Perancangan Plat Beton 2 Arah

PERENCANAAN PELAT LANTAI 2 ARAH
Diketahui denah pembalokan sebuah Rumah Sakit dengan panel-panel pelat lantai sebagai
berikut :

Langkah-langkah perencanaan :
I. Mencari tebal minimum (h)
Diambil panel eksterior J (panel dengan bentang terbesar) :
 Tidak terdapat balok pinggir
 Nilai fy = 400 Mpa ; f’c = 30 Mpa
 Dimensi kolom = 700/700
 Dimensi balok induk = 500/700
 Dimensi balok anak = 200/400
Panel eksterior :

1
Muhammad Fauzi Novrizaldy (141144022)

hasumsi =

�

−

=

=

,

~

Mencari Inersia dari pelat dan balok :
a. Inersia Balok

= ,

Ib=

b. Inersia Pelat (h asumsi)

=

Ip=
Mencari nilai αfm :
=

�
�

�

ℎ

�

�
=
��

< ,

,

�ℎ� �� berlaku rumus hminimum (SNI):

� � �

=

=

=

, +

+

Digunakan hasumsi = 115 mm

II.

III.

,

=

,

=

,

~

Pembebanan
Fungsi gedung/ruangan = Rumah Sakit, sehingga :
 Beban Hidup (LL) = 250 kg/m = 2,451 kN/m
 Beban Mati (DL) :
- Berat Sendiri = 0,115 x 24 kN/m2 = 2,76 kN/m
- Berat adukan (20 mm) = 2 x 21 kg/m2 = 0,4118 kN/m

- Ubin = 0,235 kN/m
- Plafond = 11 + 7 = 18 kg/m2 = 0,176 kN/m
- Total beban mati = 3,58 kN/m
 Wu = 1,2 DL x 1,6 LL = 8,217 kN/m
Menghitung Momen tiap Panel
1. Panel A
Tipe III
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 61 = 3,132 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 106 = -5,443 kNm
 mty = -0,001 x wu x lx2 x 77 = -3,954 kNm

2
Muhammad Fauzi Novrizaldy (141144022)

2. Panel B
Tipe VI
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 57 = 2,927 kNm

 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 100 = -5,135 kNm
 mty = -0,001 x wu x lx2 x 76 = -3,903 kNm
3. Panel C
Tipe VI
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 57 = 2,927 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 100 = -5,135 kNm
 mty = -0,001 x wu x lx2 x 76 = -3,903 kNm
4. Panel D
Tipe VI
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 57 = 2,927 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 100 = -5,135 kNm
 mty = -0,001 x wu x lx2 x 76 = -3,903 kNm
5. Panel E
Tipe VI
Nilai β=1,6

 mlx = 0,001 x wu x lx2 x 57 = 2,927 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 100 = -5,135 kNm
 mty = -0,001 x wu x lx2 x 76 = -3,903 kNm
6. Panel F
Tipe VI
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 57 = 2,927 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 100 = -5,135 kNm
 mty = -0,001 x wu x lx2 x 76 = -3,903 kNm
7. Panel G
Tipe VI
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 57 = 2,927 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
3
Muhammad Fauzi Novrizaldy (141144022)





mtx = -0,001 x wu x lx2 x 100 = -5,135 kNm
mty = -0,001 x wu x lx2 x 76 = -3,903 kNm

8. Panel H
Tipe VI
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 57 = 2,927 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 100 = -5,135 kNm
 mty = -0,001 x wu x lx2 x 76 = -3,903 kNm
9. Panel I
Tipe III
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 61 = 3,132 kNm
 mly = 0,001 x wu x lx2 x 22 = 1,129 kNm
 mtx = -0,001 x wu x lx2 x 106 = -5,443 kNm
 mty = -0,001 x wu x lx2 x 77 = -3,954 kNm
10. Panel A’

