 9,496
Balok interior 𝛼𝐹3(balok induk)
be <
1
4
Lx →
1
4
x 350 = 87,5 cm
be < bw + 8t → 30 + 8(10) = 110cm
Pilih yang terkecil, maka be = 87,5 cm
K=
1+
87,5
30
−1
10
40
[4−6
10
40
+4
10
40
2
+
87,5
30
−1
10
40
3
]
1+(
87,5
30
−1)(
10
40
)
= 1,577
Ib=
1
12
x bw x h3
x k =
1
12
x30x403
x1,577 = 252256 cm4
Is=
1
12
x l x t3
=
1
12
x350x103
= 29167 cm4
αF3 =
252256
29167
= 8,649
Balok ekterior 𝛼𝐹4 (balok induk)
𝑏𝑒 <
1
12
𝐿𝑥 →
1
12
𝑥 300 = 25 𝑐𝑚
𝑏𝑒 < 𝑏𝑤 + 6𝑡 → 30 + 6(10) = 90 cm
Pilih yang terkecil, maka be = 25 cm
K=
1+
25
30
−1
10
40
[4−6
10
40
+4
10
40
2
+
25
30
−1
10
40
3
]
1+(
25
30
−1)(
10
40
)
= 0,924
Ib=
1
12
𝑥 𝑏𝑤 𝑥 ℎ3
𝑥 𝑘 =
1
12
𝑥30𝑥403
𝑥0,924 = 147844𝑐𝑚4
Is=
1
12
𝑥 𝑙 𝑥 𝑡3
=
1
12
𝑥300𝑥103
= 25000𝑐𝑚4
αF1 =
147844
25000
= 5,914
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
2,825+9,496 +8,649+5,914
4
= 6,721 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36+9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,167 )
= 91,607 𝑚𝑚 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100𝑚

17. Pelat Lantai H
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok eksterior 𝛼𝐹1 (balok induk)
be = 20,83333 cm
be = 90 cm
Pilih be = 20,83333 cm
K = 0,856
Ib = 136905,3 cm4
Is = 20833,33 cm4
αF1 = 6,571
Balok eksterior 𝛼𝐹2 (balok induk)
be = 25 cm
be = 90 cm
Pilih be = 25 cm
K = 0,924
Ib = 147844,2 cm4
Is = 25000 cm4
αF2 = 5,914
Balok interior 𝛼𝐹3 (balok anak)
be = 62,5 cm
be = 100 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73419,72 cm4
Is = 20833,33 cm4
αF3 = 3,524
Balok interior 𝛼𝐹4 (balok induk)
be = 75 cm
be = 110 cm
Pilih be = 75 cm
K = 1,484
Ib = 237386,4 cm4
Is = 25000 cm4
αF4 = 9,5
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
6,571+5,914 +3,524 +9,5
4
= 6,376 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai K
𝛽 =
ly
lx
=
3
2,5
= 1,2 m
Balok interior 𝛼𝐹1 (balok anak)
be = 62,5 cm
be = 100 cm
Pilih be = 62,5 cm
K = 1,632
Ib = 73419,72 cm4
Is = 20833,33 cm4
αF3 = 3,524

18. Balok eksterior 𝛼𝐹2 (balok induk)
be = 25 cm
be = 90 cm
Pilih be = 25 cm
K = 0,924
Ib = 147844,2 cm4
Is = 25000 cm4
αF2 = 5,914
Balok eksterior 𝛼𝐹3 (balok induk)
be = 20,83333 cm
be = 90 cm
Pilih be = 20,83333 cm
K = 0,856
Ib = 136905,3 cm4
Is = 20833,33 cm4
αF1 = 6,571
Balok interior 𝛼𝐹4 (balok anak)
be = 75 cm
be = 110 cm
Pilih be = 75 cm
K = 1,484
Ib = 237386,4 cm4
Is = 25000 cm4
αF4 = 9,5
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
6,571+5,914 +3,524 +9,5
4
= 6,376 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,2)
= 91,025 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai P
𝛽 =
ly
lx
=
3
3
= 1 m
Balok ekterior 𝛼𝐹1 (balok induk)
be = 25 cm
be = 90 cm
Pilih be = 25 cm
K = 0,924
Ib = 147844,2 cm4
Is = 25000 cm4
αF1 = 5,914
Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 110 cm
Pilih be = 75 cm
K = 1,484
Ib = 237386,4 cm4
Is = 25000 cm4
αF2 = 9,5
Balok interior 𝛼𝐹3 (balok anak)
be = 75 cm
be = 100 cm
Pilih be = 75 cm
K = 1,74
Ib = 78278,99 cm4
Is = 25000 cm4
αF3 = 3,131
αFm =
( 𝛼𝐹3𝑋2)+𝛼𝐹1+𝛼𝐹2
4
=
(2𝑥3,131)+5,914 +9,5
4
= 5,419 > 2