Tipe VII
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 50 = 2,567 kNm
 mly = 0,001 x wu x lx2 x 18 = 0,924 kNm
 mtx = -0,001 x wu x lx2 x 80 = -4,108 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
11. Panel B’
Tipe II
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 49 = 2,516 kNm
 mly = 0,001 x wu x lx2 x 15 = 0,770 kNm
 mtx = -0,001 x wu x lx2 x 78 = -4,005 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
12. Panel C’
Tipe II
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 49 = 2,516 kNm
 mly = 0,001 x wu x lx2 x 15 = 0,770 kNm
 mtx = -0,001 x wu x lx2 x 78 = -4,005 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm

13. Panel D’
Tipe II
4
Muhammad Fauzi Novrizaldy (141144022)

Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 49 = 2,516 kNm
 mly = 0,001 x wu x lx2 x 15 = 0,770 kNm
 mtx = -0,001 x wu x lx2 x 78 = -4,005 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
14. Panel E’
Tipe II
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 49 = 2,516 kNm
 mly = 0,001 x wu x lx2 x 15 = 0,770 kNm
 mtx = -0,001 x wu x lx2 x 78 = -4,005 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
15. Panel F’
Tipe II
Nilai β=1,6

 mlx = 0,001 x wu x lx2 x 49 = 2,516 kNm
 mly = 0,001 x wu x lx2 x 15 = 0,770 kNm
 mtx = -0,001 x wu x lx2 x 78 = -4,005 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
16. Panel G’
Tipe II
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 49 = 2,516 kNm
 mly = 0,001 x wu x lx2 x 15 = 0,770 kNm
 mtx = -0,001 x wu x lx2 x 78 = -4,005 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
17. Panel H’
Tipe II
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 49 = 2,516 kNm
 mly = 0,001 x wu x lx2 x 15 = 0,770 kNm
 mtx = -0,001 x wu x lx2 x 78 = -4,005 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
18. Panel I’
Tipe VII
Nilai β=1,6
 mlx = 0,001 x wu x lx2 x 50 = 2,567 kNm
 mly = 0,001 x wu x lx2 x 18 = 0,924 kNm
 mtx = -0,001 x wu x lx2 x 80 = -4,108 kNm
 mty = -0,001 x wu x lx2 x 54 = -2,773 kNm
5
Muhammad Fauzi Novrizaldy (141144022)

19. Panel J
Tipe VI
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mly = 0,001 x wu x lx2 x 28 = 3,681 kNm
 mtx = -0,001 x wu x lx2 x 54 = -7,099 kNm
 mty = -0,001 x wu x lx2 x 60 = -7,888 kNm
20. Panel K
Tipe II
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mly = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mtx = -0,001 x wu x lx2 x 51 = -6,705 kNm
 mty = -0,001 x wu x lx2 x 51 = -6,705 kNm
21. Panel L
Tipe II
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mly = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mtx = -0,001 x wu x lx2 x 51 = -6,705 kNm
 mty = -0,001 x wu x lx2 x 51 = -6,705 kNm
22. Panel M
Tipe II
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mly = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mtx = -0,001 x wu x lx2 x 51 = -6,705 kNm
 mty = -0,001 x wu x lx2 x 51 = -6,705 kNm
23. Panel N
Tipe II
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mly = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mtx = -0,001 x wu x lx2 x 51 = -6,705 kNm
 mty = -0,001 x wu x lx2 x 51 = -6,705 kNm
24. Panel O
Tipe II
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
6
Muhammad Fauzi Novrizaldy (141144022)





mly = 0,001 x wu x lx2 x 25 = 3,286 kNm
mtx = -0,001 x wu x lx2 x 51 = -6,705 kNm
mty = -0,001 x wu x lx2 x 51 = -6,705 kNm

25. Panel P
Tipe II
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mly = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mtx = -0,001 x wu x lx2 x 51 = -6,705 kNm
 mty = -0,001 x wu x lx2 x 51 = -6,705 kNm
26. Panel Q
Tipe II
Nilai β=1
 mlx = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mly = 0,001 x wu x lx2 x 25 = 3,286 kNm
 mtx = -0,001 x wu x lx2 x 51 = -6,705 kNm
 mty = -0,001 x wu x lx2 x 51 = -6,705 kNm
27. Panel R,S,T,U,V,W,X,Y,Z sama dengan Panel A,B,C,D,E,F,G,H,I,J.
28. Panel R’,S’,T’,U’,V’,W’,X’,Y’,Z’ sama dengan Panel A’,B’,C’,D’,E’,F’,G’,H’,I’.
IV. Perhitungan Jumlah dan Jarak Tulangan
Setelah didapatkan momen-momen di tiap panelnya, maka dicari momen terbesar
berdasarkan sumbu global x. Berikut adalah momen-momen maksimalnya setelah di sort
di tabel pada excel. (Tabel momen terlampir).
Mlx (kNm) Mly (kNm) Mtx (kNm) Mty (kNm)
3,286
3,681
7,099
7,888

a) Mlx = 3,286 kNm
Dengan diameter tulangan asumsi = 8 mm (D8)
Mn =

,

= ,

,

�

= ,

4,107 x 106 = As . fy . 0,9 . d
Mencari d = h – 20 – 8/2= 91 mm
,

As =
a=

,

,

=

,

= ,

As aktual

As =

,

−

Cek Daktilitas
,
ρmin = � = ,

,9

=

�

,
,

7
Muhammad Fauzi Novrizaldy (141144022)

��

ρaktual =

.

,

=

= ,

ρmin > ρaktual (TIDAK OK)
Sehingga digunakan As
As = ρmin x b x d = 318,5 mm2
Dengan menggunakan tulangan diameter 10 D10
�

A1 =

Ʃ tulangan =

=

��

�

,

= ,

� ℎ

s (jarak tulangan per meter lebar ) =

,

=

,

~

Sehingga untuk tulangan lapangan arah memendek (mlx) digunakan tulangan
D10 - 200

diameter 10 dengan jarak 200 mm.
Menghitung ρmax
ρmax = ρb x 0,75
ρb =

,

�′

�

�

+�

akibat f’c>28 MPa

Menghitung reduksi

ρb =

= ,
,

,

,

= ,

+

; Sehingga :

= ,

ρmax = ρb x 0,75 = 0,0240
b.) Mly = 3,681 kNm

Dengan diameter tulangan asumsi = 8 mm (D8)
Mn =

,

= ,

,

�

= ,

4,601 x 106 = As . fy . 0,9 . d
Mencari d = h – 20 – 8/2= 91 mm
,

As =
a=

,

,

=

,

= ,

As aktual

As =

,

−

Cek Daktilitas
,
ρmin = � = ,

,

=

�

,
,

Muhammad Fauzi Novrizaldy (141144022)

8

��

ρaktual =

.

,

=

= ,

ρmin > ρaktual (TIDAK OK)
Sehingga digunakan As
As = ρmin x b x d = 318,5 mm2
Dengan menggunakan tulangan diameter 10 D10
�

A1 =

Ʃ tulangan =

=

��

�

,

= ,

� ℎ

s (jarak tulangan per meter lebar ) =

,

=

,

~

Sehingga untuk tulangan lapangan arah memanjang (mly) digunakan tulangan
D10 - 200

diameter 10 dengan jarak 200 mm.
Menghitung ρmax
ρmax = ρb x 0,75
ρb =

,

�′

�

�

+�

akibat f’c>28 MPa

Menghitung reduksi

ρb =

= ,
,

,

,

= ,

+

; Sehingga :

= ,

ρmax = ρb x 0,75 = 0,0240

c.) Mtx = -7,099 kNm
Dengan diameter tulangan asumsi = 8 mm (D8)
Mn =

,

=

,

,

�

= ,

8,873 x 106 = As . fy . 0,9 . d
Mencari d = h – 20 – 8/2= 91 mm
,

As =
a=

,

,

=

,

=

As aktual

As =

,

−

Cek Daktilitas

,

=

,

�

,
,
9

Muhammad Fauzi Novrizaldy (141144022)

,

ρmin = � = ,
��

ρaktual =

.