19. h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1)
= 94,67 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai F3
𝛽 =
ly
lx
=
3
3
= 1 m
Balok interior 𝛼𝐹1 (balok anak)
be = 75 cm
be = 100 cm
Pilih be = 75 cm
K = 1,74
Ib = 78278,99 cm4
Is = 25000 cm4
αF3 = 3,131
Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 110 cm
Pilih be = 75 cm
K = 1,484
Ib = 237386,4 cm4
Is = 25000 cm4
αF2 = 9,5
αFm =
( 𝛼𝐹1𝑋2)+( 𝛼𝐹2𝑋2)
4
=
(2𝑥3,131)+(2𝑥9,5)
4
= 6,315 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1)
= 94,67 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai Q
𝛽 =
ly
lx
=
3
2
= 1,5 m
Balok interior 𝛼𝐹1 (balok induk)
be = 75 cm
be = 110 cm
Pilih be = 75 cm
K = 1,484
Ib = 237386,4 cm4
Is = 25000 cm4
αF2 = 9,5
Balok eksterior 𝛼𝐹2 (balok induk)
be = 16,67 cm
be = 90 cm
Pilih be = 16,67 cm
K = 0,782
Ib = 125148,7 cm4
Is = 16666,67 cm4
αF2 = 7,5
Balok eksterior 𝛼𝐹3 (balok induk)
be = 25 cm
be = 90 cm
Pilih be = 25 cm
K = 0,924
Ib = 147844,2 cm4
Is = 25000 cm4
αF1 = 5,914

20. Balok interior 𝛼𝐹4 (balok anak)
be = 50 cm
be = 100 cm
Pilih be = 50 cm
K = 1,5
Ib = 67500 cm4
Is = 16666,67 cm4
αF4 = 4,05
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
9.5+7.5+5.914+4.05
4
= 6,741 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,5)
= 86,061 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai T
𝛽 =
ly
lx
=
2
1,5
= 1,333 m
Balok interior 𝛼𝐹1 (balok induk)
be = 37,5 cm
be = 110 cm
Pilih be = 37,5 cm
K = 1,103
Ib = 176507,4 cm4
Is = 12500 cm4
αF1 = 14,12
Balok interior 𝛼𝐹2 (balok anak)
be = 50 cm
be = 100 cm
Pilih be = 50 cm
K = 1,5
Ib = 67500 cm4
Is = 16666,67 cm4
αF4 = 4,05
Balok eksterior 𝛼𝐹3 (balok induk)
be = 12,5 cm
be = 90 cm
Pilih be = 12,5 cm
K = 0,703
Ib = 112444,1 cm4
Is = 12500 cm4
αF3 = 8,996
Balok interior 𝛼𝐹4 (balok induk)
be = 50 cm
be = 110cm
Pilih be = 50 cm
K = 1,251
Ib = 200238,1 cm4
Is = 16666,67 cm4
αF4 = 12,014
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
14,12+4,05 +8,996+12,014
4
= 9,795 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,333 )
= 88,75 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai U
𝛽 =
ly
lx
=
3,5
3
= 1,167 m

21. Balok eksterior 𝛼𝐹1 (balok induk)
be = 29,167 cm
be = 90 cm
Pilih be = 29,167 cm
K = 0,988
Ib = 158043,2 m4
Is = 29166,67 cm4
αF1 = 5,419
Balok interior 𝛼𝐹2 (balok induk)
be = 75 cm
be = 100 cm
Pilih be = 75 cm
K = 1,74
Ib = 78278,99 cm4
Is = 25000 cm4
αF3 = 3,131
Balok interior 𝛼𝐹3 (balok anak)
be = 87,5 cm
be = 100 cm
Pilih be = 87,5 cm
K = 1,831
Ib = 82389,71 cm4
Is = 29166,67 cm4
αF3 = 2,825
Balok eksterior 𝛼𝐹4 (balok induk)
be = 25 cm
be = 90 cm
Pilih be = 25 cm
K = 0,924
Ib = 147844,2 cm4
Is = 25000 cm4
αF4 = 5,914
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
5,419+3,131 +2,825 +5,914
4
= 4,322 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,167 )
= 91,613 > 90𝑚𝑚 ( 𝑜𝑘) ≈ 100 𝑚𝑚
Pelat Lantai W
𝛽 =
ly
lx
=
3,5
2
= 1,75 m
Balok interior 𝛼𝐹1 (balok induk)
be = 87,5 cm
be = 110 cm
Pilih be = 87,5 cm
K = 1,577
Ib = 252256,5 cm4
Is = 29166,67 cm4
αF1 = 8,649
Balok interior 𝛼𝐹2 (balok induk)
be = 50 cm
be = 110cm
Pilih be = 50 cm
K = 1,251
Ib = 200238,1 cm4
Is = 16666,67 cm4
αF4 = 12,014
Balok eksterior 𝛼𝐹3 (balok induk)

22. be = 29,167 cm
be = 90 cm
Pilih be = 29,167 cm
K = 0,988
Ib = 158043,2 m4
Is = 29166,67 cm4
αF1 = 5,419
Balok eksterior 𝛼𝐹4 (balok induk)
be = 16,67 cm
be = 90 cm
Pilih be = 16,67 cm
K = 0,782
Ib = 125148,7 cm4
Is = 16666,67 cm4
αF4 = 7,509
αFm =
𝛼𝐹1+𝛼𝐹2+𝛼𝐹3+𝛼𝐹4
4
=
8,649+12,014+5,419+7,509
4
= 8,398 > 2
h =
𝐿𝑛 (0,8+
𝐹𝑦
1400
)
36 +9𝛽
=
4200 (0,8+
300
1400
)
36+9(1,75 )
= 82,319 > 90𝑚𝑚 ( 𝑛𝑜𝑡 𝑜𝑘) ≈ 100 𝑚𝑚
No. Pelat αFm Tebal (mm)
1 H 6,376 100
2 K 6,376 100
3 P 5,419 100
4 F3 6,315 100
5 Q 6,741 100
6 T 9,795 100
7 U 4,322 100
8 V 6,721 100
9 W 8,398 100