,

=

= ,

ρmin > ρaktual (TIDAK OK)
Sehingga digunakan As
As = ρmin x b x d = 318,5 mm2
Dengan menggunakan tulangan diameter 10 D10
�

A1 =

Ʃ tulangan =

=

��

�

,

= ,

� ℎ

s (jarak tulangan per meter lebar ) =

,

=

,

~

Sehingga untuk tulangan tumpuan arah memendek (mtx) digunakan tulangan
D10 - 200

diameter 10 dengan jarak 200 mm.
Menghitung ρmax
ρmax = ρb x 0,75
ρb =

,

�′

�

�

+�

akibat f’c>28 MPa

Menghitung reduksi

ρb =

= ,
,

,

,

= ,

+

ρmax = ρb x 0,75 = 0,0240

; Sehingga :

= ,

d.) Mtx = -7,888 kNm
Dengan diameter tulangan asumsi = 8 mm (D8)
Mn =

,

=

,

,

�

= ,

9,860 x 106 = As . fy . 0,9 . d
Mencari d = h – 20 – 8/2= 91 mm
,

As =
a=

,

,

As aktual

,

=

=

,

�

,

10
Muhammad Fauzi Novrizaldy (141144022)

,

As =

−

Cek Daktilitas
,
ρmin = � = ,
��

ρaktual =

.

=

,

,

,

=

= ,

ρmin > ρaktual (TIDAK OK)
Sehingga digunakan As
As = ρmin x b x d = 318,5 mm2
Dengan menggunakan tulangan diameter 10 D10
A1 =

�

Ʃ tulangan =

=

��

�

,

= ,

� ℎ

s (jarak tulangan per meter lebar ) =

,

=

,

~

Sehingga untuk tulangan tumpuan arah memanjang (mty) digunakan tulangan
diameter 10 dengan jarak 200 mm.

D10 - 200

Menghitung ρmax
ρmax = ρb x 0,75
ρb =

,

�

�

�′

+�

akibat f’c>28 MPa

Menghitung reduksi

ρb =

= ,
,

,

,

+

= ,

ρmax = ρb x 0,75 = 0,0240

; Sehingga :

= ,

11
Muhammad Fauzi Novrizaldy (141144022)

Gambar detail penulangan :

12
Muhammad Fauzi Novrizaldy (141144022)

LAMPIRAN
Tabel momen-momen tiap panel

No Nama Panel Nilai β
1
Panel A
1,6
2
Panel B
1,6
3
Panel C
1,6
4
Panel D
1,6
5
Panel E
1,6
6
Panel F
1,6
7
Panel G
1,6
8
Panel H
1,6
9
Panel I
1,6
10 Panel A'
1,6
11 Panel B'
1,6
12 Panel C'
1,6
13 Panel D'
1,6
14 Panel E'
1,6
15 Panel F'
1,6
16 Panel G'
1,6
17 Panel H'
1,6
18
Panel I'
1,6
19
Panel J
1
20 Panel K
1
21
Panel L
1
22 Panel M
1
23 PaneL N
1
24 Panel O
1
25
Panel P
1
26 Panel Q
1
27 Panel R
1,6
28
Panel S
1,6
29
Panel T
1,6
30 Panel U
1,6
31 Panel V
1,6
32 Panel W
1,6
33
Panel X
1,6
34 Panel Y
1,6
35
Panel Z
1,6
36 Panel R'
1,6
37 Panel S'
1,6
38 Panel T'
1,6
39 Panel U'
1,6
40 Panel V'
1,6
41 Panel W'
1,6
42 Panel X'
1,6
43 Panel Y'
1,6
44 Panel Z'
1,6

Mlx (kNm) Mly (kNm) Mtx (kNm) Mty (kNm)
3,132
1,129
-5,443
-3,954
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
3,132
1,129
-5,443
-3,954
2,567
0,924
-4,108
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,567
0,924
-4,108
-2,773
3,286
3,681
-7,099
-7,888
3,286
3,286
-6,705
-6,705
3,286
3,286
-6,705
-6,705
3,286
3,286
-6,705
-6,705
3,286
3,286
-6,705
-6,705
3,286
3,286
-6,705
-6,705
3,286
3,286
-6,705
-6,705
3,286
3,681
-7,099
-7,888
3,132
1,129
-5,443
-3,954
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
2,927
1,129
-5,135
-3,903
3,132
1,129
-5,443
-3,954
2,567
0,924
-4,108
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,516
0,77
-4,005
-2,773
2,567
0,924
-4,108
-2,773

13
Muhammad Fauzi Novrizaldy (141144022)